Chapter-2

Measurement of Physical quantities in Physics

Introduction

You have studied in the last chapter, the fundamentals of scientific study

in relevance to the subject Physics, the science of physical phenomenon

of nature. You can now proceed to frame the hypothesis/theory/law to

explain any physical phenomenon of nature. But as we have already

discussed that there is no meaning of words until unless our theory or

law is based on certain physical parameters of study or may be called as

physical quantities, those could be expressed and measured

quantitatively. These quantities are the basis of experimental validation

of the theory propounded. Now we shall try to understand the concept of

physical quantities regarding their definition and their use in

interpreting the laws of Physics. At the same time regarding

measurement of these quantities, we shall glance over the different

standards of measurements prevailing world wide to quantitatively

measure physical quantities and how these standards make

computations involving physical quantities make so simple.

Unit One

Now in this unit of about half an hour you will be able to learn

a) Concept of fundamental and derived quantity.

b) System of measurements in practice.

c) System of measurements in S.I. system.

d) System of units and conversion of units from one system to

another.

Concept of fundamental and derived quantity

As we have discussed that to understand the seen phenomenon of nature

theories/hypothesis/laws are framed. But there is no sense and meaning

of words until unless the laws or postulates are based on measurable

quantities. The laws of physics are expressed and experimentally

validated in terms of these physical quantities. The laws expressed in

terms of these physical quantities are invariant with space and time.

For the purpose of Physics it is very essential that quantities should be

defined clearly and precisely and should have the conceptual meaning

for the postulated law. Among these are force, time, velocity, density,

temperature, charge, magnetic susceptibility and numerous others.

Now as per the practice all the physical quantities involved in the Physics

today may be grouped in two sets. One set contains the physical

quantities as a fundamental one and other set contains quantities derived

from these fundamental one. For an example length and time may be

taken as a fundamental quantities and the quantity velocity defined as

the ratio of length upon time may be termed as a derived quantity.

However which quantity will be taken as a fundamental and which one as

derived depends on system to system. For example S.I. system of units

takes length, mass and time as a fundamental quantity and the quantity

force defined as the product of mass and acceleration (length/time2) as a

derived quantity. While the F.P.S. system of units and measurement

takes force, length and time as a fundamental quantity and mass as a

derived quantity.

Now the intention of creating physical quantities is that physical laws are

expressed in terms of these quantities and which are themselves

evaluated in terms of numerical values. These values should be such that

can be characterized as intrinsic to the experiment conducted and may

be compared to the same quantity arrived in another set of experiment.

Now the comparison of physical quantities can be done only when each

physical quantity either fundamental or derived one is based on some

standard of measurement. The standard of measurement is a set of

standard where each fundamental quantity is assigned a specific unit

value. The unit value to that quantity is such that it is invariant with

space and time and is easily accessible. For example in the S.I. system of

measurement the physical quantity mass has been given a unit value as

one kilogram and defined as mass of the international standard body

preserved at Severes, France. However it certainly involves the

procedure to account the unit value of the fundamental quantity, where

different objects having different values in the same set of conditions can

be compared. For example by the procedure to obtain mass of a standard

body, the spring balance in terms of stretch of spring may be used. The

stretch of spring is directly proportional to the mass of the body. By

comparing the different stretches of spring, the masses of different

bodies, may be compared.

So in final words the operational definition of fundamental quantity

involves two steps, first choice of a standard and second the

establishment of procedures for measuring the quantity in terms of

standard so that a number and unit are determined as a measure of

quantity. But very important aspect of choice of standard is that it should

be accessible and invariant. For example we have selected our standard

for length to measure the distance between two points as one meter,

then by a comparison of this length with a second object three times in

length as standard, we say that second bar has a length of three meters.

However most quantities cannot be measured directly in comparison to

standard and indirect approach using some involved procedure is

required and also certain assumptions are made to ascertain the matter.

For example measuring the time of sending and receiving the

electromagnetic pulse with known speed, the distance can be measured

as product of speed and one half of time interval. Here we have set half

time of sending and receiving the signals as our standard of

measurement and different distances may be compared with different

times of observations. However the speed of pulse is to be determined

through the other acclaimed procedure.

Similarly, we use an indirect method to measure very small distances

between atoms and molecules by particle scattering method.

System of measurements in practice

Now so far we have studied that quantities in physics are either

fundamental one or derived one depending upon the system of

measurement we are using and shall study the different systems of

measurement prevailing and used all over the world. The fundamental

quantities are defined in terms of a standard of measurement devised in

that particular system of measurement and procedure to measure the

quantity so that comparison of different objects may be done in terms of

that quantity. The derived quantities are derived from these fundamental

one and for the complete descriptions will have dimensions showing the

fundamental quantities involved and units of these dimensions involved

depending upon the system of measurement used.

Now we shall discuss the different system of measurements prevailing all

over the world and their merits and demerits.

Imperial System of units measurement

Before S.I. system of units adopted around the world, the British

systems of English units and later Imperial system of units were used in

Britain, the Commonwealth and the United States. The system came to

be known as U.S. Customary units in the United States and is still in use

there and in a few Caribbean countries. These various systems of

measurement have at times been called foot-pound-second systems

after the Imperial units for distance, weight and time. Many Imperial

units remain in use in Britain despite the fact that it has officially

switched to the SI system. Road signs are still in miles, yards, miles per

hour, and so on. People tend to measure their own height in feet and

inches and beer is sold in pints, to give just a few examples. Imperial

units are used in many other places, for example, in many

Commonwealth countries, which are considered metricated, land area

is measured in acres and floor space in square feet, particularly for

commercial transactions (rather than government statistics). Similarly,

the imperial gallon is used in many countries that are considered

metricated at gas/petrol stations, an example being the United Arab

Emirates.

Metric System of Measurement

The metric system is a decimalised system of measurement based on

the metre and the Kilogram. It exists in several variations, with

different choices of base units, though these do not affect its day-to-day

use. Since the 1960s the International System of Units (SI), explained

further below, is the internationally recognized standard metric system.

Metric units of mass, length, and electricity are widely used around the

world for both everyday and scientific purposes. The main advantage of

the metric system is that it has a single base unit for each physical

quantity. All other units are powers of ten or multiples of ten of this

base unit. Unit conversions are always simple because they will be in

the ratio of ten, one hundred, one thousand, etc. All lengths and

distances, for example, are measured in meters, or thousandths of a

metre (millimeters), or thousands of meters (kilometres), and so on.

There is no profusion of different units with different conversion factors

as in the Imperial system (e.g.inches, feet, yards, fathoms, rods).

Multiples and submultiples are related to the fundamental unit by

factors of powers of ten, so that one can convert by simply moving the

decimal place: 1.234 metres is 1234 millimetres or 0.001234

kilometres. The use of fractions, such as 2/5 of a meter, is not

prohibited, but uncommon.

S.I system of measurement

The International System of Units (abbreviated SI from the French

language name Système International d'Unités) is the modern, revised

form of the metric system. It is the world's most widely used system of

units, both in everyday commerce and in science. The SI was developed

in 1960 from the metre-kilogram-second (MKS) system, rather than the

centimetre-gram-second (CGS) system, which, in turn, had many

variants. At its development the SI introduced several newly named

fundamental units that were previously not a part of the metric system.

The S.I. system of units has the following set of standards for

measurement:

Standard of Length

The first standard of length measurement conceived was a bar of

platinum-iridium alloy kept at International Bureau of weights and

measures near Paris. The distance between two lines engraved on gold

plugs near the ends of the bar (when the bar was at 0.00 degree

centigrade and supported in a certain mechanical system) was defined

as one meter. But the main disadvantage of the meter bar is that it is

not accurately producible at everywhere. In 1961 an atomic standard of

length was accepted by International agreement. The wavelength of

orange radiation emitted by atoms of Krypton (Kr36) in electrical

discharge was chosen. One meter is now defined to be 1,650,763.73

wavelengths of this light. The choice of atomic standard has offered a

great advantage that there is enough precision in length measurements

and all atoms generate light of same wavelength, at everywhere

therefore accessible and invariant with respect to space and time.

Standard of Time

For the development of the time standard there are two different aspects

for the purpose. One is for civil and another for scientific work according

to the desired accuracy in the work. We define the time to know the

duration between the start and end of events and classify them in such a

sequence where precedence of start or end of one event can be

compared with other by counting number of repetitions as the division of

time phase. An oscillating pendulum, quartz crystal, oscillating spring or

digital counter can be used for the purpose. Of the many repetitive

phenomenon occurring in the nature, the rotation of earth on its own

axis, which in time span is length of day, has been used as time standard

and still the basis of determining civil time standard. One mean solar

second being defined as 1/86,400 of a mean solar day and time expressed

in terms of Earth’s rotation about its own axis is called universal time

(UT)

In 1956, The International Congress of Weight and measures redefined

the second for the scientific purposes requiring high precision, in terms

of the earth orbital motion about the sun and found it to be the fraction

1/31,556,925.9747 of the tropical year 1900 and time defined in terms of

the earth orbital motion is called ephemeris time (ET).

But the main disadvantage of the above time standards is that both UT

and ET must be determined by astronomical observations extending over

several weeks (for UT) or several years (for ET) and a secondary

terrestrial clock, calibrated by the astronomical observations is needed

for the purpose. Quartz crystal clocks, based on the electrically sustained

natural periodic vibration of a quartz crystal serve as a time standard

and have measured time with a maximum error of 0.02 sec in a year.

One of the most common uses of a time standard is the determination of

frequencies and for the purpose the atomic clocks using the periodic

atomic vibrations as a standard have been developed which gives

accurate time estimates to a accuracy of fraction of micro seconds and

are invariant with space and time.

As per the S.I. system of units the second (s) is the duration of

9,192,631,770 periods of the radiation corresponding to the transition

between the hyperfine levels of the ground state of the Cs133 atoms.

Standard of Mass

The kilogram (kg) is the mass of the international standard body

preserved at Severes, France.

Standard of amount of Substance (Mole)

The amount of substance that contains as many elementary entities

(Avogadro number 6.02x1023) like atoms if substance is monoatomic or

molecules is called a mole. For example 0.012 kg of carbon-12, called as

one mole substance contains Nos of atoms of carbon-12.

Standard of Current

The ampere (A) is the current in two very long parallel wires 1m apart

that gives rise to a magnetic force of N/m.

Standard of Temperature

The Kelvin (K) is of the thermodynamic temperature of the triple

point of water.

Standard of Luminous Intensity

The candela (cd) is the luminous intensity in the perpendicular direction

of a surface of a area sq meter of a black body at a temperature

of freezing platinum at pressure of 1 atm.

Units and their conversion

We have studied earlier that all derived quantities are dependent upon

the fundamental quantities and fundamental quantities on certain

standards. The choice of standard units for these fundamental

quantities determines the system of units for all physical quantities

used all over the world. For example the M.K.S. system of units

classifying fundamental quantities mass, length and time as Kilogram,

meter, second. Therefore once the choice of the system of units has

been made the derived or dependent quantities follow the same system

of units and will have both number and unit in its notification.

When such quantities are added, subtracted, multiplied, or divided in an

algebraic equation the unit can be treated as an algebraic quantity. For

example we wish to find out the distance traveled by a person traveling

at a speed of 5 km per hour in five hours than

Distance traveled = Speed x Time= =25 km

Ex 1. 5 litre of benzene weigh more in summer or winter?

(A) Summer (B) Winter (C) Equal in both (D) None of these

Sol: Since volume increases and density decreases after rise in

temperature so for a given volume of benzene say 5 litre it will weigh

more in the winter for higher density.

Ex 2. The SI unit of length is the meter. Suppose we adopt a new unit of

length which equal to x meters the area 1m2 expressed in terms of the

new unit has a magnitude

(A) x (B) x2 (C)1/x (D)1/x2

Sol: Area = 1m x 1m = (1 unit/x)(1 unit/x) = 1/x2 unit2

So magnitude=1/x2

Ex 3. If the unit of length is micrometer and the unit of time is microsecond, the unit of

velocity will be

(A) 100 m/s (B) 10 m/s (C) micrometer/s (D) 1 m/s

Sol: Velocity=L/T = 1 m/s

Ex 4. If the units of length and force are increased four times, then the units of energy will

(A) Increase 8 times (B) increase 16 times (C) decrease 16 times (D) increase 4 times

Sol: E0 = F

0 d

0

En = 16 f0 d

0

En = 16 E0

Hence each unit of new energy dimension is 16 times each unit of energy in old dimension

Ex 5. In a particular system, the unit of length, mass and time are chosen to be 10 cm.,10 gm,

and 0.1 s respectively. The unit of force in this system will be equivalent to

(A) 1/10 N (B) 1 N (C) 10 N (D) 100 N

Sol: unit of mass = 10 gm

Length = 10 cm.

Time = 0.10 s

Unit force in the new system with dimensions M L T –2 have 10 (gm)(10 cm) (0.1 s)-2

And so equals to 1/10 N

OR

Since 1N = 1kg m.s –2

= 103gm.(102cm)( s -2)

Hence each unit of force = 1N/10

Ex 6. What will be the unit of time in that system in which the unit of length is ‘metre’ unit

of mass ‘kg’ and unit of force ‘kg.wt.’?

(A) 19.8 sec (B) (9.8)2sec (C)9.8 sec (D) 9.8 sec

Sol: Force F = MLT-2

kg (9.80 m/s2) = kgmT-2

T = 1/9.8 s

OR

F= M L T –2

T= ML/F = kg . m/ kg wt

So T= kg . m / kgwt = 1/ 9.8 sec

Ex7. If the velocity of light c, acceleration due to gravity g , and the atmospheric pressure p

are taken as the fundamental units , then the unit of mass will be

(A) 1 kg (B) 81 kg (C) 9x1018 kg (D) 81x1034 kg

Sol: M= c x g y p z = M z L x+y-z T –x –2y –2z ; so z=1, x=4, y=-3

M=pc4/g3 =81x1034 Kg

Ex8 Is the time variation of position as shown in the fig is observed in nature? Explain.

Position o f Partic le (x )

Tim

e in

Sec

. (t)

Sol: The answer is clearly NO because at any instant of time the particle can’t have two

positions and time doesn’t decreases.

Ex9. The normal duration of physics practical period in Indian colleges is 100 minutes.

Express this period in micro century. 1 micro century =10-6 x 100 year.

Sol: T= 100 minutes =100/60x24x365x10 –4 = 1.9 Microcenturies;

Ex 10. The SI and C.G.S. units of energy are joule and erg respectively. How many erg are

equal to one joule.

Sol: In M.K.S. system of units Energy =Kg m2 /s 2 = 1 joule= 107 grams cm 2 /s 2 = 1 Erg in

C.G.S. system

Ex 11. Young’s modulus of steel is 19x1010 N/m2. Express it in dyne/cm2. Here dyne is the

C.G.S. unit of force.

Sol: 19 x1011 Dyne /cm2;

Ex 12. The density of a substance is 8 g/cm3. Now we have a new system where unit of

length is 5 cm and unit of mass 20 g. Find the density in the new system.

Sol: Density of substance = 8 g/cm3; Unit of length = 5 cm ; Unit of mass = 20 g.

Since dimension of density = ML-3T0

2/

1 = (M

2/M

1) (L

2/L

1)-3 = 20 (5)-3= 4/25; Density of substance in new system = 50 units

Exercise 1

Q1. If the units of force, energy and velocity in new system be 10 N, 5 J and 0.5 m s-1

respectively, find the units of mass, length and time in the new system.

Which gives M=E/v2=5J/(0.5)2 = 20 Kg; L=E/F=0.5 m; T=E/FV=1 sec

Q2. The radius of proton is about 10-9 microns and the radius of the universe is about 1028 cm.

Name a physical object whose size is approximately half way between these two on a

logarithmic scale.

Q3 Assuming that the length of the day uniformly increases by 0.001s in a century. Calculate

the cumulative effect on the measure of time over twenty centuries. Such a slowing down of

the earth‘s rotation is indicated by observations of the occurrences of solar eclipses during

this period.

Q4. The unit of length convenient on the nuclear scale is fermi ; 1 f=10-15 m. Nuclear sizes

obey roughly the following empirical relation; r=r0A10 where r is the radius of the nucleus.

A its mass number and r0 is a constant equal to 1.2 f. Find out whether mass density is nearly

constant for different nuclei.

Solutions Exercise 1

Ans1. Let M = F x E y v z and so x=0, y=1 and z=-2

Ans2: On logarithmic scale the exponent value= -15+(15+26)/2= -15 + 41/2=5.5

Hence the size of object=105.5 1x106 ( Radius of moon)

Ans3: Increase in length of day in 20 centuries = 20 x 0.001s

Average increase in length of day = 0.001x 20/2 = 0.01s

Cumulative error in time measurement = cumulative error in day length increase in 20

centuries = 20 x100 x 365 x 0.01/3600 h= 2.03 h

Ans4:Mass density of nucleus=A(mp)/(4/3 r3) 1017 Kg/m3

Introduction

So far we have defined the physical quantities in terms of their interdependence and evolved

procedures to measure the physical quantities. Now we shall try to understand the way the

physical quantities appear in the laws of physics and the rules of mathematical manipulations

followed.

Each physical quantity may be attributed a set of dimensions according to the base units

involved in the quantity and laws of physics involve mathematical manipulations of these

physical quantities. As each statement should follow certain rules involving the physical

quantities regarding their mathematical manipulations, is governed by the dimensional

analysis.

Beside that physical quantity regarding their representation numerically should follow certain

rules so that degree of accuracy of the measurement could be ascertained at a glance of the

presentation of measurement. It is governed by the rules of significant digits of measurement

of physical quantities. These two are the scope of study now.

So at the end of this unit of around one hour you will be able to learn

Dimensions of physical quantities and dimensional analysis.

Concept of significant digits for the presentation of measurement of quantities in Physics.

Dimensions of Physical Quantities

As we have stated earlier that physical quantities can be added, subtracted, multiplied or

divided in the same manner as any other algebraic quantity. The addition or subtraction of

quantities is meaningful only if the quantities have followed same standard of units and so

are dimensionally homogeneous. The dimensional consistency is must for equation to be

correct. The correctness of equation can be checked by comparing the physical dimensions of

each term in the equation under discussion. A list of various physical quantities with their

usual convention and dimensions are

:

S.No Quantity Symbol Physical formula

S.I unit Dimension formula

1 Acceleration a a=v/t m/s2 M0LT-2

2 Angular acceleration

= /t Rad/sec2 M0 L0 T -2

3 Angular Displacement

=arc/radius Radian M0 L0 T0

4 Angular Momentum

L L=m v r Kg m2/s M L2 T -1

5 Angular Velocity

=/t Rad/sec M0 L0 T -1

6 Area A l x b (Metre)2 M0L2T0

7 Capacitance C Q = CV Farad M-1L-2 T 4A2

8 Charge q q=I t Coulomb M0 L0 T A9 Current I ---- Ampereor A M0L0T0A10 Density d d =M/V Kg/m3 M L-3 T 0

11 Dielectric constant

r

r=/0 ___ M0L0T0

12 Displacement S ___ Metre or m M0LT0

13 Electric dipole

moment

P P=q 2l Coulomb metre

M0L T A

14 Energy KE/U KE=1/2mv2 Joule M L2 T -2

15 Force F F=ma Newton or N M L T -2

16 Force Constant

K K=F/x N/m M L0 T -2

17 Frequency f f=1/T Hertz M0 L0 T -1

18 Gravitational Constant

G G=F r2/m1m2 N m2/kg2 M-1L3 T -2

19 Heat Q Q=mst Joule/calorie M L2 T -2

20 Impulse - F x t N.sec M L T -1

21 Intensity of electrical field

E E = F/q N/coul M L T -3A-1

21 Intensity of electrical field

E E = F/q N/coul M L T -3A-1

22 Latent Heat L Q = mL Joule/kg M0 L2 T -2

23 Magnetic dipole

moment

M M=NIA Amp.m2 M0 L2A T 0

24 Magnetic flux E=d/dt Weber M L2 T - 2A-1

25 Magnetic intensity

H B =H A/m M L-1T 0A

26 Moment of Inertia

I I=mr2 Kg m2 ML2T0

27 Momentum P P = mv Kg m/s M L T -1

28 Mutual inductance

L E=L.di/dt Henery M L 2 T -2A-2

29 Power P P=W/t Watt M L2 T -3

30 Pressure P P =F/A Pascal M L-1 T -2

31 Resistance R V=IR Ohm M L2 T-3 A-2

32 Specific heat S Q=m s t Joule/kg.kelvin

M0 L2T -2-1

33 Strain ___ l/l; A/A; v/v

___ M0L0T0

34 Stress ___ F/A N/m2 M0 L0 T -2

35 Surface Tension

T F / l N/m M L0 T -2

36 Temperature ___ Kelvin M0 L0 T0 1

37 Torque F.d N.m. M L2 T -2

38 Universal gas constant

R PV = nRT Joule/mol.k M L 2 T -2-1

39 Velocity v v = s/t m/s M0LT -1

40 Volume V V=l x b x h(Metre)3

M0L3T0

41 Work W F.d N.m M L2 T-2

42 Young Modulus

Y Y = (F/A)/l/l

N/m2 M L-1 T -2

Application of dimensional analysis

1. To find the unit of a physical Quantity.

2. To convert units of a physical Quantity from one system to another.

3. To check the dimensional correctness of a given relation. It is to be further noted that

every dimensionally correct relation does not mean to have physically correct relationship. If

we have some idea or can make an educated guess as to how one physical quantity relates to

another we can use dimensions to derive the form of the equation.

As an example, consider the equation for the period of pendulum bob. We might suppose that

the period depends on the mass of the bob, the length of the pendulum and the acceleration

due to gravity

We can express this as T=mx l y gz. Where x, y and z are as yet undetermined indices.

To find the values of x, y and z we convert the formula into its dimensions. On the left-hand

side the dimension of the period is [T], the dimension of mass is [M], the dimension of the

length of the pendulum is [L] and the dimension of g is [LT-2].

[T]=[M]x[L]y[LT-2]z.

Equating left-hand indices with matching dimensions on the right-hand side.

[M]: x=0

[L]: y+z=0

[T]: 2z=-1

From this we can deduce that z=-1/2, while y=1/2 and x=1/2

Substituting these values into the original equation we obtain. T=

k(l/g)1/2. Where k is a constant of proportionality. Compare this with the

equation for the period of a pendulum

The form of the equation is correct, but it cannot determine the constant of proportionality.

Limitations of application of dimensional analysis

1. If the dimensions are given, physical quantity may not be unique as

many physical quantities have same dimensions.

2. Since numerical constants have no dimensions, can’t be deduced by

dimensional analysis. For example 1, etc.

3. The dimensional analysis can’t be used to derive relationship other

than the product of power functions. However the dimensional

correctness of the relation may be checked. For e.g. S=u t+1/2 a t2 , y= a sin t

4. The method of dimension can’t be applied to derive relationship when a physical quantity

depends on more than three quantities. For e.g. T=2 I/(m g l)

Ex1. If velocity, force and time are taken to be fundamental quantities find dimensional

formula for quantity mass

(A) V-1

FT-1

(B) V-1

FT (C) VF-1

T-1

(D) V-1

F-1

T

Sol: M=K (V)x (F)y (T)z ; x= -1, y =1, z =1; M = K V-1 F T

Ex2. You may not know integration. But using dimensional analysis can check or prove

results .in the integral

dx/(2ax-x2)1/2 = an sin-1 [x/a-1]

The value of n should be

(A) 1 (B) –1 (C) 0 (D) 1/2

Sol: The dimension of a is that of x and for dimensional consistency of the equation n should be equal to zero.

Ex3. In the cauchy’s formula for the refractive index n = A+B/2 the dimensions of A and B

are

(A) Both are dimensionless (B) A is dimension less , B has dimension M0 L-2 T0

(C) A is dimension less, B has dimensions M0L2T0 (D) Both A and B have dimensions

M0L2T0

Sol: Since refractive index is dimensionless hence A should be dimensionless and B should have dimensions of Length 2 that is M0L2T0

Ex4. A highly rigid cubical block A of small mass M and side L is fixed rigidly on the

another cubical block B of same dimensions and of low modulus of rigidity such that

lower face of A completely covers the upper face of B. The lower face of B is rigidly held

on a horizontal surface. A small force F is applies perpendicular to one of the side face of

A. After the force is withdrawn, block A executes small oscillation, the time period of

which is given by

(A) 2 (mL) (B) 2 (M/L) (C) 2 (ML/) (D) 2 (M/L)

Sol: Check the dimensions of the left and right side quantities.

Ex5. The frequency of oscillation of an object of mass m suspended by means of spring of

force constant k is given by f = c m x ky, where c is a dimensionless constant the value of

x and y are

(A) x = ½, y = ½ (B) x = - ½, y = ½ (C) x = ½, y = - ½ (D) x = - ½, y = - ½

Sol: Putting the dimensions on the two sides and equating the powers of the base quantities we get x= -1/2, y =1/2

Multiple Choice Type of Questions

Ex1. Suppose a quantity x can be dimensionally represented in terms of M, L and T i.e.[X] =

Ma Lb T c the quantity mass

(A) Can always be dimensionally represented in terms of L, T and x.

(B) May be represented in terms of L,T and x if a = 0

(C) May be represented in terms of L, T and x if a = 0

(D) May be represented in terms of L, T and x if a 0

Sol: Any quantity may be represented in terms of other quantities as base quantities such that the quantities are dependent upon each other with a certain relationship. Here I the case the quantity M may be represented in terms of L, T, x if a0 so that M remains related with the rest of quantities.

Ex2. Choose the correct statement (s)

(A) A dimensionally correct equation may be correct

(B) A dimensionally correct equation may be incorrect

(C) A dimensionally incorrect equation must be correct

(D) A dimensionally incorrect equation must be incorrect

Sol: A dimensionally correct equation may be correct or incorrect. But a dimensionally incorrect equation can never be correct.

Ex3. Choose the correct statement (s)

(A) All quantities may be represented dimensionally in terms of the base quantities.

(B) All base quantities cannot be represented dimensionally in terms of the rest of base

quantities.

(C) The dimension of a base quantity in other base quantity is always zero.

(D) The dimension of a derived quantity is never zero in any base quantity.

Sol: ( A ), (B),( C )

Descriptive Type of Questions:

Ex.1 (a) Can a physical quantity have no unit and dimensions? If yes give an example.

(b) Can a physical quantity have units without dimensions? If yes give an example.

Ans41: (a) Strain has neither unit nor dimensions but it is a defined physical quantity.(b) Angle measured in radians is physical quantity, which has unit of radian but no dimension. The angle measurement in a plane that is radian and angle of a solid that is steradian are supplementary units to supplement the physical quantities like angular displacement and angular velocity. No dimension has been assigned to these quantities.

Ex2. Find the dimensional formulae of e0& m0 (Where e0 is the absolute permittivity and m0

is the permeability of vacuum or free space respectively ).

Sol: The dimensions of 0= M-1 L-3 T 4A2 ; The dimensions of

0 = MLT-2A-2

Ex3. Assuming that the largest mass that can be moved by a flowing river depends on

velocity of flow density of river water and on gravity, find out how the mass varies with

velocity of flow.

Sol: M=K(v)x(d)y(g)z ; x=6, y=1, z=-3 and so M= K v6 d g –3

Ex4. The gas constant R depends upon pressure of the gas P, volume of the gas V,

temperature of the gas T and number of moles n. Derive an expression for gas constant, R

Sol: Let P= K V x n y R z T m , on solving for dimensions of base quantities M,L,T and n we

get x=-1, y=1, z=1, T=1 and so PV=n RT

Exercise 2

Matching Type of Questions:

Q1. In the following table, there are two lists A and B, but the list B is not in order of list A.

write down the list B in order of list A in each table.

(a) Moment of inertia (i) Newton /Meter 2

(b) Surface tension (ii) kg/ (metre-s)

(c) Angular acceleration (iii) kg – meter2

(d) Coefficient of viscosity (iv) Newton/meter

(e) Coefficient of elasticity (v) radian/s2

(f) Momentum (vi) MLT-1

(g) Gravitational Constant (vii) ML2T-1

(h) Plank Constant (viii) M-1 L3 T-2

Q2. Column I gives three physical quantities. Select the appropriate units for these from

choices given in column II. Some of the physical quantities may have more than one choice :

I II

Capacitance Ohm second

Inductance Coul2 joul-1

Magnetic inductance Coulomb (Volt)1

Newton(ampere-m)-1

Volt-sec (Ampere)-1

Q3. Match the physical quantities given in column I with dimensions expressed in column if

in tabular form

(a) Angular momentum (a) ML2T-2

(b) Latent heat (b) ML2Q-2

(c) Torque (c) ML2T-1

(d) Capacitance (d) ML3T-1Q-2

(e) Inductance (e) M-1L-2T2Q2

(f) Resistivity (f) L2T-2

Q4. The position of a particle at any time is given by S(t) = V0/a (1-e-at), where a>0 and V0

are constants. What are the dimensions of a and V0 ?

Q5. The equation of a wave traveling along the x axis by y = A e[x/a-t/T] 2, write the

dimensions of A , a and T.

Q6. Suppose an attractive nuclear force acts between two protons which may be written as F

= Ce-kt/r2

Write down the dimensional formula and appropriate SI units of C and K.

Q7. Find the dimensional formula of L/R, where R is the resistance and L is the coefficient of

self-inductance.

Q8. Find out the dimensions of electrical conductivity.

Q9. The equation of state of a real gas is given by [p+a/v2](v-b) = RT, where p, v and T are

pressure , volume and temperature respectively and R is the universal gas constant. Find out

the dimensions of the constant a in the above equation.

Q10. The heat produced in a wire carrying an electric current depends on the current, the

resistance and the time; assuming that the dependence is of the product of powers type, guess

an equation between these quantities using dimensional analysis. The dimensional formula

of resistance is ML2T-3 A-2 and heat is a form of energy.

Q11. The frequency of vibration of a stream depends on the length L between the nodes, the

tension F in the string and its mass per unit length M. Guess the expression for its frequency

from dimensional Analysis.

Q12. The kinetic energy E of a rotating body depends on its moment of inertia I and its

angular speed w. Assuming the relation to be E =K Ia wb where K is a dimensionless

constant, find a and b. Moment of Inertia of the sphere about its diameter is 2/5Mr2.

Q13 (a) In the formula X=3YZ2, X and Z have dimensions of capacitance and magnetic

induction respectively. What are the dimensions of Y in MKSQ system?

(b) Calculate the dimension(s) of VCR/L, where V=supply voltage, C=capacitance,

R=resistance and L=inductance.

Q14. If velocity of light c, the gravitational constant G and plank constant h be chosen as the

fundamental units, then find the dimensions of mass, length and time in the new system.

Concept of Significant digits

As we have seen every measurement pertains to the standard we are

going to use and its numerical value is read from the calibrated scale

based on that standard of measurement. The value of measurement

contains two parts

(i) One part with all digits read directly from the scale by the smallest

subdivision of the scale called as certain digits of measurement

(ii) and the second part contains doubtful digits at end corresponding to

the eye judgment within the least sub-division of the scale. For example

the length measured by the meter scale having least count of 1cm as

50.35 cm contains digits 5, 0 as certain 3 is doubtful and 5 is

insignificant. The digits 5, 0, 3 are termed as significant digits and 5 as

insignificant digit. There may be some confusion while dealing with the

measurements containing zero at their end but rule is the same.

As a general practice we report only the significant digits and magnitude

of any physical quantity is represented by proper power of 10. For

example if only 5 is significant in 500 cm then we report it as 5 x 102 cm,

with 5 as a significant digit. If 5 and first zero that is two digits are

significant then we write 5.0 x 102 cm. and if all the digits are significant

we report it as 5.00 x 10 2 cm.

If the integer part of the digit is zero then all the continuous zeros after

the decimal are treated as insignificant digits as the number 0.00003

having first digit as zero then all the continuous digits are also

insignificant. For this reason it is important to keep the trailing zeros to

indicate the actual number of significant figures.

However if first digit is nonzero as in 1.0005 then all zeros will be

counted for significant digits containing five significant digits as 1,0,0,0

and 5 respectively.

For numbers without decimal points, trailing zeros may or may not be

significant. Thus, 400 indicate only one significant figure. To indicate

that the trailing zeros are significant a decimal point must be added. For

example, 400 has three significant figures, and has one significant figure.

Exact numbers have an infinite number of significant digits. For example,

if there are two oranges on a table, then the number of oranges is

2.000... . Defined numbers are also like this. For example, the number of

centimeters per inch (2.54) has an infinite number of significant digits, as

does the speed of light (299792458 m/s).

Significant digits in arithmetical calculation

As per internationally accepted practice for finding out the significant digits in the

arithmetical calculation say division or multiplication of two physical quantities following

rule has been formulated for determination of significant:

1. In multiplication or division of two or more quantities the number of significant digits in

the answer is equal to the number of significant digits in the quantity, which has least number

of significant digits. Thus 60.0/2.0 will have only two significant digits. The insignificant

digits are dropped from the result if appear after the decimal point, and replaced to zero if

appear to the left of the decimal point. The least significant digit is rounded according to the

following rules:

(A) If the digit next to one rounded is more than 5, then the digit to be rounded is increased

by 1

(B) If next digit is less than 5 then rounding digit is left unchanged.

(C) If the digit next to be rounded is equal to 5 then rounding digit is increased by one if it is

odd otherwise left unchanged.

For example,

(2.80) (4.5039) = 12.61092

should be rounded off to 12.6 (three significant figures like 2.80).

2. For addition or subtraction of quantities all the numbers are written with the decimal point

in one line up to the number, which has the first doubtful digit counted from left. All digits

are written to that digit after rounding off at that level and addition/subtraction is performed.

For example,

89.332 + 1.1 = 90.432

should be rounded to get 90.4 (the tenths place is the last significant place in 1.1).

Ex1. Evaluate 25.2x1374/33.3 All the digits in this expression are significant.

Sol: Expression has value =1039.7838Since the expression has number with lowest number of significant digits as 3 with number

25.2 or 33.3 hence the expression will also have three significant digits and number will be

written as per rule 1.04x103

Ex2. Evaluate 24.36+0.0623+256.2

Sol: As per rule write the numbers with decimal point in a line and check where the first

doubtful digit occurs between these numbers, which in our case occurs at 256.2. Now change

the other numbers with proper rounding and than add as 24.4 + 0.1 + 256.2 = 280.7

Ex3. Given P=0.0030 m; Q=2.40 m and R=3000 m. Find out the number of significant

figures in P, Q, R respectively.

Sol: No of significant figures in P=2(3,0) since the first digit is zero so all zeros after that are insignificant;No of significant figures in Q=3(2,4,0); No of significant figures in R=4(3,0,0,0)

Ex4. The volume of one sphere is 1.76 c.c. Find out the volume of 25 such spheres

(according to idea of significant figures).

Sol: The volume of 25 spheres=25 x1.76 = 44. Since the number 1.76 has three significant

digits and so result should also be written to the three significant digits as 44.0 c.c,

Exercise 3

Q1. The length breadth and thickness of a block are measured as 12.5 cm, 5.6 cm, and 0.32

cm respectively. Which measurement is least accurate?

Q2. The radius of the circle is stated as 2.12 cm. Find out its area written as in appropriate

number of significant digits.

Q3. Round off the following numbers to three significant digits (a) 15462 (b) 14.745 (c)

14.750 (d) 14.650x1012.

Hints and Solutions Exercise 2

Ans4: The power of e should be dimensionless so the dimension of a = M0L0T-1; The

dimension of V0=M0LT-1;

Ans5: The dimension of a and T should be of such that x/a and t/T are dimensionless and

dimensions of A should be of y.

So the dimension of T= M0L0T; A= M0L T0 ; a=M0 L T 0;

Ans6: The dimension of C = M L1T -2 ; The dimension of K=M0L2T-1

Ans7: The dimension of L=V/(dI/dt)=W/I 2 =ML2T-2A-2; The dimension of R=W/I 2 t

=M L2 T-3A-2

Ans8: Electrical conductivity=1/R= M-1 L-2 T 3 A2

Ans9: a=pv2=ML5T-2;

Ans10: H= I x R y t z which on solving for dimensions of two side for base quantities M, L, T

and I we get H=I2 R t

Ans11: Let = K L x F y m z

Put the dimensions of the quantities on two sides and solve for powers x,y, and z we get = K/L (F/m)Ans12: E.= K I x y

Put the dimensions of E, I and and we get E=K I 2

Ans13: Y=X/3Z2 ; Y= M-3L-6T8A6= M-3 L-6 T2 Q6

Ans14: M=C h/G ; L=h1/2 G1/2/C3/2, T= h1/2 G1/2/C5/2

Exercise 3

Ans1: The breadth of the block is only up to two significant digits hence supposed to be least accurate.Ans2: Area of circle= R2= 14.112416 cm2=14.1 cm2 (three significant figures)

Ans3: (a) 1.55x 104 ; The last number has been rounded to five since digit next to 4 is six greater

than 5

(b) 1.47 x 101 The last number has been rounded to 7 only since digit next to it is less than 5

(c) 1.48 x 101 The last number has been rounded to 8 since digit next to it is 5, and itself odd number.

(d) 1.46 x 1013 The last number has been rounded to 6 only since digit next to it is equal to 5 and

itself even number.

Introduction

Now let us continue our discussion about measurement about physical quantities. We know,

that all sort of measurements are arrived at by taking measurements in some set of

experiments of the physical quantities. The authenticity of the results of experiments is

totally dependent upon the precision of the measurements taken, which itself dependent upon

the certain factors like how the instrument is calibrated with respect to reference one, least

sub division of instrument, skill of the person making measurements, secondary effect of

environment or errors in the instrument etc. Although we try to make accurate measurements

but it is also true that it is very difficult to arrive at the fictitious true value of measurement.

One measurement of the same quantity taken many times will differ from each other. So it is

a very difficult task to arrive the true value of the measurement. Now in this unit we shall try

to understand the different causes of errors in measurement and will classify the errors

according to their origination. Next we shall determine the standard procedures to arrive at

the true value and mode of representing them for all purposes.

Now in this next unit of around one hour you will be able to learn

Types of errors in measurements and level of uncertainty.

Presentation of magnitude of quantities in Physics.

Errors and Uncertainty

Errors are always the part of measurements and nothing can be done about. If a measurement

is repeated, the values obtained will differ and none of the results can be preferred over the

others. Although it is not possible to do anything about such error, it can be characterized.

For instance, the repeated measurements may cluster tightly together or they may spread

widely. This pattern can be analyzed systematically.

When we measure something the measurement is meaningless without knowing the

uncertainty in the measurement. This leads us to the idea of errors in measurement. Other

factors such as the conditions under which the measurements are taken may also affect the

uncertainty of the measurements. Thus when we report a measurement we must include the

maximum and minimum errors in the measurement.

For example, measuring the height of a person, the measure may be accurate to a scale of 1

mm. But depending on how the person being measured holds himself during the

measurement we might be accurate in measuring to the nearest cm.

Generally, errors can be divided into two broad and rough but useful classes: systematic and

random.

Systematic errors

Systematic errors are errors, which tend to shift all measurements in a systematic way so

their mean value is displaced. This may be due to such things as incorrect calibration of

equipment, consistently improper use of equipment or failure to properly account for some

effect. In a sense, a systematic error is rather like a blunder and large systematic errors can

and must be eliminated in a good experiment. But small systematic errors will always be

present. For instance, no instrument can ever be calibrated perfectly.

Other sources of systematic errors are external effects which can change the results of the

experiment, but for which the corrections are not well known. In science, the reasons why

several independent confirmations of experimental results are often required (especially

using different techniques) is because different apparatus at different places may be affected

by different systematic effects. Aside from making mistakes (such as thinking one is using

the x10 scale, and actually using the x100 scale), the reason why experiments sometimes

yield results, which may be far outside the quoted errors, is because of systematic effects,

which were not accounted for.

Random errors

Random errors are errors, which fluctuate from one measurement to the next. They yield

results distributed about some mean value. They can occur for a variety of reasons.

They may occur due to lack of sensitivity. For a sufficiently a small change an instrument

may not be able to respond to it or to indicate it or the observer may not be able to discern

it.

They may occur due to noise. There may be extraneous disturbances that cannot be taken

into account.

They may be due to imprecise definition. For example taking measurements with a

magnetic compass, the effect of improper leveling of instrument during observations.

They may also occur due to statistical processes such as the roll of dice.

Random errors displace measurements in an arbitrary direction whereas systematic errors

displace measurements in a single direction. Some systematic error can be substantially

eliminated (or properly taken into account). Random errors are unavoidable and must be

lived with.

So the systematic errors are to be removed from the measurements by rectification of the

cause or my taking measurements by several instruments otherwise the results will remain

shifted from the true value. However the random errors are very uncertain and it is very

difficult to account for them in our measurements. Random errors will always remain in our

measurements how precisely we have taken our measurements. By taking repetitive number

of measurements and taking average of large number of measurements, the average value

will be close to the true value. But still there is some uncertainty associated with the true

value.

The uncertainty associated with the mean value is determined by the standard deviation of

the measurements as detailed below:

Let x1 x2 x3 x4 x5…xN be the results of an experiment repeated N times then standard

deviation is defined as

Where is the average of all values of x. It is supposed to be the best value of x

derived from the experiments and the true value is likely to occur within a range .

However the interval of is quite often taken as the interval in which the true

value lies with 95% probability. And if the interval is chosen to be than the

probability of occurrence of true value in that interval is 99 %.

Probabilities of occurrence the true value in any range say is given by

But it is fully acceptable only if the numbers of experiments are large. In general the value of

N should be greater than 8 for a good approximation.

Fractional and percentage errors

If x is the error in the measurement in the value x then fractional and Percentage errors are

defined as :

Fractional error=x/x

Percentage error=x/x100 %

Propagation of errors (addition and subtraction)

Let error in quantity x is x and error in quantity y is y then the error in x + y or x - y

is (x+y) that is the errors add.

Prefixes and Magnitudes

To make sense of the vast range over which physical quantities are

measured, prefixes are used as a short-cut to writing the magnitude

using scientific notation. Other prefixes which are commonly used but

are not strictly part of the SI system.

Q1.When a current of 2.50.5 ampere flows through a wire, it develops a potential

difference of 201 volt. Find the resistance of the wire-

(A) 6.0 3 (B) 7.0 2(C) 8.0 2 (D) 9.0 3

Ans1: R=V/I=8Also R/R=V/V-I/IFor max R/R all terms to be positive and thereforeR/R=V/V + I/I=0.25; R=2 Ohmand so R= V/I R = 8.0 2

Q2. In an experiment the value of two resistance were measured to be as given below R1 =

5.0 0.2 ohm and R2 = 10.0 ± 0.1 ohms. Find there combined resistance in (i) series (ii)

parallel.

Ans2: When resistance are in series R=(R1+ R2)

R=R1+R2= 0.3 and R=15 0.3 Ohm

and when in parallel R=(R1R2)/(R1+ R2)

R/R=R1/R

1+R

2/R

2-R

2/(R

1+R

2)- R

2/(R

1+R

1)

For max R all terms must be positive and R/R =7%; and R=3.3 7%

Q3. In an experiment to determine acceleration due to gravity by simple pendulum, a student

commits 1% positive error in the measurement of length and 3% negative error in the

measurement of time period. Find the percentage error in the value of measurement of g.

Ans3: We have T=2 L/g; or T2= K L / g2 ln T= ln K +ln L- ln g; (2/T) dT=1/L dL-1/g dg; So actual percentage error in measurement of g value (dg/g) =7%

Q4. A naval destroyer is testing five clocks. Exactly at noon as determined by the wwv signal

on the successive days of a week the clocks read as follows

Clock Sun Mon Tue Wed Thru Fri Sat

A 12:36:40 12:36:56 12:37:12 12:37:27 12:37:44 12:37:59 12:38:14

B 11:59:59 12:00:02 11:59:57 12:00:07 12:00:02 11:59:56 12:00:03

C 15:50:45 15:51:43 15:52:41 15:53:39 15:54:37 15:55:35 15:56:33

D 12:03:59 12:02:52 12:01:45 12:00:38 11:59:31 11:58:24 11:57:17

E 12:03:59 12:02:49 12:01:54 12:01:52 12:01:32 12:01:22 12:01:12

Justify your choice.

Ans4: The standard deviations of the five clocks are increasing in order of C, D, A, B, E. Hence the clocks with minimum standard deviation is the most consistent one. So the clocks placed in the same order may be kept in terms of consistency as good timekeepers.

Q5. A wire has a mass 0.3±0.003 g, radius 0.5±0.005 mm and length

6±0.06 cm. Find out the maximum percentage error in the measurement

of its density.

Ans5: d=M/r2 L; For Max, d/d all terms should be positive so Maximum error =M/M+2/r r+L/L=0.04=4%

Vernier Calliper

The meter scale enables us to measure the length to the nearest millimeter

only. Engineers and scientists need to measure much smaller distances

accurately. For this a special type of scale called Vernier scale is used.

Vernier Calliper

The Vernier scale consists of a

main scale graduated in

centimeters and millimeters.

On the Vernier scale 0.9 cm is

divided into ten equal parts.

The least count or the smallest

reading which you can get with

the instrument can be

calculated as under:

Least count = one main scale (MS) division - one vernier scale (VS) division.

= 1 mm - 0.09 mm

= 0.1 mm

= 0.01 cm

The least count of the vernier

= 0.01 cm

The Vernier calliper consists of a main scale fitted with a jaw at one end.

Another jaw, containing the vernier scale, moves over the main scale. When the

two jaws are in contact, the zero of the main scale and the zero of the vernier

scale should coincide. If both the zeros do not coincide, there will be a positive

or negative zero error. After calculating the least count place the object

between the two jaws. Record the position of zero of the vernier scale on the

main scale (3.2 cm in figure below).

Principle of Vernier

You will notice that one of the vernier scale divisions coincides with one of the

main scale divisions. (In the illustration, 3rd division on the vernier coincides

with a MS division).

Reading of the instrument = MS div + (coinciding VS div x L.C.)

= 3.2 + (3 x 0.01)

= 3.2 + 0.03

= 3.23 cm

To measure the inner and outer diameter of a hollow cylinder or ring, inner and

outer callipers are used. Take measurements by the two methods as shown in

figure below.

Micrometer Screw-Gauge

Micrometer screw-gauge is another instrument used for measuring accurately

the diameter of a thin wire or the thickness of a sheet of metal.

It consists of a U-shaped frame fitted with a screwed spindle which is attached

to a thimble.

Screw-gauge

The screw has a known pitch such as 0.5 mm. Pitch of the screw is the distance

moved by the spindle per revolution. Hence in this case, for one revolution of

the screw the spindle moves forward or backward 0.5 mm. This movement of

the spindle is shown on an engraved linear millimeter scale on the sleeve. On

the thimble there is a circular scale which is divided into 50 or 100 equal parts.

When the anvil and spindle end are brought in contact, the edge of the circular

scale should be at the zero of the sleeve (linear scale) and the zero of the

circular scale should be opposite to the datum line of the sleeve. If the zero is

not coinciding with the datum line, there will be a positive or negative zero

error as shown in figure below.

Zero error in case of screw gauge

While taking a reading, the thimble is turned until the wire is held firmly

between the anvil and the spindle.

The least count of the micrometer screw can be calculated using the formula

given below:

Least count

= 0.01 mm

Determination of Diameter of a Wire

The wire whose thickness is to be determined is placed between the anvil and

spindle end, the thimble is rotated till the wire is firmly held between the anvil

and the spindle. The rachet is provided to avoid excessive pressure on the wire.

It prevents the spindle from further movement. The thickness of the wire could

be determined from the reading as shown in figure below.

 

 

Reading = Linear scale reading + (Coinciding circular scale x Least count) 

= 2.5 mm + (46 x 0.01) 

= (2.5 + 0.46) mm = 2.96 mm

 

Relationship in the Metric system of length  

1 kilometer (km) = 103 m 

1 centimeter (cm) = 10-2 m 

1 millimeter (mm) = 10-3 m

Q1. The pitch of a screw gauge is 1mm and there are 100 divisions on the circular scale.

While measuring diameter of a wire the linear scale reads 1mm and 47 th division on the

circular scale coincides with the reference line. The length of wire is 5.6 cm. Find the curved

surface area (in cm2) of the wire in appropriate number of significant digits.

Q2. In a Searle’s experiment the diameter of the wire as measured by a screw gauge of least

count 0.001 cm is 0.050 cm. The length measured by a scale of least count 0.1 cm is 110.0

cm. When a weight of 50 N is suspended from the wire the extension is measured to be 0.125

cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of

Young’s modulus of the material of the wire from these data.

Solution:

Ans 1: Dia of wire=1+47/100=1.47 mm=0.147 cm; Length of wire =5.6 cm

Hence curved surface area = D L=2.6 cm2.

Ans 2: Y=F.L/A.F.L/ r 2 On differentiating two sides and dividing with Y on two sides

of the equation we get, dY/Y=dL/L-2/rdr-d/

For maximum error all terms should be positive

dY/Y=dL/L+2/rdr+d/=0.0489; dY=1.09x1010

Motion in One Dimension

Introduction

We have studied till now that the laws of physics are expressed in terms

of physical quantities. These quantities are either fundamental or derived

one from these fundamental quantities so that law of nature could be

best expressed in terms of involved quantities. These quantities are

meaning less until unless we have set certain standards for quantifying

the physical quantity. The set of standards and the units involved in

totality denotes the true value of the physical quantity and then only can

participate in mathematical manipulations of laws of physics. Beside that

these physical quantities don’t take part in mathematical operations like

ordinary numeric values but are assigned certain predefined properties

according the role to play to best describe the fundamental law of nature

and are so called as scalar, vector or tensor quantity.

Now we shall start our expedition to understand the very fundamental

aspect of physical observation of nature involving the motion of particles

or rigid bodies. These all aspects of motion are covered in the study of

classical mechanics; the oldest branch of Physics, which is further,

subdivided namely Statics and Dynamics.

(A) The statics is the branch of Physics that deals with the study of physical objects or

system of objects that are at rest.

(B) The Dynamics is the branch of Physics that deals with the study of physical objects or

system of objects that are in motion. When we are not concerned with the cause of motion

and limit our study to the involved parameters of particulate motion only then the dynamics

of particles may be termed as Kinematics.

As we have come across with two terms rest and motion and for the purpose of Physics it is

not so easy to call the term rest and motion as we usually do with. The terms rest and motion

denotes the state of motion of the object under consideration with respect to frame of

reference of observer. We have stated earlier in our previous discussions that measurement

of physical quantities depends upon the frame of reference of observer. For an example an

object at rest with respect to the observer attached to a moving train, is in motion with respect

to an observer attached to the Earth. Hence the complete description of a physical quantity

desires a proper set of standard units and complete information of frame of reference for

observance.

The mechanics of motion of objects we are dealing here is a part of Classical mechanics, also

known as Newtonian mechanics (1860), doctrine of Sir Alexander Newton. As physics is not

a static tool for explaining all the phenomenon of nature but itself a developing one to face

the challenges posed by the incident results. The Newtonian mechanics was developed for

understanding the observations regarding motions of objects in the nature and are perceived

through the naked eyes. It proves all experimental results when the speeds of objects are slow

enough in comparison to the speed of light. However it is helpless in describing the

collisions, decay and interactions of elementary particles of atom like proton, electron and

neutron moving at high speed of the order of speed of light, regarding deviation from the

results for the relative velocity of particles observed from the different frame of reference and

prediction of position and velocity at a time for the high speed moving particles which is an

essential requirement for describing any physical quantity. For explaining the above

phenomenon the new theories like Einstein theory of relativity (1905) and Quantum

mechanics (1925) been developed which satisfies all experimental results involving particles

of small mass and high speed (vc). These theories are considered as a more general theory

and Newtonian mechanics is considered as a special case of application for particles moving

with velocity very very less comparable to the light.

We now return to the classical mechanics to study the slow motions, which can be perceived

through our common sense without any intuitive effort. But before taking a leap for

understanding laws of physics involved in motion of objects, we shall introduce the physical

quantities involved in various types of motion and start our expedition with the simplest kind

of motion that is motion in a single direction also called as one-dimensional motion.

At the end of this chapter of around one hour, you will be able to learn

Concept of point object

Motion in One Dimension

Distance

Displacement

Average Speed

Average Velocity

Instantaneous Velocity

Average acceleration

Instantaneous acceleration

Motion with constant acceleration

Motion with variable acceleration

Time dependent acceleration

Position dependent acceleration

Velocity dependent acceleration

Relative motion in one dimension

Concept of point object

When we think of a motion there may be different possibilities of motion either in the choice

of path or the choice of the body itself. But to limit our discussion to the beginners we have

simplified our choices. We start our discussion with motion of objects that have assumed

physically zero dimensions, called particles or point objects. One tends to think of a particle

as a tiny object, e.g., a piece of shot, but actually no size limit is implied by the word

“particle”. If we are not interested in the rotational motion of an object, any object can be

considered as a particle. For example, sometimes we consider the motion of earth around the

sun. In this case we consider the motion of the center of the Earth in the circular path and

ignore the rotational motion of the earth on its own axis then for our treatment the Earth may

be considered as a particle. In some astronomical problems the solar system or even a whole

galaxy is treated as a particle. In other words when the size of the object is very small in

comparison to the distance it moves then the object may be treated as point object. There is a

specific nature of point object is that all the points on the object undergo same displacement,

hence the displacement of any of the points may be treated in the experiment. Hence in all

the concept of point object has its significance with reference to the type of motion of the

object is under consideration and displacement that undergo in comparison to its own

dimensions.

Motion in One Dimension

To describe the motion of the particle, we are now in a stage to develop the concepts of displacement,

velocity and acceleration. In the general motion of a particle in three dimensions, these quantities are

vectors, which have direction as well as magnitude. However at this stage we have confined our

discussion to the movement of a particle in a straight linear path, with only two possible directions,

distinguished by designating one positive and the other negative. A simple example of one-

dimensional motion is a vehicle moving along a straight, aligned road. We can choose any convenient

point on the vehicle for the location of the point mass for the discussion of motion. We shall now

define various physical quantities

associated with rectilinear motion of a

point object and then try to understand

the need of their development.

Distance

The distance is defined as the length

of actual path the particle traverses in

its motion. In our case when particle is

restricted its motion in a one

dimension then the distance is defined

as the length of actual path the particle

has traversed irrespective of its

direction of motion. For example the

vehicle traveling in one dimension in

the fig say East-West direction,

moving 6 miles East ward and 4 miles

back west ward will be defined to

have moved a distance 10 miles.

Hence it is a scalar quantity and always positive and increasing with time. Its unit are (m) in

MKS and (cm) in CGS system.

Displacement

The displacement of the particle or point object in general is defined as the actual

displacement the particle has undergone with respect to its original position in the time

interval under consideration. In other words it is the change in position vector of the object in

a scheduled time. The displacement of particle is not concerned with the path it has elapsed

and journey details. It is a unique value defined with a vector having direction as the

movement from its initial to final position and magnitude equal to the straight distance

between initial and ending positions. In our previous example for a motion in one dimension

the displacement will be 2 miles east ward since the vehicle has traversed a total distance of

10 miles but has been displaced with net amount of 2 miles from it’s initial position say at

origin. Hence it is a vector quantity and units are same as of distance.

Average Speed

The average speed of a particle is defined as the ratio of the total distance moved to the time

taken up to that instant of motion. So it is simply the time rate at which the distance moved

by the particle. So

Average speed = total distance / total time

For example, if the vehicle in the previous example moves a total distance 10 miles in what

ever direction in 0.2 h, than its average speed is

which suggest us that the driver might have moved with this speed uniformly elapsing total

distance of 10 miles in the given time interval of 0.2 hours. It is therefore a scalar quantity

having SI units of meters per second, and written as m/s.

Test your understanding

Ex1. A car covers a distance of 2 km in 2.5 minute. If it covers half of the distance with

speed 40 km/hr, the rest distance it will cover with speed-

(A) 56 km/hr (B) 60 km/hr (C) 50 km/hr (D) 48 km/hr

Sol : Time taken to cover the first half distance t1 =Distance traveled/Speed

=(1/40)60 = 1.5min;

Time taken to cover remaining half distance t2=T-t1= 1 min

Speed of the car in the next half journey=

Distance traveled/Time taken =1km/min=60km/hr (B)

Ex2. Mark the wrong statement-

(A) The time displacement graph of a particle cannot be parallel to the time axis.

(B) The time displacement graph of a particle cannot be perpendicular to the time axis.

(C) The time velocity graph of a particle cannot be perpendicular to the time axis

(D) The area of the time velocity graph gives the displacement

Sol : The time displacement graph of a particle can be parallel to the time axis for a particle at rest. (A)

Ex3.The position vector of a particle is determined by the expression r = 3 t2 i^+ 4 t2 j^ + 7

k^. The distance traveled in the first 10s is-

(A) 100 m (B) 150 m (C) 500 m (D) 300 m

Sol: r = 3t2 i^ +4t2 j^ +7k^; v = dr/dt =dx/dt i^+dy/dt j^+dz/dt k^= 6ti^+8tj^ ;

S =0v dt = 0t dt= 500 m

Ex4.What is the displacement of the point of the wheel initially in contact with the ground

when the wheel roles forward half a revolution? Take the radius of the wheel R and x-axis in

forward direction.

(A) R/2+4 (B) R2+16 (C) 2R (D) R

Sol: Displacement in x direction ( x )= R; Displacement in y direction (y) = 2R; Resultant

displacement = x2+y2 = R2+4

Average Velocity

The concept of velocity is similar to that of speed but differs in respect of that here net

displacement is accounted in place of total distance moved by the particle. So average

velocity is defined as the ratio of the net vector displacement of the particle up to that instant

and the total time of motion up to that instant. Since the displacement is a vector quantity so

the average velocity is also a vector because it includes the direction of motion. We have

defined earlier the term displacement as a vector showing change in position of a particle

with magnitude of direct distance measured between initial and final position and directed to

the line joining the two points. Hence the average velocity is also a vector quantity, as a rate

of change of displacement or in other words displacement per unit time, having the same

direction as displacement. Now in the case of previous example the vehicle has elapsed a

net displacement of 2 miles in 0.2 hours in the direction toward east so the average velocity

in the given time frame may be given as

Now consider the motion of the vehicle as discussed in the previous example with direction

of motion as along x axis denoting the east ward direction and its position on the x axis at

any instant (t) may be shown as in fig . The value of x depends on the unit chosen as the

measure of distance and sign depends on its position relative to the origin O; if it is to the

right, it is positive; to the left, negative.

Suppose that our vehicle under

consideration is at position x1 at time t1

and at point x2 at time t2 . The change in

the position of the vehicle x2- x1 is called

the displacement in time t2- t1. It is

customary and easy to use the Greek letter

(capital delta) to indicate the change in a

quantity. Thus the change in x as x

The average velocity of the vehicle is defined to be the ratio of the displacement x and the

interval t =t2–t1:

It is to be noted that average velocity like displacement vector can be positive or negative

according as displacement in the direction of positive/negative x-axis of the chosen

coordinate system.

Observation of fig depicts the displacement of particle as difference of ordinate on vertical

axis in the said time interval. The line joining two points P1, P2 is the hypotenuse of the

triangle with sides x and t. The ratio is called the slope of the hypotenuse. In geometric

terms it indicates the steepness of the line and as per definition it is a measure of average

velocity of the particle. The steeper the slope indicates that particle has traversed greater

displacement in a given time interval and has got higher average velocity. Since in our case

the vehicle has moved along the +ve x-axis therefore there is no doubt about the direction of

velocity that is also along +ve x-axis.

Test your understanding

Ex5. A body covers a distance AB of 2 km with speed of 2.5 km/h, while going from A to B

and comes back from B to A with speed 0.5 km/hr, his average speed will be-

(A) 1.5 km/hr (B) 0.83 km/hr (C) 1.2 km/hr (D) 1.8 km/hr

Sol : Vav

=4/(2/2.5+2/0.5)=0.83 km/h

Ex6. A particle moves with constant speed v along a regular hexagon ABCDEF in same

order (i.e. A to B, B to C to D, D to E to F, F to A) then magnitude of average velocity for its

motion from A to C is-

(A) v (B) v/2 (C)3v/2 (D) None of these

Sol : AC= 3a; t = 2a/v; Av. Velocity = 3v/2

Ex7. A car travels for time t with a uniform velocity of 108 km/h on a straight road and then

immediately reverses gear and travels for time t on the same road with a uniform velocity of

72 km/h. then the average velocity of the car in this time interval 2t is-

(A) 90 km/h (B) 86 km/h (C) 36 km/h (D) 18 km/h.

Sol : Average velocity Vav

=Total displacement /Total time of motion=(108+72)t/2t=90 Km/h

Ex8. A person walks along an east-west street,

and a graph of his displacement from home is

shown in figure. His average velocity for the

whole time interval is-

(A) 0 m/s (B) 23 m/s (C) 8.4 m/s (D) None of

the above

Sol : (A) Since net displacement in whole time interval

is zero.

Ex9. A point traveling along a straight line traverses one-third the distance with a velocity

v0. The remaining part of the distance was covered with velocity v1 for half the time and

o

t ( s )

2 51 0 1 55

- 3 0 Q . N o . 6 4

S

4 0

with velocity v2 for the other half of the time. The mean velocity of the point averaged over

the whole time of motion will be-

(A) v0(v1+v2)/3(v1+v2+v0) (B) 3v0(v1+v2)/(v1+v2+v0) (C) v0(v1+v2)/(v1+v2+4v0) (D)

3v0(v1+v2)/(v1+v2+4v0)

Sol : Time of motion of one third distance (T1) =d/3 v

0,

Time of journey of rest 2/3 distance (T2) = 4d/3(v

1+v

2);

Vav

=d/(T1+T

2)= 3v

0(v

1+v

2)/(v

1+v

2+v

0)

Ex10. A table clock has its minute hand 4.0 cm long. The average velocity of the tip of the

minute hand between 6.00 AM to 6.30 AM and 6.30 PM will respectively be- (in cm/s)

(A) 4.4 x 10-3, 1.8 x 10-4 (B) 1.8 x 10-4, 4.4 x 10-3 (C) 8 x 10-3, 4.4 x 10-3 (D) 4.4 x 10-3,

8 x 10-4

Sol : Average velocity=displacement/time;

Av. Velocity (6 AM to 6:30 AM)=8/30x60= 4.4 x 10-3

Av. Velocity (6 AM to 6:30 PM)=8/12.5x3600=1.8x10-4

Instantaneous Velocity

At a first glance, it seems impossible to define

the velocity of a particle at a single instant,

i.e. , at a specific time t1, the particle is at a

single specific position say x1 then the

question arises that if we were talking about

velocity of a particle at a single point then as

per definition, what about the displacement at

that instant? It seems to be a paradox, which

can be resolved when we realize motion as a whole at different instants and limit our

discussion of motion to a infinitesimal time interval (t) tending to zero then in the limiting

condition displacement is termed as instantaneous velocity of the particle at that particular

instant. The basic requirement for finding out instantaneous velocity of the particle is that

one should have a complete mathematical algorithm available for position of particle at any

instant and time.

Figure shows x (t) curve indicating various sequence of time intervals t1, t2, t3, each

one smaller than the previous one. For each time interval t, the average velocity is the slope

of the dashed line appropriate for that interval. This figure shows that as the time intervals

becomes smaller, the dashed lines get steeper but never incline more then the line tangent to

the curve at point t1. We define the slope of this tangent line to be the instantaneous velocity

at the time t1. The instantaneous velocity is then defined as the limit of the ratio x/t as t

approaches zero. In the limiting notation this is defined as derivative of x (t) with respect to t

at time t1 and is written as dx/dt and its value could be found by differential calculus.

= Slope of line tangent to x (t) at time t=t1

So instantaneous velocity of the particle at any instant say t

Test your understanding

Ex11. At an instant, the coordinates of a particle are x=at2, y=b t2 and z=0,then its velocity at

the instant t will be.

(A) ta2+b2 (B) 2ta2+b2 (C) a2+b2 (D) 2t2a2+b2

Sol : velocity component in x direction vx=dx/dt=2at

Velocity component in y direction vy=dy/dt=2bt

Resultant Velocity (v) = vx2+v

y2 =2ta2+b2

Ex12.The displacement of a body is given by x=a2-t2)+ t cos t2, where t is time and a is

constant. Its velocity is.

(A) a-t2-t sint2 (B) 2 t/a2-t2+cost2-t sin t2 (C) -t/a2-t2 +cost2-2t2sint2 (D) –a/(a2-

t2)+cost2-t sint2

Sol : Velocity component in x direction v=dx/dt=-t/a2-t2 +cost2-2t2sint2

Ex 13. If the velocity–time diagram for the rectilinear motion of a particle

is as shown in figure (representing half wave of a sine curve). Find the

distance traveled by the particle in a time T/2 seconds.

(A) T v max/ (B) V max T/3 (C) V max T (D) V max T/2

Sol : S=2

Vm

sin 2tT dt=Vm

T/

Average acceleration

Average acceleration is defined as a quantity measuring the rate of change of instantaneous

velocity in a certain time interval. The change in velocity vector may be either in magnitude

or direction or both and change in either way will be termed as the average acceleration.

Since instantaneous velocity is a vector quantity and we are interested in change in vector

quantity so the acceleration can only be a vector quantity and will follow the laws of vector

algebra.

Since for a particle moving in one dimension will have velocity vector having orientation

either on +ve or -ve x-axis therefore change in velocity vector will also be associated in the

two possible directions. If the change in velocity vector is directed along the direction of

velocity vector at the instant of starting point of motion under consideration than the resultant

velocity vector gets added up and so the acceleration vector is termed as positive acceleration

and if converse is true than the acceleration is designated as negative acceleration.

So in final words average acceleration is the ratio of change in instantaneous velocity vector

(v) in a given time interval (t).

t

vaav

The dimensions of acceleration in S.I unit system are meter per Second Square.

Test your understanding

Ex14. A particle is moving eastward with a velocity of 5 ms-1 in 10 s the velocity change to

5 ms-1 northward. The average acceleration in this time is-

v

o t

v m

T /2

(A) Zero (B) 1/2ms-2 towards north–west (C) 1/2ms-2 towards north–east (D) 1/2ms-2

towards north–west

Sol : Average acceleration=v/t=50/10=1/2ms-2 towards north–west

Ex15.A rifle bullet loses 1/20 of its velocity in passing through a wooden plank. The least

number of planks required stopping the bullet is-

(A) 5 (B) 10 (C) 11 (D) 20

Sol : Let x0 be the thickness of one plank and a be the retarding acceleration produced due to

friction of the plank than

x0=u2-v2/2a; x

0=u2-(19u/20)2/2a;

Number of planks=Total path length/thickness of one plank= u2/2ax0=400/(39) 11

Instantaneous acceleration

If the instantaneous velocity of the particle is varying continuously with position and time

then instantaneous acceleration at that particular instant is defined as the change in velocity

vector for a time interval t infinitesimally small tending to zero. It is to be noted that change

in velocity vector is also a vector term and the law of vectors has to be followed. We shall

see later on that such type of situation may be handled easily when vector is considered to be

resolved into its components in the respective directions and the changes are also considered

in the respective components directions individually.

For the case of motion in one dimension the particle moves in a straight line with two

possible directions of motion say either in +ve or -ve x-axis. Now if the particle moves such

that velocity at every instant of motion is defined and there is no abrupt change in velocity

vector than the instantaneous acceleration vector is just the rate of change in magnitude of

velocity vector and direction will be in the direction of velocity vector if velocity time curve

is increasing and opposite to the velocity vector if it is a decreasing curve.

So if we plot graph between instantaneous velocity of the particle moving in one dimension

and time then as per the definition the instantaneous acceleration is defined as slope of

tangent line at that instant.

t

vta

tLim

0

)(

= Slope of v (t) curve

The instantaneous acceleration is therefore derivative of velocity with respect of time and is

written as dv/dt. Since velocity is also derivative of x with respect to t therefore acceleration

is also referred as second derivative of x with respect to t. We shall study in the later stage

how acceleration vector is useful in defining Newton’s second law of motion.

Test your understanding

Ex16. Relationship between the distance traveled by a body and the time is described by the

equation S=A + Bt + Ct2 + Dt3, where C=0.14 m/sec2, D= 0.01 m/sec

3. In what time after

motion begins, will acceleration be 1 m/sec2 and what is the average acceleration during this

time?

(A) 12 seconds, 0.64 m/sec2 (B) 6 seconds, 0.5 m/sec2 (C) 3 seconds, 0.4 m/sec2 (D) 8

seconds, 0.6 m/sec2.

Sol : v=B+2C t+3 D t2; a=2C+6 D t=1 at t=12 s

aav

=v(12)-v(0) /t=0.64 m/s2

Motion with constant acceleration along the axis of initial velocity vector

Now let us consider the motion of a particle moving with constant acceleration vector

directed along the axis of the initial velocity vector. Since the acceleration vector is directed

along the axis of initial velocity vector than it will lead to the change in magnitude of

velocity vector. If the acceleration is positive or directed along the direction of velocity

vector at any instant increases the magnitude, otherwise decreases the magnitude. Also if the

direction of initial velocity vector is different from the acceleration vector than particle

moves in the plane containing both the vectors in a parabolic trajectory, which we shall

discuss in the next chapter. The motion of a particle with constant acceleration is quite

common in nature. If air resistance is neglected than all the particles irrespective of their

masses fall with constant acceleration in the influence of gravity that is attraction of the

Earth. This acceleration due to gravity is designated by ‘g’ and has approximate value of 9.81

m/s2 or 32.2 ft/s2

For a particle moving with a constant acceleration in one dimension the velocity changes

linearly with time. The positive acceleration means it adds the velocity vector and negative

acceleration subtracts the velocity vector or in other words, for the case of one dimensional

motion, the acceleration vector will be in the direction of velocity vector if it is positive other

wise against the velocity vector.

The both cases have been illustrated in

Fig and effect on motion are

summarized here below:

If velocity of the particle is v0 at time t=0 then it’s value at time t as per definition is given

by

tatv v 0

If the particle starts at x0 at time t=0 and it’s position is x at time t then as per definition of

displacement x is given by

tx vav

Also

22 0

0 at

t

xvvvv t

av

2

2

0tv atx

2

2

00tvxa

ttx

We are sometimes interested in finding the final velocity of particle at a particular instant

when its initial velocity, constant rate of acceleration and distance traveled or displacement is

known to us. Since for constant acceleration

20 vvv t

av

and

Now after eliminating t from the above equation yields

xat vv 2022

If we are interested in finding out the displacement of particle in the nth

second of its motion

then

2

1201

navSSS nnnth

These equations are known as Newton’s equations of motion at constant acceleration.

Graphical presentation of one-dimensional motion at constant

acceleration

(1) The graph between positions of particle at

different instances vs. time is parabolic curve given

by equation as shown in fig for +ve and -ve

acceleration respectively.

(2) The graph between velocities

of particle with respect to time of

motion at constant +ve

acceleration given by equation as

shown in figure (a), (b), (c) for

different initial velocities.

(3) The graph between velocities

of particle with respect to time of

motion at constant -ve

acceleration given by equation as

shown in fig (a), (b), (c) for

different initial velocities.

The important and well-observed example of motion in one dimension at constant

acceleration is the motion of a body under the influence of gravity. At normal heights above

the earth surface the force/ acceleration remains constant and its magnitude denoted by g may

be taken as 9.80 m/sec2. It is directed radially towards the earth center and for our purpose it

may be assumed acting vertically downward that is towards the Earth surface and all the

motions in the azimuth may be considered as linear motions without influenced by the air

drag and all viscous resistance.

Let us consider motion of a body falling freely under the influence of gravity. If we choose

point of projection as origin and taking downward direction as positive, we have

u=0 and a=g as body starts from rest

that is falling freely under the influence

of gravity only, hence as per Newton’s

equation of motion at any instant ;

These equations will yield to all

unknowns at any instant of motion. It is to be noted that all the equations are vector equations

and gives vector quantities on substitution of known parameters.

These equation are presented graphically as in figure 9 (a), (b), (c)

Similarly if we consider motion of a body projected vertically upward under the influence of

gravity and choosing point of projection as origin and vertical up direction as positive. We

have

These equation are presented graphically as in figure (a), (b), (c)

Test your

understanding

Ex17. An iron ball and a wooden ball of same radius are released from a height h in vacuum.

The time taken by both of them to reach the ground is

(A) Roughly equal (B) exactly equal (C) Unequal (D) equal only at equator

Sol : (B) Since both have same initial velocity and fall with the same acceleration due to gravity (g) in the absence of air resistance.

Ex18. A body moving with uniform acceleration describes 4m in 3rd second and 12 m in the

5th second, then distance described in next three second-

(A) 100 m (B) 80 m (C) 60 m (D) 20 m

Sol : Distance traveled in nth second Sn=u+a/2(2n-1);

a=4 m/s2 and u=-6 m/s

V5=-6+4x5=14 and V8=-6+4x8=26

S=(V5+ V

8)t/2=60 m

Ex19. A particle P is at the origin and starts with velocity 2i-4j (m/s) and constant

acceleration (3i+5j) m/s2 after it has traveled for 2 seconds. Its distance from the origin is-

(A) 10 m (B) 10.2 m (C) 9.8 m (D) 11.7 m

Sol : x = uxt+1/2 a

xt2 =10m; y= u

yt+1/2 a

yt2 = 2; R = x 2 + y 2 =10.2

Ex20. A body moves from rest with constant acceleration then the variation of its K.E. with

the distance (S) traveled is represented by:

(A) Straight line (B) Parabola (C) Hyperbola (D) None of these

Sol : K.E. = 1/2mv2= 1/2m(u2+2as) = m a s which is a equation of straight line.

Ex21. A car accelerates from rest at a constant rate for sometime after which it decelerates

at constant rate to come to rest. If the total time elapsed is t sec, the maximum velocity of

car will be-

(A) /t (+B) (B) (+B) t/b (C) /(+B) t (D)t/(+)

Sol : vm = t

1; v

m = t

2; t = t

1+ t

2= v

m/ + v

m/v

m=t/(+B)

Ex22. A stone is dropped into a well and the sound of impact of stone on the water is heard

after 2.056 sec of the release of stone from the top. If acceleration due to gravity is

980-cm/sec2 and velocity of sound in air is 350 m/s, calculate the depth of the well

(A) 1.96 m (B) 19.6 m (C) 6.91 m (D) 69.1 m

Sol : Let d be the depth of well and t be the time taken to reach the stone on the water

surface then d = 1/2gt2;

t = 2d/g;

Also time taken by the sound of splash to reach the projector= d/vs

Total time taken since the release of stone to sound of splash to reach the projector=

d/vs+2d/g = 2.05;

d = 19.81m

Ex23. A particle travels for 40 seconds under the influence of a constant force. If the distance

traveled by the particle is s1 in the first twenty seconds and s2 in the next twenty seconds.

Then

(A) s2=s1 (B) s2=2s1 (C) s2=3s1 (D) s2=4s1

Sol : Let T be the total time of motion then distance moved in half of time period as say 20s

is s1 = ½ a (T/2)2=1/8a T2;

Total distance moved in time T is given as s2 = ½ aT2

Distance moved in second time interval of 20s (s2) =1/2 aT2-1/8 aT2=3/8aT2

And so s2/s

1 = 3

Ex24. A bullet fired into a fixed block of wood loses half its velocity after penetrating 60 cm.

Before coming to rest it penetrates a further distance of (Assume constant frictional

resistance)

(A) 60 cm (B) 30 cm (C) 20 cm (D) 10 cm

Sol : Let s be the distance moved by the bullet in first part of motion when loses half of its

speed as

u2/4 = u2 – 2as1;

a = 3u2/8s1;

Let s2 be the total distance moved up to the instant it comes to rest is given as

s2 = u2/2a = 4/3s1;

Distance moved in the second time interval (s2)= s

2-s1= s1/3 = 20 cm

Ex25. A particle is projected vertically upwards and it reaches the maximum height H in time

T seconds. The height of the particle at any time t will be

(A) g(t-T)2 (B) H-1/2g(t-T)2 (C)1/2g(t-T)2 (D) H-g(t-T)

Sol : h= H-1/2g(t-T)2

Ex26. At time t=0, an object is released from rest at the top of a tall building. At the time t0

a second object is dropped from the same point, ignoring air resistance the time at which the

objects have a vertical separation of l is-

(A) t=l/gt0+t0/2 (B) t= l/gt0-t0/2 (C) t=l/gt0 (D) none of the above

Sol : Total displacement up to time t, S = l = ½ g t0 2 + g t0 (t- t0)

2l/gt0 = 2t-t

0;

t = l/gt0+t

0/2

Ex27. A steel ball is dropped from the roof of a building. A man standing in front of a 1 m

high window in the building notices that the ball takes 0.1s to fall from the top to bottom of

the window. The ball continues to fall and strikes the ground. On striking the ground, the ball

gets rebounded with the same speed with which it hits the ground. If the ball appears at the

bottom of the window 2 s after passing the bottom of the window on the way down, then the

height of the building is

(A) 12.40 m (B) 21.0 m (C) 24.0 m (D) 42.0 m

Sol : Let u and v be the velocities of the ball at the instant it appears at the top and bottom of window during the downward journey. So {(u+v)/2}(0.1)=1; Or u+v=20Also v=u+g(0.1)Or v=u+1; u=19/2 m/s

Height above window (h1)= u2/20=(9.50)2/20 =4.50 m;

Height below window h3 = v+1/2g= 15.5 m;

Total height of the building H = 4.50+1+15.50= 21.0 m

Ex28. The vertical height of P above the ground is twice that of Q. A particle is

projected downward with the speed of 9.8 m/s from P and simultaneously another

particle is projected upward with the same speed of 9.8 m/s from Q both particles reach

the ground simultaneously. The time taken to reach the ground is-

(A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec.

Sol : PG = 9.8t{1+0.5t}; QG = 9.8t{0.5t-1}; PG/QG = 2; t = 6 sec

Ex29. A particle is projected vertically upwards from a point x on the ground. It takes a time

t1 to reach a point A at a height h above the ground. It continues to move and takes a time t2

to reach the ground. The velocity of the particle at half the maximum height is

(A) 22/g (t1+ t

2) (B) g/22 (t

1+ t

2) (C) 22g/(t

1+ t

2) (D) (t

1+ t

2)/ 22g

Sol : Tm= (t2-t

1)/2+t

1; Tm = (t

1+t

2)/2; u = g(t

1+t

2)/2; H = u2/2g = g2(1+t

2)2/8g; V

m= g(t

1+t

2)/22

Ex30. A juggler keeps on moving four balls in the air throwing the balls after regular

intervals. When one ball leaves his hand (speed=20 m/s) the position of the other balls

(height in meter) will be (take g=10 m/s2)-

(A) 10, 20, 10 (B) 15, 20, 15 (C) 5, 10, 20 (D) 5, 10, 20

Sol : The ball should be placed taking equal time interval of replacement and for that should be taking

time interval 2/2=1 sec and respective positions of 15, 20,15m from the hand.

Ex31. A man standing on the edge of a cliff throws a stone straight up with initial speed u

and then throw another stone straight down with the same initial speed and from the same

position. Find the ratio of the speed the stones would have attained when they hit the ground

at the base of the cliff

(A)2:1 (B) 1:2 (C) 1:1 (D) 1:2

Sol : ( C ) The stones have the same velocity at the point of throw in the downward direction.

Motion with variable acceleration

So far we have discussed the motion of a particle moving in one dimension under constant

acceleration. Now we shall extend the motion of a particle with variable acceleration in one

dimension. Since motion is in one dimension then variation in acceleration is limited to

change in magnitude only. We shall discuss the variation in three categories as:

Time dependent acceleration

Under this category of one-dimensional motion the acceleration of particle is a function of

time or in other words it may be written as a=f (t). Then velocity may be obtained as integral

of f(t) with respect to dt and displacement may be obtained as further integral with respect to

dt.

tfdt

dva

tFdttfdt

dsv

dttFs

Position dependent acceleration

Under this category of one-dimensional motion the acceleration of particle is dependent upon

the displacement of particle from origin. Then Integrating both sides of equation results

velocity as a function of displacement of particle and may be further integrated to find

displacement as a function of time.

xfdx

dvv

dt

dva

Velocity dependent acceleration

Under this category of one-dimensional motion the acceleration of the particle is dependent

upon the instantaneous velocity of particle. Then Integrating both sides of equation results

velocity as a function of time, which may be further, integrated with respect to time to find

displacement as a function of time.

vfdt

dva

Test your understanding

Ex32. The acceleration (a) of moving particle varies with displacement (x) according to the

following relation a=x2+3x, then correct relation between velocity and displacement is-

(A) v=x3+3x2+c (B) v=2/3 x3+3x2+c ( C ) v=[2/3x3+3x2+c] (D) v=2x+3 where c is a

constant

Sol : a = x2+3x; a= vdv/dx = x2+3x; v = 2/3x3+3x2+c

1

Ex33. A particle moving along straight line has a velocity v ms-1, when it cleared a distance

of x metres. These two are connected by the relation v= 49+x. When its velocity is 1ms-1,

its acceleration (in ms-2) is-

(A) 2 (B) 7 (C) 1 (D) 0.5

Sol : a= v dv/dx=0.5 m/s2

Relative motion in one dimension

As we have stated earlier that

measurement of certain physical

quantities like displacement, velocity and acceleration depends upon the reference frame of

observation. The reference frame of observation may be defined as a frame from which

observations are made. It may be either a frame fixed to the earth or a moving one. Now let

us investigate how the measurements may differ when physical parameters like displacement,

velocity and acceleration of the same moving object are measured from different reference

frames.

Now let us consider the simple one-dimensional motion of an object P. Initially at time t=0,

the object P, earth frame and moving frame are all positioned at origin of the coordinate

system as shown in figure.

Now after a lapse of time t the object P has made a displacement a with respect to the frame

attached to the earth and the moving frame has also elapsed a displacement b with respect to

the earth frame. It is most worthy to note here that the displacement of the object P as

measured by the moving frame will not be the same as measured by the earth frame, because

of the displacement of the moving frame in the same interval and would be measured as a-b.

It is now clear that the displacement and hence velocity measurements are dependent upon

the frame of reference from where the physical quantities are observed. How ever the

acceleration of the object as measured by the moving frame differs from the earth frame only

when the moving frame is also accelerating. In other words the acceleration of an object is

same as observed by the earth frame and a frame moving with constant velocity. For

example if an object is accelerating with an acceleration a1 with respect to the earth frame

and at the same time the moving fame is also accelerating with an acceleration a2 with

respect to the earth frame than the acceleration of the object with respect to the moving frame

will be given by a1-a2.

The reference frames at rest or moving with constant velocity are termed as inertial reference

frame and accelerating frames are known as non inertial frames. We shall discuss in detail

the inertial and non inertial frames during the study of applications of Newton’s laws of

motion, which are applicable in inertial reference frames where true value of acceleration is

measured and inertial state of particle under observation is defined. However it is clear from

the so far discussion that measured value of any physical quantity involving displacement

depends upon the frame of reference of observation and for correct interpretation of physical

quantity appropriate reference frame should be ascertained.

Test your understanding

Ex34. A boat P is moving at 40 km/hr and another boat Q is moving at 20 km/hr. Which one

of the following is not a possible value for their relative velocity –

(A) 10 km/hr. (B) 20 km/hr. (C) 30 km/hr. (D) 40 km/hr.

Sol : (A) The relative velocity may have values v1-v2 to v1+v2.

Ex35. Two trains along the same straight rails moving with constant velocities 60 km/hr and

30 km/hr towards each other. If at time t=0, the distance between them is 90 km, the time

when they collide is-

(A) 1 hr (B) 2 hr (C) 3 hr (D) 4 hr

Sol : (A) Time of collision (t)= Initial Displacement/(Relative velocity of approach)= 1 hr

Ex36. A boat moves relative to water with a velocity which is 1/n times the river flow

velocity. At what angle to the stream direction must the boat to move for minimizing

drifting?

(A) /2 (B) sin-1(1/n) (C) /2 + sin-1 (1/n) (D) /2- sin-1 (1/n)

Sol : Horizontal drift (D)= {v-v/n sint={v-v/n sinLv/n cos

sin = 1/n; = sin-1 (1/n);

= /2+sin-1(1/n)

Ex37. A thief walking slowly along a road sees a policeman at a perpendicular distance L

from him and starts running at a constant speed u along the road. The policeman also starts

running simultaneously with speed v always aiming at him. Find how soon the policeman

will catch the thief-

(A) t= v L/v2-u2 (B) t= vL/v2-u2 (C) t= uL/(v2-u2) (D) t= v2/L2(v2-u2)

Sol :

(u cos-v)dt=l; u 0

cos dt- vT=l

0

v cos dt=uT

cos dt=uT/v ; T=lv/(v2-u2)

Ex38. Two points move in the same straight line starting at the same moment from the same

point. The first moves with constant velocity u and the second with constant acceleration f.

During the time that elapses before second catches the first, the greatest distance between the

particles is –

(A) u/f (B) u2/2f (C) f/2u

2 (D) f/u

2

Sol : At max separation ft=u or t=u/f ; Xm = u2/f-1/2fu2/t2 = u2/2f

Ex39.A truck starts from rest with an acceleration of 1.5 m/s2 while a car 150 m behind starts

from rest with an acceleration of 2 m/s2. How long will it take before both the truck and car

side by side?

(A) 5.24 sec (B) 24.5 sec (C) 2.45 sec. (D) 52.4 sec.

Sol : (B) 150= ½ (2-1.5) t2; t=24.5 sec

Ex40. Three points are located at the vertices of an equilateral triangle whose side equals a.

They all start moving simultaneously with velocity v constant in magnitude. With the first

point heading continuously for the second, the second for the third and the third for the first.

How soon will the points converge?

(A) 3v/a (B) 2a/3v (C) a/3v (D) a/v

Sol : t = a/(v+vcos60)= 2a/3v

Ex41. A motor cycle and a car start from rest at the same place at the same time and travel in

the same direction .The cycle accelerates uniformly at 1m/s2 up to a speed of 36 km/h and

the car at 0.5 m/s2 up to a speed of 54 km/h then the distance at which the car overtakes the

cycle

(A)100 m (B) 200 m (C) 300 m (D) 400 m

Sol : Vcy = 10 m/s; Vca = 15 m/s; Time of acceleration t1=10 sec and t

2=30 sec; Scy = 50 m

t2 – 40t+200=0; t=6 sec >30 sec; 50+10(t-10)=225+15(t-30); t=35 sec and Scy=Sca=300 m

Ex42. A passenger is standing d m away from a bus. The bus begins to move with constant

acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus.

The minimum speed of the passenger so that he may catch the bus will be-

(A) 2ad (B)ad (C)2ad (D) ad

Sol : d = v.t – 1/2at2; at2 - 2vt+2d = 0; t = (2v4v2 – 8ad)/2a; t = v/a1/av2-2ad

for a real t; v2 – 2ad>0; vmin

2 = 2ad; vmin

= 2ad

Ex43. A ship sails in still water at the rate of 5 m/s. it is sailing northwards in a river flowing

eastwards with a velocity of 3 m/s. A monkey is climbing a vertical pole on the ship at the

rate of 2 m/s. A person is walking along the bank of the river at the rate of 1 m/s. To him the

monkey will appear climbing at the rate of-

(A) 10 m/s (B) 11 m/s (C) 33 m/s (D) 45 m/s

Sol : Velocity of monkey relative to man Vrel

= 29+4 = 33 and Vrel

= 41+4 = 45

Ex44. A string passes over a fixed pulley. Two boys P and Q of the same mass hang at the

same height at each end. Both start to climb upward at the same time to reach the pulley. The

velocity of P relative to the string is v and that of Q is 3v, then the time taken by P to reach

the pulley is equal to

(A) 1/3rd of the time taken by Q (B) 3 times the time taken by Q (C) the time taken by Q

(D) twice the time taken by Q

Sol : (C) Velocity of Q relative to pulley = 3v-v= 2v; Velocity of P relative to pulley =

v+v= 2v

Single Choice Type Objective Questions

Q1. At a certain moment of time the angle between velocity vector v and the acceleration a of

a particle, is greater than 900. What can be inferred about its motion at the moment?

(A) It is curvilinear and decelerated (B) it is rectilinear and accelerated.

(C) It is curvilinear and accelerated (D) it is rectilinear and decelerated

Q2. The motion of a particle of mass 1kg is confined to a plane and is determined by x=3t2,

y=2t3where x and y are its coordinates at time t, then magnitude and direction of the force on

the particle at t=1/2 second is-

(A) 2 N at 300 with the x-axis (B) 62N at 450 with the x-axis (C) 6 N at 300 with the x-axis

(D) 2N at 450 with the x-axis

Q3. For a particle moving along a straight line, the displacement x depends on time t as

x=t3+t2+t+. The ratio of its initial acceleration to its initial velocity depends-

(A) Only on and (B) only on and (C) only on and (D) only on

Q4. Which

one of the

following

curves does

not represent motion in one dimension?

Q5. The length of second’s hand in a watch is 1cm. The change in velocity of its tip in 15

second is

(A) Zero (B) (302) cm/sec (C) /30 cm/sec. (D) (2)/30 cm/sec.

Q6. The modulus of the acceleration vector is constant. The trajectory of the particle is a/an-

(A) Parabola (B) ellipse (C) hyperbola (D) circle

Q7. A body A is thrown vertically upward with the initial velocity v1. Another body B is

dropped from a height h. Find how the distance x between the bodies depends on the time t,

if the bodies begin to move simultaneously.

(A) x=h-v1t (B) x=(h-v1)t (C) x= h – v1/t (D) x= h/t-v1

Q8. Which one of the following equations represents the motion of a body with finite

constant acceleration in these equations y denotes the displacement of the body at time t and

a, b and c are the constant of the motion

(A) y= a/t + bt (B) y=at (C) y=at + bt2 (D) y=at + bt2 + ct3

Q9. The motion of particle is defined by x=a cos t and y=a sin t. The acceleration of

particle is –

(A) a(B) a2(C) a2 (D) a22

v

t

v

t

x

t

v

t

Q10.The displacement- time relationship for a particle is given by x=a0 + a1t + a2t2. The

acceleration of the particle is-

(A) a0 (B) a1 (C) a2 (D) 2a2

Q11. The acceleration vector of a particle is a constant. The trajectory of the particle is a/an-

(A) parabola (B) ellipse (C) hyperbola (D) circle

Q12. Which

of the

following

distance- time

graphs represents one-dimensional uniform motion?

Q13. A horse rider is moving towards a big mirror with velocity v. The velocity of his image

with respect to him is

(A) 0 (B) 4v (C) 2v (D) v

Q14. Figure shows the displacement- time graphs for two boys going home

from the school. Which of the following statements is correct about their

relative velocity-

(A) first increases and then decreases

(B) first decreases and then increases (C) is zero (D) is non zero but constant

Q15.A particle is

confined to move along

the x-axis between

reflecting walls at x=0

and x=a between these

x

t

x

t

x

t

x

t(A ) (B ) (C ) (D )

x

t

AB

tim e0

a

a/v

dis

plac

emen

t

2 a /v 3a /v

tim e

a/v 2a /v 3a /v0

a

dis

pla

cem

e nt

a

tim e0

a

a/v

disp

lac e

men

t

2 a /v 3a /v

tim e0

a

a/v

disp

lace

men

t

2 a /v 3a /v

(A ) (B )

(C ) (D )

two limits, moves freely at constant velocity v. If the walls are perfectly reflecting than its

displacement time graph is-

Q16. The acceleration time graph of a particle

moving along a straight line is as shown in figure. At

what time the particle acquires its initial velocity?

(A) 12 s (B) 5 s (C) 8 s (D)16 s

Q17. The displacement of a particle as a function of time is

shown in fig. The fig indicates that-

(A) The particle starts with a certain velocity, but the motion

is retarded and finally the particle stops

(B) The velocity of particle is constant throughout

(C) The acceleration of the particle is constant throughout

(D) The particle starts with a constant velocity, the motion is accelerated and finally the

particle moves with another constant velocity.

Q18. A ship sailing south-east sees another ship which is steaming at the same rate as itself

and which always appears to be in a direction due east and to be always coming nearer. Find

the direction of motion of second vessel.

a(m /s )2

10

4 t (s )o

Q .28

s (m )

4 t (s )o

20

Q .29

(A) sails south-west (B) sails south-east (C) sails 300 south of west (D) sails 300 south of

east

Q19. An particle travels along the inside of straight hollow tube, 2.0 meter long, of a

particle accelerator under uniform acceleration. How long is the particle in the tube, if it

enters at a speed of 1000 m/s and leaves at 9000 m/s.

(A) 4 x 10-4 sec (B) 2 x 10-7

sec (C) 4 x 10-3

sec (D) 2 x 10-6

sec

Q20. The adjoining cure represents the

velocity–time graph of a particle, its

acceleration value along OA,AB and BC in

metre/sec2 are respectively-

(A)1, 0, -0.5 (B)1, 0, 0.5 (C)1, 1, 0.5 (D)1, 0.5,

0

Q21. The v-t graph of a linear motion is shown in adjoining figure.

The distance from origin after 8 seconds is

(A) 18 meters (B) 16 meters (C) 8 meters (D) 6 meters

Q22. The displacement-time graph of a moving particle is

shown in the fig, the instantaneous velocity is negative at the

point

(A) D (B) F (C) C (D) E

Q23. The graph between the displacement x and time t for a particle moving

in a straight line is shown is figure. During the interval OA, AB, BC and

CD the acceleration of the particle is-

0 10 20 30

Veloc ity (m /s)

A B10

5

tim e(sec)

1 30

4

-2

4 8t(s )

v (m /s)

5 7

y

D

C x

Tim e

Di s

pla c

emen

t

x

A

D

CB

to

(A) +, 0, + , + (B) - , 0 , + , 0 (C) + , 0 , - , + (D) - , 0 , - , 0

Q24. A particle starts out at t=0 from the point x0=10 m with an initial velocity v0=15 m/s

and a constant acceleration a = -5m/s2. Then its velocity – time graph is-

Q25. A particle starts out at t=0 from the point x0=10 m with an initial velocity v0= 15 m/s

and a constant acceleration a= - 5m/s2. Then its displacement–time graph is –

Q26. The displacement time graph for a one

dimensional motion of a particle is shown in figure.

Then the instantaneous velocity at t=20 sec is

(A) 0.1 m/s (B) -0.1 m/s (C) -0.05 m/s (D) 1.0 m/s

00

2 4 6 (s)

10

20

v

(m /s)

t0

02 4 6 (s)

10

20

v

(m /s)

t0

02 4 6 (s)

10

20

v

(m /s)

t0

02 4 6 (s)

10

20

v

(m /s)

t(A ) (B ) (C ) (D )

x(m )40

302010

00 2 64 t(s )

x (m )40302010

00 2 64 t(s )

x (m )40

302010

00 2 64 t(s )

x (m )40

302010

00 2 64 t(s )

(C ) (D )(A ) (B )

20 30 40

t (s)

o 60

1

S (m

)

2

Q .40

Q27. The acceleration versus time graph of a particle is as shown

figure.The respective v-t graph of the particle is-

v

tO

v

O t

v

tO

v

tO

(A ) (B ) (C ) (D )

Q28. The displacement- time graph of a moving

particle with constant acceleration is shown in the

figure. The velocity-time graph is given by-

10

0

-10 (a)

2 t (s)

10

0

-10 (b )

2t (s)

v (m /s)v (m /s)

10

0

-10 (c)

2t (s)

v (m /s)

10

0

(d)

2 t (s)

v (m /s)

1111

Q29. Two balls are dropped from the top of a high tower with a time interval t0 second,

where t0 is smaller than the time taken by the ball to reach the floor, which is perfectly in

elastic. The distance’s’ between the two balls, plotted against the time lapse t from the instant

of dropping the second ball is represented by

tO

a

2o

s

x (m )

(t/s)

1

s

t (s)o

(a)

s

t (s)o

(b )

s

t (s)o

(c)

s

t (s)o

(d )

Q30. The graph below describes the motion of a ball

rebounding from a horizontal surface being released from a

point above the surface. The quantity represented on the y-axis

is the ball’s –

(A) Displacement (B) velocity (C) acceleration (D)

momentum.

Q31. The acceleration of a particle as a function of time is a= 1.5 t-0.15 t2. The particle starts

motion from rest at a time t = 0 sec. Then the maximum velocity in the forward direction is.

(A) 10 m/s (B) 25 m/s (C) 50 m/s (D) none of the above

Q32. A balloon going upward with a velocity of 12 m/sec is at a height of 65 m from the

Earth at any instant. Exactly at this instant a packet drops from it. How much time will the

packet take in reaching the Earth? (g=10 m/sec2)

(A) 7.5 sec (B) 10 sec (C) 5 sec (D) None

Q33. A body of mass 3 kg falls from the multistoried building 100 m high and buries itself 2

meters deep in the sand. The time of penetration will be.

(A) 9 sec (B) 0.9 sec (C) 0.09 sec (D)10 sec.

Q34. A car moving with constant acceleration covers the distance between two points 60 m

apart in 6 sec. Its speed as it passes the second point is 15 m/sec. At what prior distance from

the first point was the car at rest?

(A) 7.5 m (B) 15 m (C) 20 m (D) 25 m

Q35. A body starts from rest with constant acceleration a. It’s velocity after n second is v.

The displacement of body in last two seconds is-

(A) 2v(n-1)/n (B) v(n-1)/n (C) v(n+1)/n (D) 2v(n+1)/n

t(s)

y

o

Q . 44

Q36. A rocket is fired vertically from the ground. It moves upward with a constant

acceleration 10m/s2 for 30 seconds after which the fuel is consumed. After what time from

the instant of firing the rocket will attain the maximum height? Take g=10 m/s2

(A) 30 s (B) 45 s (C) 60 s (D) 75 s

Q37. A ball is thrown vertically upward with a velocity of 30 m/s. If the acceleration due to

gravity is 10 m/s2, what will be the distance traveled by it in the last second of motion?

(A) 5 m (B) 10 m (C) 25 m (D) 30 m.

Q38. A stone is released from an elevator going up with an acceleration a. The acceleration

of the stone after the release is-

(A) a upward (B) (g-a) upward (C) (g-a) downward (D) g downward

Q39. The water falls at regular intervals from a tap 5 m above the ground. The third drop is

leaving at instant when first one touches the ground. How far above the ground is the second

drop at that instant?

(A) 1.25 m (B) 2.50 m (C) 3.75 m (D) 4.00 m

Q40. An object is thrown

upward with a velocity u,

then its displacement time

graph is-

Q41. A car moves with uniform acceleration along straight line PQR. Its speeds at P and R

are 5 m/s and 25 m/s respectively. If PQ: QR=1:2, the ratio of the times taken by car to travel

distance PQ and QR is-

(A) 1:2 (B) 2:1 (C) 1:1 (D) 1:5

Q42. The greatest acceleration or deceleration that a train may have is a. The minimum time

in which the train can get from one station to the next at a distance s is-

s

o u /g

u /g2

(C )

s

o u /gt

u /g2

(B )

s

o u /gt t

u /g2

(D )

s

o u /gt

u /g2

(A )

E x 63

(A)s/a (B)2s/a (C) 1/2s/a (D) 2s/a

Q43. A particle moves with a constant acceleration such that in the successive time intervals

t1, t2 , t3 its average velocities are v1, v2 and v3. The ratio of v1-v2 and v2-v3 is

(A) t1- t2 : t2+ t3 (B) t1+ t2 : t2+ t3 (C) t1- t2 : t2- t1 (D) t1- t2 : t2- t3

Q44. A body is thrown up in a lift with a velocity u relative to the lift and the time of flight is

found to be ‘t’. The acceleration with which the lift is moving up will be –

(A) u-gt/t (B) u+gt/t (C) 2u-gt/t (D) 2u+gt/t

Q45. A ball is dropped from a height of 20 m and rebounds with a velocity, which is 3/4th of

the velocity with which it hits the ground. What is the time interval between the first and

second bounces (g=10 m/s2)

(A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec.

Q46. A pebble is thrown vertically upwards from bridge with an initial velocity of 4.9 m/s. It

strikes the water after 2s. If acceleration due to gravity is 9.8 m/s2. The height of the bridge

will be-

(A) 4.9 m. (B) 19.6 m (C) 9.8 m (D) 24.5 m

Q47. A man in a balloon rising vertically with an acceleration of 4.9 m/sec2, releases a ball 2

seconds after the balloon is let go from the ground. The greatest height above the ground

reached by the ball is-

(A) 14.7 m (B) 19.6 m (C) 9.8 m (D) 24.5 m

Q48. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of

the elevator in a time t1, if the elevator is stationary and in time t2 if it is moving with

constant velocity. Then

(A) t1=t2 (B) t1<t2 (C) t1>t2 (D) t1<t2 or t1>t2 depending whether lift is going up or down

Q49. A river has width 0.5 km and flows from west to East with a speed 30 km/hr. If a

boatman starts sailing his boat at a speed 40 km/hr. normal to bank, the boat shall cross the

river in time-

(A) 0.6 minute (B) 0.75 minute (C) 0.45 minute (D) 3 minute

Q50. Two trains take 3 seconds to pass one another when going in opposite direction but

only 2.5 second if the speed of one is increased by 50%. The time one train would take to

pass the other when going in the same direction at their original speed is –

(A) 10 sec (B) 12 sec (C) 15 sec (D) 18 sec

Q51. A steamer takes 12 days to reach from port A to B. Everyday only one steamer sets out

from both the ports. How many steamers does each boat meet in the open sea

(A) 12 (B) 13 (C) 23 (D) 24

Q52. A body of mass 0.5 kg is found to be moving 30 m away from the starting point during

the fourth second and 70 m towards the starting point during the ninth second of its motion.

Identify the correct statements appropriate to the motion from the following.

(A) The initial kinetic energy is 2500 J

(B) It is uniformly accelerated motion

(C) It is uniform acceleration to begin with and then uniform retardation

(D) It will be at the starting point after 10 seconds.

Q53. A freely falling object crosses T.V tower of height 102.9 m in three seconds. Find the

height above the top of the tower from which it would have started falling.

(A) 122.5 m (B) 102.9 m (C) 19.6 m (D) 82.3 m

Q54. A boatman could row his boat with a speed 10 m/sec. He wants to take his boat from P

to a point Q just opposite on the other bank of the river flowing at a speed 4 m/sec. He should

row his boat

(A) at right angle to the steam

(B) at an angle of sin-1 (2/5) with PQ up the stream

(C) at an angle of sin-1 (2/5) with PQ down the stream

(D) at an angle cos-1 (2/5) with PQ down the stream

Q55. A river is flowing from west to east at a speed of 5 meters/minute. A man on the south

bank of the river capable of swimming at 10 meters/minute in still water wants to swim

across the river in shortest time. He should swim in a direction-

(A) Due North (B) 300 east of North (C) 300 west of North (D) 600 east of North

Q56. Rahul hits a ball along the ground with a speed u in a direction, which makes an angle

300 with the line joining him and the fielder Rhodes. Rhodes runs to intercept the ball with a

speed 2u/3. At what angle should he run to intercept the ball-

(A) sin-13/2 (B) sin-12/3 (C) sin-1 ¾ (D) sin-1

4/5

Q57. A bus moves over a straight level road with an acceleration a. A boy in the bus drops a

ball outside. The acceleration of the ball with respect to the bus and the Earth respectively-

(A) a and g (B) a+g and g-a (C) a2+g2 and g (D)a2+g2 and a

Q58. During a rainstorm, raindrops are observed to be striking the ground at an angle of

with the vertical. A wind is blowing horizontal at the speed of 5.0 m/s, the speed of

raindrops is

(A) 5 sin (B) 5/sin (C) 5 cos (D) 5/cos

Q59. Two spheres of equal masses but radii R and 2R are allowed to fall in a liquid. The ratio

of there terminal velocities is.

(A) 1:4 (B) 1:2 (C) 1:1 (D) 2:1

Q60.A particles of mass m moves on the x-axis as follows: it starts from rest at t=0 from the

point x=0 and comes to rest at t=1 at the point x=1. No other information is available about

its motion at intermediate times (0<t<1). If denotes the instantaneous acceleration of the

particle, than –

(A) cannot remains positive for all in the interval 0t1

(B) cannot exceed 2 at any point or points in its path

(C) must be 4 at some point or in its path

(D) must change sign during the motion, but no other assertion can be made with the

information given.

Q61. Two particles A and B are dropped from heights of 5 m and 20 m respectively. Then

the ratio of time taken by A to that by B, to reach the ground is-

(A) 1:4 (B) 2:1 (C) 1:1 (D) 1:2

Q62. A ball weighing 0.01 kg hits a hard surface vertically with a speed of 5 m/s and

rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The

average force exerted by the surface on the ball in newtons is-

(A) 10.0 (B) 1.0 (C) 5.0 (D) 0.1

Q63. A body is projected vertically upwards with velocity of 60 ms-1. Two seconds later

another body is similarly projected with velocity of 30 ms-1. Which of the following is (are)

true?

(A) The bodies meet when both are moving upwards

(B) The bodies meet when the former is coming downward and the later is moving upwards.

(C) The bodies meet at a height of 45 m

(D) The bodies do not meet

Q64. The displacement of a moving particle in a straight line is proportional to the square of

the time than for this particle-

(A) The velocity is constant (B) The velocity is variable (C) The acceleration is constant.

(D) The acceleration is variable.

Q65. Suppose you are in a closed box, which is falling freely under gravity. If you release a

ball from your hand then to you the ball would appear to be-

(A) Falling downwards (B) going upwards (C) stationary (D) oscillating.

Q66. Two particles are projected vertically upwards with the same velocity on two different

planets with acceleration due to gravities g1 and g2 respectively. If they fall back to their

initial points of projection after lapse of times t1 and t2 respectively. Then-

(A) t1 t2= g1 g2 (B) t1 g1= t2 g2 (C) t1 g2= t2 g1 (D) t12+ t2

2= g1

2+ g2

2

Q67. The speed of a swimmer in still water is 5 m/min. He crosses a river of width 24 m

flowing with a speed 4 m/min. To reach the opposite point on the other bank, the time taken

by him is-

(A) 8 min (B) 9 min (C) 19 min (D) 20 min

Q68. A ball is thrown vertically upwards from the ground. It crosses a point at the height of

25 m twice at an interval of 4 sec. The ball was thrown with the velocity of-

(A) 20 m/sec (B) 25 m/sec (C) 30 m/sec (D) 35 m/sec

Q69. A 40 m wide canal is flowing with velocity 50 m/min. A man reaches the opposite bank

swimming 50 m. His minimum speed should be-

(A) 40 m/min (B) 50 m/min (C) 60 m/min (D) 70 m/min

Q70. A bee flies in a line from a point A to another point B in 4 seconds with a velocity of

t-2m/s. The distance A and B in metre is-

(A) 2 (B) 4 (C) 6 (D) 8

Q71. A 30 m wide canal is flowing at the rate of 20 m/min. A man can swim at the rate of 25

m/min in still water. The time taken by him to cross the canal perpendicular to the flow is-

(A) 1.0 min (B) 1.5 min (C) 2.0 min (D) 2.5 min

Q72. A man crosses a 320 m wide river perpendicular to the current in 4 minutes. If in still

water he can swim with a speed 5/3 times that of the current, then the speed of the current, in

m/min,is-

(A) 30 (B) 40 (C) 50 (D) 60

Q73. A particle is

thrown vertically

upwards. The graph

between its speed v and

time t is given by (neglecting the air resistance) the following figure-

v

t

v

t

v

t

v

t

(A ) (B ) (C ) (D )

Q74. Two particles one with constant velocity 50 m/s and the other with uniform acceleration

10 m/s2 start moving simultaneously from the same place in the same direction. They will be

at a distance of 125 m from each other after-

(A) 5 sec (B) 5(1+2) sec (C) 10 sec (D) 10(2+1) sec

Q75. A ball is dropped

vertically from a height d

above the ground. It hits

the ground and bounces up

vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v

varies with the height h above the ground as-

Q76. A particle suffers three displacements by 4 m in the northward, 2 m in the southeast and

2 m in the southwest directions. What is the displacement of the particle and what is the

distance covered by it?

(A) 4-22 m, 8 m (B) 4+22 m, 6 m (C) 16 m, 8 m (D) None of these

Q77. A body covered a distance of L m along a curved path of a quarter circle. The ratio of

distance to displacement is-

(A) /22 (B) 22/ (C)/2 (D)2/

Q78. Which of the following

graph (s) is/are not possible

Q79. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,

followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long

and required 1 seco

nd to cover. How long the drunkard takes to fall in a pit 13 m away from the start?

(A) 9 s (B) 21 s (C) 32 s (D) 37 s

v

ho

(a)

v vv

hh

hoo

o

(b) (d )(c)

S (m )

0

S (m )

t (s )

(C )

t ( s )

(D )

0t (s )0

(A )

S (m )

t (s )0(B )

S (m )

Q80. Mark the wrong statement –

(A) Nothing is in the state of absolute rest or state of absolute motion

(B) Magnitude of displacement is always equal to the distance traveled

(C) Magnitude of displacement can never be greater than the distance traveled

(D) Magnitude of instantaneous velocity is equal to the instantaneous speed.

Q81. The following graph can be seen in nature-

(A) Yes (B) No (C) Sometime (D) At a particular instant

Q82. Can a body have uniform speed but non-uniform velocity?

(A) Yes (B) No (C) Depends on direction (D) Unpredictable

Q 83. The position vector r of a particle varies with time t as r = at2 i^+ bt j^ , then the

magnitude of the instantaneous velocity of the particle at time t will be-

(A) 2at+b (B) 4a2t2+b2 (C) 2a+b (D)4a2+b2

Q84. From the adjoining displacement time graph for two

particles A & B the ratio of velocities VA:VB will be-

(A) 1:2 (B) 1:3 (C) 3:1 (D) 1:3

Q85. From the adjoining graph, the distance traversed

by particle in 4 sec is

(A) 60 m (B) 25 m (C) 55 m (D) 30 m

t (s)

-ve

+ve

O

speed (m /s)

3 0 o 6 0 o

o t(s )

s (m )

B

A

o t (s )4321

2 0

1 0

v (m /s ) Q . N o . 5 2

Q86. A car travels first half distance between two places with a speed of 40 km/h and the rest

half distance with a speed of 60 km/h. The average speed of the car will be :-

(A) 100 km/hr (B) 50 km/hr (C) 48 km/hr (D) 200 km/hr

Q87. A body moves along the sides AB, BC, and CD of a square of side 10 meter with

velocity of constant magnitude 3 meter/sec. Its Average velocity will be-

(A) 3 m/sec (B) 0.87 m/sec (C) 1.33 m/sec (D) None

Q88. The displacement of a particle moving in one-dimensional direction under a force at

time t is given by t= x+3, where x is in m and t in sec. The displacements of the particle,

when its velocity is zero, will be-

(A) 0 m (B) 3 m (C) -3 m (D) 2 m

Q89. A body moves in a straight line along, x-axis. Its distance x (in meter) from the origin

is given by x=8t-3t2. The average speed in the interval t=0 to t=1 second is-

(A) 5 ms-1 (B) -4 ms-1 (C) 6 ms-1 (D) zero

Q90. If the displacement of a particle varies with time according to the relation x=k/b[1-

exp(-bt)],then the velocity (v) of the particle is.

(A) v=k exp(-bt) (B) v=k/b exp(-bt) (C) v=k2/b exp(-bt) (D) v= k/b2 exp(-bt)

Q91. The displacement x of a particle along a straight line at time t is given by x=a0 - a1 t +

a2 t2. The acceleration of the particle is-

(A) a0 (B) a1 (C) 2a2 (D) a2

Q92. A truck traveling due to North at 20 m/s turns East and travels at the same speed. The

change in its velocity is-

(A) 202 m/s North-East (B) 202 m/s South-East (C) 402 m/s North-East (D) 20 2 m/s

North-west

Q93. Which of the following statement is not correct?

(A) A body may have zero instantaneous velocity but finite acceleration

(B) A body may have zero instantaneous acceleration but finite velocity

(C) Magnitude of instantaneous velocity is equal to instantaneous speed

(D) Magnitude of average velocity is equal to average speed.

Q94. The position–time (x-t) graphs for two students A and B

returning from their school O to their homes P and Q

respectively are shown in figure. Choose the correct statements

from below.

(A) B lives closer to the school than A

(B) A/B starts from the school earlier than B/A

(C) B walks faster than A

(D) A and B reach home at the different time.

Q95. Figure shows the displacement time graph of a particle

moving on the x-axis-

(A) The particle is continuously going in positive x direction

(B) The particle is at rest

(C) The velocity increases up to a time t0, and then becomes

constant

(D) The particle moves at a constant velocity up to a time t0 and then stops.

Q96. A car is moving with a velocity of 20 m/sec.The driver sees a stationary truck at a

distance of 100 m ahead. After some reaction time t applies the brakes, produces a

retardation of 4 m/s2. The maximum reaction times to avoid collision will be-

(A) 5 sec (B) 2.5 sec (C) 4 sec (D) 10 sec

Q97. Which one of the following represents uniformly acceleration motion? a and b are

constant and x is the distance described.

(A) x=(t-a)/b (B) x=(t-a)/b (C) t=(x-a/b) (D) x=t+a

Q98. A particle starts from rest and moving along a straight line travels 19 m in the tenth

second. The acceleration of the particle is given by-

o

t ( s )

x ( m )

B

A

P

Q

o t o t

x

(A) 1.9 m/s2 (B) 2 m/s2 (C) 3.8 m/s2 (D) 1 m/s2

Q99. A ball dropped from the top of a building takes 0.5 sec to clear the window of 4.9 m

height. What is the height of building above the window?

(A) 2.75 m (B) 5.0 m (C) 5.5 m (D) 4.9 m

Q100. If a body travels half its total path in the last second of its fall from rest. The time and

height of its fall, will respectively be- (g= 9.8 m/s2)

(A) 0.59 s, 57 m (B) 3.41 s, 57 m (C) 5.9 s, 5.7 m (D) 5.9 s, 34.1 m

Q101. A stone is dropped from the top of the tower and travels 24.5 m in the last second of

its journey. The height of the lower is

(A) 44.1 m (B) 49 m (C) 78.4 m (D) 72 m

Q102. From the foot of a tower 90 m high, a stone is thrown up so as to reach the top of

tower. Two second later another stone is dropped from the top of the tower. The two stones

will meet at height

(A) 83.6 m (B) 38.6 m (C) 63.8 m (D) 68.3 m

Q103. Two bodies are thrown vertically upward, with the same initial velocity of 98

metre/sec but 4 sec apart. How long after the first one is thrown will they meet?

(A) 10 sec (B)11 sec (C)12 sec (D) 13 sec

Exercise 2

Multiple Choice Type Questions

Q1. Mark the correct statements-

(A) The magnitude of the velocity of a particle is equal to speed

(B) The magnitude of average of velocity in an interval is equal to its average speed in that

interval

(C) It is possible to have a situation in which speed of a particle is zero but the average speed

is not zero

(D) It is possible to have a situation in which the speed of the particle is never zero but the

but the average speed in an interval is zero

Q2. A particle starts moving in a straight line with a constant acceleration a. At a time t1

seconds after the beginning of motion, the acceleration changes sign, remaining the same in

magnitude. Determine the time from the beginning of motion, till it returns to the starting

point.

(A) t1(2+2) s (B) t

1(1+2) s (C) t

12 s (D) 22 t

1 s

Q3. A person walks up a stationary escalator in 90 seconds. If the escalator moves with

person, first standing on it, take 1 minute to reach the top from ground. How much time

would it take him to walk up the moving escalator?

(A) 24 s (B) 48 s (C) 36 s (D) 40 s

Q4. The displacement of a particle moving along the x-axis is given by x=1+5(t-2)+2(t-2)2

where x is in meter and t is in second.

(A) The particle starts its motion from the origin

(B) The velocity of the particle is 5 ms-1 at t= 2 sec.

(C) The acceleration of the particle is 4 ms-2

(D) The particle starts its motion at t=2 s

Q5. The velocity time plot for a particle

moving on a straight line is shown in the

figure-

(A) The particle has a constant acceleration

(B) The particle has never turned around.

(C) The particle has zero displacement

(D) The average speed in the interval 0 to 10 s is the same as the average speed in the interval

10 s to 20 s.

Q6. The velocity of a particle is zero at t=0, then

(A) The acceleration at t=0 must be zero.

(B) The acceleration at t=0 may be zero

(C) If the acceleration is zero from t=0 to t=10 s, the speed is also zero in this interval.

V (m /s )

1 0

1 0 t (s)o

Q .5

20

(D) If the speed is zero from t=0 to t=10 s, the acceleration is also zero in this interval.

Q7. A lift performs the first part of its ascent with uniform acceleration a and remainder with

uniform retardation 2a. If t is the time of ascent, the depth of the shaft is-

(A) at2/4 (B) at

2/3 (C) at

2/2 (D) at

2/8

Q8. A particle moves along the x-axis as x= u (t-2)+a (t-2)2-

(A) The initial velocity of the particle is u

(B) The acceleration of the particle is a

(C) The acceleration of the particle is 2a

(D) At t=2s particle is at the origin.

Q9. Body A begins to move with initial velocity 2 m/sec and continues to move at a constant

acceleration a. In t = 10 seconds after the body A begins to moves a body B departs from

the same point with a initial velocity 12 m/sec and moves with the same acceleration a. What

is the maximum acceleration a, at which the body B can overtake A?

(A) 1 m/sec2 (B) 2 m/sec2 (C) 1/2 m/sec2 (D) 3 m/sec2

Q10. N particles moving in a straight line have initial velocities of 1, 2, 3,....N m/s and

acceleration of 1, 2, 3,...N m/s2 respectively. If the initial spacing between any two

consecutive particles is same then, select the correct alternatives(s).

(A) The distance between any two particles remains constant

(B) The distance between any two consecutive particles is same for all particles and increases

with time.

(C) The distance between any two consecutive particles is different and increases with time.

(D) The distance between any two consecutive particles increases periodically with time.

Q11.A man throws a stone vertically up with a speed of 20 ms-1 from top of a high-rise

building. Two second later, an identical stone is thrown vertically downward with the same

speed 20 ms-1 then

(A) The relative velocity between the two stones remains constant till one hits the ground

(B) Both will have the same kinetic energy, when they hit the ground

(C) The relative acceleration between the two is equal to zero initially

(D) The time interval between their hitting the ground is 2 seconds.

Q12. Mark the correct statements for a particle going on a straight line –

(A) If the velocity and acceleration have opposite sign, the particle is slowing down.

(B) If the position and velocity have opposite sign, the particle is moving towards the origin

(C) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

(D) If the velocity is zero for a time interval, the acceleration is zero at any instant within the

time interval.

Q13. A particle move with an initial velocity v0 and retardation v, where v is its velocity at

any time then-

(A) The particle will cover a total distance v0/

(B) The particle will come to rest after time 1/

(C) The particle will continue to move for a very long time

(D) The velocity of the particle will become v0/2 after time 1/

Q14. The instantaneous velocity of a particle is related to its displacement x according to the

relation v=a x + b, where a>0 and ba/7. Which of the following statement(s) is (are) true if

x=0 at t=0

(A) The displacement of the particle at the time t is x=b/a (eat-1)

(B) The particle will experience a retardation if b<0

(C) The particle will be at origin at time t=0

(D) The acceleration of the particle is constant.

Q15. A particle initially at rest moves from a fixed point in a straight line so that at the end of

t seconds its acceleration is sin t + 1/(t+1)2. The distance from the fixed point at the end of

seconds will be.

(A) 2-log (+1) (B) 2-log (C) 2+log (+1)

Q16. The speed v of a particle moving along a straight line, when it is at a distance, x from a

fixed point on the line is given by v2=108-9x

2, all quantities are in c.g.s.units.

(A) The motion is uniformly acceleration along a straight line.

(B) The magnitude of the acceleration at a distance 3cm from the fixed point is 27ms-2

(C) The motion is S.H.M. about a given fixed point.

(D) The maximum displacement from the fixed point is 4cm

Q17. An observer moves with a constant speed along the line joining two stationary objects.

He will observe that the two objects-

(A) Have the same speed (B) have the same velocity (B) Move in the same direction (D)

moves in opposite directions.

Q 18. Out of the following graphs which is/are not possible-

Q19. The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y(t)= b

sin(pt), where a, b (<a) and p are positive constants of appropriate dimension. Then-

(A) The path of the particle is an ellipse

(B) The velocity and acceleration of the particle are normal to each other at t= /(2p)

(C) The acceleration of the particle is always towards a a focus.

(D) The distance traveled by the particle in time interval t=0 to t=/(2p), is a2+b2

Q20. If the velocity of a body is constant-

(A)Velocity = speed (B)Average velocity= speed (C) Velocity = average velocity (D)

speed= average speed

Q21. If a particle travels a linear distance at speed v1 and come back along the same track at

speed v2-

v

o x

( A )

v

o t

( B )

v

o t

( C )

v

o t

( D )

Q . N O . 4 7 E x 2

(A) Its average speed is arithmetic mean (v1+v2)/2

(B) Its average speed is geometric mean v1v2

(C) Its average speed is harmonic mean 2v1v2/(v1+v2)

(D) Its average velocity is zero

Descriptive Question Type

Q1. The acceleration of a particle depends on the velocity as a =kv.at t=0, x=0 and v=1m/s,

find the position of the particle as a function of time.

Q2. A student argues that the mean velocity during an interval of time can also be expressed

as

(v)= (vf + vi)/2 and this should always be the equal to (rf - ri)/(t2-t1) . Is he right?

Q3.What is meant by dv/dt and dv/dt? Can these be equal?

(i) dv/dt= 0 while dv/dt0

(ii) dvdt 0 while dv/dt=0

Q4. Is it possible to be accelerating if you are traveling at constant speed? Is it possible to

round curve with zero acceleration? With constant

acceleration? With variable acceleration?

Q5. x/t graphs for two cars A and B are B are as shown:

(i) Which car is moving faster?

(ii) When will they meet?

Q6. How long does it take for a train to increase its velocity in a uniformly accelerated

motion from 12 km/h to 60 km/h over a distance of 600 m? What is its acceleration?

Q7. A car is moving along a straight line with retardation 2 m/s2. After time t=3 s, its

velocity reduces to 5 m/s. Find the distance traveled in time t=3s.

o t o t (s )

x ( m ) Q .N o . 75

1 .0

6 0 o

6 0 o

B

A

Q8. Two trains, each having a speed of 30 miles/hr, are headed at each other on the same

straight track. A bird that can fly 60 miles/hr flies off one train when they are 60 miles apart

and heads directly for the other train. On reaching the other train it flies directly back to the

first train, and so forth (a) how many trips can the bird make from one train to the other

before they crash? (b) What is the total distance the bird travels?

Q9. A particle starts moving along a straight line with acceleration a=kt. (i) Find its velocity

after time t=n sec.

(ii) Find the distance traveled in t=n sec.

Q10. The acceleration of a particle is given as a = 9x. At t=0, x=1 m and v=3 m/s. Find the

velocity of the particle at t=2 sec.

Q11.The acceleration of a cart started at t=0,

varies with time as shown in fig. Find the distance

traveled in 30 second and draw the position–time

graph.

Q12. An arrow while being shot from a bow was accelerated over a distance of 2.0 ft. If its

speed at the moment it left the bow was 200 ft/s, than what was the average acceleration

imparted by the bow? Justify any assumptions you need to make.

Q13. A man can swim at a speed of 3 km/h in still water. He wants to cross a 500 m wide

river flowing at 2 km/h. He keeps himself always at an angle of 1200 with the river flow

while swimming.

(a) Find the time he takes to cross the river.

(b) At what point on the opposite bank will he arrive?

Q14. Usually “average speed” means the ratio of the total distance covered to the time

elapsed. However some time the phrase “average speed” can mean the magnitude of the

average velocity. Are the two same?

0

0 .5

-0 .5Acc

eler

atio

n

i n f

t/s2

1020 30

Tim e in Second

Q15. Consider a particle initially moving with a velocity of 5 m/s starts deceleration at a

constant rate of 2 m/s2. Determine (i) The time at which the particle becomes stationery. (ii)

The distance traveled in the 2ndsecond.

(iii) The distance traveled in the third second.

Q16. A ball is thrown up. If the air resistance is taken into account and is supposed to be

constant. Will the time of ascent be longer or shorter than the time descent?

Q17. State whether this assertion is right or wrong. “Two balls of different masses are thrown

vertically upwards with the same speed. They reach through the point of projection in their

downward motion with the same speed”.

Q18. Two balls are dropped from the same point after an interval of 1s. If acceleration due to

gravity is 10 m/s2, what will be the separation 3 seconds after the release of first ball?

Q19. Let the distance traveled by an object, as a function of time is given by s=2+3t2 where s

is in metre and t is sec. Then calculate (i) instantaneous speed at t1=1 sec and at t2=5 sec. (ii)

average speed between the time interval t1=1 sec to t2=5 sec.

Q20. A rod of length l leans by its upper end against a smooth vertical wall, while its other

end leans against the floor. The end leans against the wall moves uniformly downward. Will

the other end move uniformly too?

Q21. Whether a driver was exceeding a 30 mile/h speed limit before he made an emergency

stop. The length of skid marks on the road was 19.2 ft. Policeman made the reasonable

assumption that the maximum deceleration of the car would not exceed the acceleration of a

freely falling body and arrested the driver for speeding. Was he speeding? Explain.

Q22. A stone is dropped from a balloon going up with a uniform velocity of 5.0 m/s. If the

balloon was 50 m high when the stone was dropped, find its height when the stone hits the

ground. Take g=10 m/s2.

Q23. A particle moves in the x-y plane with a constant acceleration 1.5 m/s2 in the direction

making an angle of 370 with the x-axis. At t=0 the particle is at the origin and its velocity is

8.0 m/s along the x-axis. Find the velocity and the position of the particle at t=4.0 s.

Q24. (A) A train moving at an essentially constant speed of 60 mile/h moves eastward for 40

min, then in a direction 450 east of North for 20 min, and finally westward for 50 min. What

is the average velocity of the train during this run?

(B) A point traversed half a circle of radius R=160 cm during time interval =10.0 s.

Calculate the following quantities averaged over that time:

(a) The mean speed (v)

(b) The modulus of the mean velocity vector <v>

Q25. A train traveling at 72 km/h is checked by track repairs. It retards uniformly for 200 m

covering the next 400 m at constant speed and accelerates uniformly to 72 km/h in a further

600 m. If the time at constant lower speed is equal to the sum of the times taken in retarding

and accelerating. Find the total time taken.

Q26. A boy standing on a long railroad car throws a ball straight upwards. The car is moving

on the horizontal road with an acceleration of 1m/s2 and the projection velocity in the

vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Q27. At a distance L=400 m from the traffic light, brakes are applied to a locomotive

moving at a velocity v = 54 km/h. Determine the position of the locomotive relative to the

traffic light 1 minute after the application of brakes if its acceleration a= - 0.3 m/s2

Q28. An ant runs from an anthill in a straight line so that its velocity is inversely

proportional to the distance from the center of the anthill. When the ant is at point A at a

distance L1=1m from the center of the ant-hill, its velocity v1=2 cm/s, what time will it take

ant to run from point A to point B, which is at a distance L2=2 m from the center of the ant

hill?

Q29. A point moving in a straight line with uniform acceleration describes distances a, b

meters in successive intervals of t1

, t2 seconds respectively. What is the acceleration if a=b?

Q30. Two trains A and B are moving along parallel rails. A particle is thrown from train A

vertically upwards. What will be the shape of the path as seen by (a) an observer standing on

the ground, (b) an observer in train B moving with same speed as train A in the same

direction, (C) an observer in train B moving with twice the speed that of A in the same

direction?

Q31. Two particles A and B move in a uniform gravitational field where the acceleration due

to gravity is g. Initially the particles located at a point O move with velocities v1 and v

2

horizontally in opposite directions. At what time from the start of motion the velocities will

be perpendicular to each other and what will be the distance between them at this instant?

Q32. The current velocity of a river grows in proportion to its distance from the bank and

attains the maximum value v0 at the middle of the stream. Near the bank its velocity is zero.

A boat is moving along this river in such a manner that it is always perpendicular to the

current. It’s speed in still water is u. The width of river is c. By how much distance it will be

drifted by the current?

Q33. A truck has to carry a load in the shortest time from one station to another station

situated at a distance L from the first. It can start up or slowdown at the same acceleration or

deceleration a. What maximum velocity must the truck attain to satisfy this condition?

Q34. The speed of a train increases at a constant rate from zero to V, then remains constant

for an interval and finally decreases to zero at a constant rate . If l is the total distance

described compute the total time taken.

Q35. A motorcar can acquire in one minute by uniform acceleration a speed 90 m/sec. When

it is halting at a place in a straight narrow road it sees another car approaching it from behind

with uniform speed 60 m/sec. Find out whether it will be possible to avoid collision if the

first car starts in full force before the second car has approached it within 1200 meters.

Q36. Two points A and B move with speeds v and 2v in two concentric circles, with center at

O and radii 2r and r respectively in the same sense and if OAB= find the value of when

the relative motion is along AB.

Q37. A person traveling with velocity u towards northeast finds that the wind appears to

come from north. But on doubling his speed it seems to come from a direction inclined at an

angle cot–12 on the east of north. Find the true velocity and the direction of the wind.

Q38. The slider block B starts from rest and moves to the

right with a constant acceleration. After 4s the relative

velocity of A with respect to B is 0.06 m/s. Determine (a) the acceleration of A and B (b) the

velocity of B after 3s.

Q39. A motorist is traveling at 365/6 m/s. when he observes that a traffic light 800 m ahead

of him turns red. The traffic light is timed to stay red for 15 second. If the motorist wishes to

pass the light without stopping just as it turns green again, determine (a) the required

uniform deceleration of the car, (b) the speed of the car as it passes the light.

Q40. A train A and an automobile B travel at the constant

speeds 120 km/hr and 96 km/hr respectively as shown. Three

seconds after the train A passes under the highway bridge the

automobile crosses the bridge. Determine the velocity of train

relative to the automobile and the change in position of the

train relative to the automobile during four-second interval.

Q41. From point A located on a highway as shown in fig, one

has to get by a car as soon as possible to point B located in the

field at a distance l from the highway. It is known that the car moves in the field n times

slower than on the highway. At what distance from point D one must turn of the highway?

Q42. A point travels along the x-axis with a velocity

whose projection v(x) is presented as a function of

time by the plot in fig. Assuming the co-ordinate of

A B

CA

L

B

D

Q . N o E x . 3 ( 2 2 )

Q . N o E x .3 (2 3)

o

V (x )

1 3 4 6 7 t (s )

- 2

1

A6 0 0

1 2 0 k m /h r90 k m /h rB

the point x=0 at the moment t=0, draw the approximate time dependence plot for the

acceleration wx, the x co-ordinate, and the distance covered s.

Q43. Find out the direction of of shortest route for a boat rowing in the river when the

velocity of boat is greater or lower than of the river. Where v is velocity of boat and u is the

velocity of the river.

Q44. A bullet traveling horizontally pierces in succession three screens placed at equal

distance ‘a’ apart. If the time from the first screen to the second be t1 and from the second to

the third t2,

(a) Find the retardation assuming it to be uniform and (b) also calculate the velocity at the

middle screen. Only the screens cause the retardation and medium between the screens offers

no resistance.

Q45. Car A is traveling along a straight highway.

While B is moving along a circular exit ramp of

150 m radius. The speed of A is increased at the

rate of 1.5m/s2 and the speed of B is being

decreased at the rate of 0.9 m/s2 for the position

shown. Determine (a) the velocity of A relative to

B. (b) the acceleration of A relative to B.

Q46. Three blocks are connected by pulleys as shown in figure. Find

the velocity of each block given that relative velocity of block A with

V km hrA 75 /

V km hrB 40 /

30 0150 m

A

B

AB

C

respect to C is 0.3 m/s upwards and that the relative velocity of block B with respect to block

A is 0.2 m/s downwards.

Q47. A dog sees a flowerpot sail up and then back down past a window 5.0 ft. high. If the

total time the pot is in sight is 1.0 s, find the height above the window that the pot rises.

Q48. A particle is traveling with a uniform acceleration. If a, b and c were the distances

covered by it during x th, y th and z th second of its motion respectively, then find out the

value of a(y-z)+b(z-x)+c(x-y)

Q49. Find out whether for the particle moving with uniform acceleration the distances

described in consecutive equal interval of time are in A.P.

Q50. Three points are located at the vertices of an equilateral triangle whose side equals a.

They all start moving simultaneously with velocity constant in modulus, with the first point

heading continually for the second, the second for the third, and the third for the first so on.

How soon the particles collide each other?

Q51. Point A moves uniformly with velocity so that the vector is continually “aimed “ at

point B which in its turn moves rectilinearly and uniformly with velocity u<. At the initial

moment vu and the points are separated by a distance l. How soon will the points converge?

Q52. A train of length l = 350 m starts moving rectilinearly with constant acceleration w =

3.0x10-2 m/s2; t = 30 sec after the start the locomotive headlight is switched on (event 1),

and T=60 sec after the event the tail signal is switched on (event 2). Find the distance

between these events in the reference frame fixed to the train and to the earth. How and at

what constant velocity V, relative to the earth the frame K should move for the two events to

occur at the same point.

Q53. A particle moving with uniform retardation is found to cover three successive equal

distances. The average velocities during the first and third parts of the journey are 20 m/s and

12 m/s respectively. Determine its average velocity in the middle part of the journey.

Q54. A train passes a station A at 40 km/h and maintains this speed for 7 km and is then

uniformly retarded; stopping at B which is 8.5 km from A. A second train starts from A at the

instant the first train passes and being accelerated for the part of journey and uniformly

retarded for the rest, stops at B the same time as the first train. What is the greatest speed of

the second train?

Q55. A bus is beginning to move with an acceleration of 0.5 m/s2. A man standing 20 meters

behind the bus runs at a constant speed of 4.5 m/s.

(a) Find the time in which man will overtake the bus running at a constant speed of 4.5 m/s.

(b) Find out whether he will be able to catch the bus if he is 20.25 meters behind,.

Q56. Two trains A and B leave the same station on parallel lines. A starts with uniform

acceleration of 1/6 m/s2 and attain a speed of 24 km per hour when steam is reduced to keep

the speed constant. B leaves 40 seconds after with uniform acceleration of 1/3 m/s2 to attain

a maximum speed of 48 km/hr. When will B overtake A?

Q57. A particle moving in a straight, line with uniform retardation leaves point O at a time

t=0 and comes to an instantaneous rest at D. On its way to the point D the particle passes

point A, B and C at time t=T, 2T and 4T respectively after leaving O. Given that AB = BC

=L, find (i) the length CD and (ii) the length OA.

Q58. A particle is moving in a straight line and is observed to be at a distance ‘a’ from a

marked point initially to be at a distance ‘b’ after an interval of n seconds, to be at a distance

c after 2n seconds and to be at a distance ‘d’ after 3n seconds. If the acceleration is uniform,

then find the acceleration of the particle.

Q59. State whether the following assertion is right or wong. “ If a body moving with uniform

acceleration in a straight line describes successive equal distances in time intervals t1, t2, t3;

then 1/t1-1/t2+1/t3= 3/(t1+t2+t3)”

Q60.A railway train goes from one station to another moving during the first part of the

journey with uniform acceleration f, when the steam is shut off and breaks are applied it

moves with uniform retardation g. If ‘a’ be the distance between the stations, then find out

the time that the train takes in moving one station to another.

Q61. A particle starts from rest with acceleration f, at the end of time t; it becomes 2f, at the

end of 2t; it becomes 3f at the end of time 3t and so on. Find the velocity and distance

described at the end of time nt.

Q62. Two cars start off to race with velocities u and v and travel in a straight line with

uniform acceleration p and q respectively. If the race ends in a dead heat, find out the length

of the course.

Q63. For 1/m of the distance between two stations, a train is uniformly accelerated and for

1/n of the distance it is uniformly retarded. It start from rest at one station and comes to rest

at other station. Find out the ratio of its greatest velocity to its average velocity.

Q64. A driver has a definite reaction time and is capable of stopping his car over a

distance of 30 m on seeing a red traffic signal ,when the speed of the car is 72 km/h and

over a distance of 10m , when the speed is 36 km/h. Find the distance over which he can

stop the car if it were running at a speed of 54 km/h. Assume that his reaction time and

its deceleration of the car remains same in all three cases .

Q65. A ,B, C and D are points in a vertical line such that AB=BC=CD. If a body falls from

rest at A, find out the ratio of times taken for distances AB, BC and CD.

Q66. From an elevated point A, a stone is projected vertically upwards. When the stone

reaches a distance h below A, its velocity is double of what is at a height h above A. Find out

the greatest height attained by the stone above A.

Q67. A particle is dropped from the top of a tower h meter high and at the same moment

another particle is projected upward from the bottom. They meet when the upper one has

descended a distance h/n. Find out the ratio of the velocities of the two when they meet and

the initial velocity of the particle projected.

Q68. Two steel balls fall freely on an elastic slab. The first ball is dropped from a height

h1=44 cm and the second from the height h2=11cm,s after the first ball. After the passage

of the time t, the velocities of the balls coincide in magnitude and direction. Determine the

time and the time interval during which the velocities of the two balls will be equal,

assuming that the balls do not collide.

Q69. A radius vector of a particle varies with time t as r = a t(1-t), where a is a constant

vector and is a positive factor. Find:

(a) The velocity v and the acceleration w of the particle as functions of time.

(b) The time interval t taken by the particle to return to the initial points, and the distance s

covered during that time.

Q70. A lift performs the first part of its ascent with uniform acceleration f and the remainder

with uniform retardation 2f. If t is the time of ascent, find the depth of the shaft.

Q71. A particle moves in a straight line with the

velocity shown in the figure. Knowing that x=-

16 m at t=0, draw the acceleration–time and

displacement-time curves for 0<t<30 s and

determine the maximum value of the position

coordinate of the particle.

Q72. How long will a plane take to fly around a square with side a, when the wind is blowing

at a speed u parallel to the diagonal of the square? The velocity of plane in still air is v>u.

Q73. A man rows directly across a flowing river in time t1 and rows an equal distance down

the stream in time t2. If u be the speed of man in still water and v that of the stream then find

out t1/t2.

Q74. Two particles. 1 and 2, move with constant velocities v1 and v2 along two mutually

perpendicular straight lines towards the intersection point O. At the moment t=0, the particle

were located at the distance l1 and l2 from the point O. How soon will the distance between

the particles become the smallest? What is it equal to?

Q75. A solid ball of density half that of water falls freely under gravity from a height of 19.6

m and then enters water. Up to what depth will the ball go? How much time will it take to

t(s)

30241810

2

-6

m /sV

come again to the water surface? Neglect air resistance and viscosity effects in water. (g=9.8

m/s2)

Q76. A ball of density d is dropped onto a horizontal solid surface. It bounces elastically

from the surface and returns to its original position in time t1. Next, the ball is released and it

falls through the same height before striking the surface of a liquid of density dL,. (a) If

d<dL, obtain an expression (in terms of d. t1 and dL) for the time t2 the ball takes to come

back to the position from which it was released (b) Is the motion of the ball simple

harmonic? (c) If d=dL how does the speed of the ball depend on its depth inside the liquid ?

Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.

(g=10 m/s2)

Q77. A motorboat going downstream overcame a raft at a point A; t = 60 min later it turned

back and after some time passed the raft at a distance l =6.0 km from the point A. Find the

flow velocity assuming the duty of the engine to be constant.

Q78. Two particles, 1 and 2, moves with constant velocities v1 and v2. At the initial moment

their radius vectors are equal to r1 and r2 , how must these four vectors be interrelated for the

particles to collide?

Q79. Two particles move in a uniform gravitational field with an acceleration g. At the initial

moment the particles were located at one point and moved with velocities 1 = 3.0 m/s and

2 = 4.0 m/s horizontally in opposite directions. Find the distance between the particle at the

moment when their velocity vectors become mutually perpendicular.

Q80. A helicopter takes off along the vertical with an acceleration a=3 m/s2 and zero initial

velocity at a certain time t1, the pilot switches off the engine. At the point of take-off, the

sound dies away at a time t2=30s.

Determine the velocity v of the helicopter at the moment when its engine is switched off.

Assuming that the velocity c of sound is 320 m/s. (Time is measured from the instant of take

off).

Q81. A motorboat travels the distance between two spots on a river in 8 hours and 12 hours

down stream and upstream respectively. Calculate the time required by the boat to cover this

distance in still water.

Q82. A person sitting on the top of a tall building is dropping balls at regular intervals of one

second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.

Q83. A point mass starts moving in a straight line with a constant acceleration a. At a time t1

after the bigining of motion, the acceleration change sign, remaining the same in magnitude.

Determine the time t from the beginning of motion in which the point mass returns to the

initial position.

Q84. A pilot is taking his plane towards north with a velocity of 100 km/h. At that place the

wind is blowing with a speed of 60 km/h from east to west. Calculate the resultant velocity of

the plane. How far the plane will be after 20 min. from the starting point?

Q85. The velocity of a particle moving in the positive direction of the x axis varies as

v=x , where is a positive constant. Assuming that at the moment t=0, the particle was

located at the point x=0, find mean velocity of the particle averaged over the time that the

particle takes to cover the first s meter of the path.

Q86. An elevator car whose floor to ceiling distance is equal to 2.7 m starts ascending with

constant acceleration 1.2 m/s2; 2.0 s after the start a bolt begins falling from the ceiling of the

car. Find:

(a) The bolt’s free fall time;

(b) The displacement and the distance covered by the bolt during the free fall in the reference

frame fixed to the elevator shaft.

Q87. A truck starts from rest with an acceleration of 1.5 metre/sec2 while a car 150 metre

behind starts from rest with an acceleration of 2 metre/sec2. How long will it take before

both the truck and car side by side, and how much distance is traveled by each?

Q88. A point traversed half the distance with a velocity v0. The remaining part of the

distance was covered with velocity v1 for half the time, and with velocity v2 for the other

half of the time. Find out mean velocity of the point averaged over the whole time of motion.

Q89. The graph of x versus t as shown in

the fig for a particle in straight-line

motion. State for each interval whether

the velocity v( x) is +, -, or 0, and

whether the acceleration a(x) is +, -, or 0. The intervals are OA, AB, BC, and CD. From the

curve is there any interval over which the acceleration is obviously not constant?

Q90. A ball is dropped from a height of 19.6 m above the ground. It rebounds from the

ground and raises itself up to the same height. Take the starting point as the origin and

vertically downward as the positive x-axis. Draw approximate plots of x versus t, v versus t

and a versus t. Neglect the small interval during which the ball was in contact with the

ground.

Q91. A tennis ball is dropped onto the floor from height of 4.0 ft. It rebounds to a height of

3.0 ft. If the ball was in contact with the floor for 0.010 s, what was its average acceleration

during contact?

Q92. Distance between two points A and B is 33 m. A particle P starts from B with a

velocity of 1 m/s along AB with an acceleration of 2 m/s2. Simultaneously another particle Q

starts from A with a velocity of 9 m/s in the direction AB and has an acceleration 1 m/s2 in

the direction AB. Find whether Q will be able to catch P.

Q93. Two cars A and B having velocities of 72 km/h and 18 km/h running in the same

direction. The car B being ahead of the A. The distance between the cars is 150 m. If the car

now starts retarding at a uniform rate of 1 m/s2 while the car B moves along at a uniform

velocity. Will the car A over take the car B?

AB C

D

t

x

O t

x

O t

(a) (b )

A BC

D

Q94. A point moving with constant acceleration from A to B in the straight line AB has

velocities u and v at A and B respectively. Find its velocity at C, the mid-point of AB and the

ratio of v and u when time from A to C is twice of C to B.

Q95 A car starts moving rectilinearly a=5.0 m/s2 (the initial velocity is equal to zero), then

uniformly, and finally decelerating at the same rate a, comes to a stop. The total time of

motion equals T=25 s. The average velocity during that time is equal to v = 72 km per hour.

How long does the car move uniformly?

Q96.From the velocity–time graph of a particle given

in figure, describe the motion of the particle

qualitatively in the interval 0 to 4 s. Find

(a) The distance traveled during first two seconds, 2 s

to 4 s, and 0 to 4 s,

(b) displacement during 0 to 4s,

(c) acceleration at t=1/2 s and at t=2 s.

Q97. A particle beginning from rest, travels a distance s with uniform acceleration and

immediately after travels a distance of 3 s with uniform speed followed by a distance 5 s with

uniform decleration and comes to rest. Find the ratio of average speed to the maximum speed

of the particle.

Q98. When a particle is projected up ward with speed u from the top of a tower, it reaches

the ground in time t1. When it is projected downward with the same speed, it reaches the

ground in time t2.

How long does it take to reach the ground if it is just dropped.

Q99. From the foot of a lower 90 m high a stone is thrown up so as to reach the top of the

tower. Two second later another stone is dropped from the top of the tower. When and where

two stones meet.

Q . N o E x.5 (13)

o

10

V (m /s)

Q . N o E x. (13)

1 2 3 4 5 6

F

t (s)

-1 0

EA

C

B D

Q100. A rocket is fired vertically up from the ground with a resultant vertical acceleration of

10 m/s2. The fuel is finished in 1 minute and it continues to move up.

(a) The maximum height reached

(b) After how much time from the instant of finishing fuel, the maximum height be reached

(take g=10 m/s2)

Q101. A car, starting from rest, starts moving with an acceleration a= t. At the same instant

a truck passes that point with a velocity 4 m/s. in the same direction. After how much

distance, the car overtakes the truck?

Q102. An engine driver of a passenger train traveling at 40 m/s sees a goods train, whose

last compartment is 250 m ahead on the same track. The goods train is traveling in the same

direction as the passenger train, with a contant speed of 20 m/s. The passenger train driver

has reaction time of 0.5 sec. He applies the brakes which causes the train to declerate at the

rate of 1 m/sec2, while the goods train continues with its constant speed. Can the driver avoid

a crash?

Q103. Two boats A and B move away from a buoy anchored at the middle of a river along

mutually perpendicular straight line. The boat A moves along the stream and the boat B

across the river and after moving off an equal distance of 500 meter from the buoy both the

boats returned to their original position. Find out the ratio of the time taken by boat A to boat

B if the velocity of each boat with respect to water is 20 m/s and the stream velocity is 10

m/s.

Q104. A motorboat , with its engine on a running and blown over by a horizontal wind is

observed to travel at 20 km/hr in a direction 530 east of North. The velocity of the boat with

its engine in still water and blown over by the horizontal winds is 4 km/hr East ward and the

running river, in the absence of wind is 8 km/hr due south. Find out the velocity of the boat

in magnitude and direction over still water in the absence of wind.

Q105 A ship moves along the equator to the east with velocity 30 km/hr. Wind blows from

south-eastern direction at an angle 600 to the equator with velocity 15 km/hour. Find out the

wind velocity relative to the ship and angle between the equator and the wind direction as

observed by a person in the ship.

Q106. The position of a particle moving along the x-axis depend on the time according to the

equation x=at2-bt3. where x is in feet and t in second. For the following, let the numerical

values of a and b be 3.0 and 1.0, respectively. (a) At what time does the particle reach its

maximum positive x position? (b) What total length of path does the particle cover in the first

4.0s? (c) What is its displacement during the first 4.0s? (d) What is the particle’s acceleration

at the end of each of the first four second?

Q107. Two trains of Length 180 m are moving on parallel tracks. If they move in the same

direction then they cross each other in 15 s, and if they move in opposite direction then they

cross in 7½ s, then calculate their velocities.

Q108. A person walks up a stalled escalator in 90 s. When standing on the same escalator,

now moving, he is carried up in 60 s. How much time would it take him to walk up the

moving escalator?

Q109. Two bodies move in a straight line towards each other at initial velocities v1 and v2

and with constant accelerations a1 and a2 directed against the corresponding velocities at the

initial instant. What must be the maximum initial separation lmax between the bodies for

which they meet during the motion?

Hints and Solutions

Single Choice Type Objective Question

Ans1: Since the acceleration vector has two components, one in opposite direction to the

velocity vector and one perpendicular to the velocity vector. The vector perpendicular to

velocity vector causes only change in direction while the vector component in the direction

opposite to the velocity vector causes the deceleration motion.

Ans2: Force component in x-direction Fx=Ma

x=Md2x/dt2=6 N,

Force component in y-direction Fy=Ma

x=Md2y/dt2=12t ,

Hence F=Fx2 + Fy

2 = 62N

tan= Fy /Fx=1 so acts in a direction 450 with the x-axis

Ans3: Acceleration a=d2x/dt2=6t+2 which is at t=0 as a=2

Velocity v=dx/dt=3 t2 +2t + which is at t=0 as v=

So a/v=2/

Ans4: The motion in fig (B) has two values of velocity at a particular instant, which is not

possible in motion in one dimension.

Ans5: The angular velocity of second hand of watch= 2/60

Velocity of tip= r

Magnitude of change in velocity vector = 2v = 2(2/60) cm/s = 2/30

Ans6: The particle moving in a circle has acceleration vector constant in magnitude with the

direction radially inwards. The motion in a parabolic path has both magnitude and direction

of acceleration vector constant.

Ans7: x=(h-1/2 g t 2)-(v t –1/2 g t 2 )=h-vt

Ans8: For a constant acceleration the second derivative of displacement with time should be

a constant.

Ans9: a= d2x/dt2 i + d2y/dt2 jd2x/dt2)2 + (d2y/dt2)2

Ans10: a= d2x/dt2 = 2a2

Ans11: In a parabolic motion the acceleration vector is constant in magnitude and direction

and directed along one of the axes and defined by equations x=k1t and y=k

2t2

Ans12:In a one-dimensional uniform motion the speed is constant in magnitude. The fig (D)

has speed =dx/dt is constant.

Ans13: The mirror image of the object forms at a far distance away from the mirror which is

equal to the distance of the object from the mirror.

Ans14: The velocities of the two objects are constant so the relative velocity of the one object

with respect to the other is also constant.

Ans15: Since the particle moves with a constant velocity than displacement time graph will

be as per fig (A) showing the displacement varying linearly with time to a maximum value of

a at a time a/v then it decreases linearly to value zero at time 2a/v and so on.

Ans16: After t = 8s, +ve area = -ve area so total change in velocity is zero therefore particle will attain

initial velocity at t=0.

Ans17: The rate of change of displacement with time that is velocity is decreasing means

there is retardation of motion and finally the velocity is zero.

Ans18: (A) sails 450 south-west;

Ans19:Sm = (u+v)t; t= 4x10-4sec

Ans20:(A)

Ans21: d=Area under curve=6 m; (D)

Ans22:(D)

Ans23:(B)

Ans24:(B)

Ans25:(D)

Ans26:(A)

Ans27:(A)

Ans28:(A)

Ans29: (D) In the first part the distance decreases linearly and in the second part when first

particle is at rest the distance decreases parabolically finally to zero when second collides.

Ans30 (A)

Ans31: dv/dt=1.5t-0.15t2

dv/dt=0 at t=10 sec

v=1.5t2/2-0.15t3/3; vmax

=25 m/s

Ans32: -65=12 t-5 t2; t=5 sec

Ans33: u2 = 2x9.8x100= 1410 m/s; a = 490 m/s2; t = 0.09 s

Ans34: (v+15/2)6=60 v=5 m/s ; a=5/3 m/s2 ;t=3s ; S=7.5 m

Ans35: v=an; Sn = n(0+v)/2=an2/2; Sn-2

= a(n-2)2/2; Sl2= 2v/n(n-1)

Ans36: v=300 m/s ,t=30 sec and T=60 sec

Ans37: t=3 sec and S3=30-1/2x10x5=5 m

Ans38: (D)

Ans39: (C)

Ans40:(A)

Ans41: (5+v)t1/2=PQ; (25+v)t

2/2=QR; 15(t

1+t

2)=3PQ; t

1/t

2=1:1

Ans42: The train will take minimum time when it travels distance of S/2 at maximum

velocity at acceleration

S/2 = 1/2at2; t = S/a; Tmin

= 2S/a

Ans43: v1=v

0+at

1/2; v

2=v

0

+a(t1+t

2/2); v

3=v

0+a(t

1+t

2+t

3/2); v

2-v

1/v

3-v

2 =( t

1+t

2)/(t

2+t

3)

Ans44: (U+u)t – 1/2gt2 = u.t+1/2at2; a = (2U/t – g)

Ans45: V = 20m/s; Vr = 3/4x20 = 15m/s; 0 = 15-10t; t = 1.5s; Time interval = 2t = 3s

Ans46: S = -4.9x2+1/2x9.8x = 9.8m; v = -4.9+9.8x2 = 14.7 m/s

Ans47: S=1/2(4.9)22+(0+9.8)/2=9.8 m

Ans48: t1=t

2=2s/g

Ans49:(B)

Ans50: (v1+v

2)3 = L; (v

1+1.5v

2)2.5 = L; v

1 = 3/2 v

2; (v

1-v

2)t = 3(v

1+v

2); t = 15 sec

Ans51: (C)

Ans52: Distance travelled in the nth second of motion Sn=u +a/2(2n-1); 30=u+7a/2; -

70=u+17 a/2 ;

u=100 m/s and a= -20 m/s2 ; and at t=10 sec S10

=0

Ans53: (C)

Ans54:(B)

Ans55: (A)

Ans56: sin30/(2u/3t) = sin/ut; sin=¾; = sin-1(3/4)

Ans57: (C)

Ans58: sin = 5/v; v = 5/sin

Ans59: v = kr2; v1:v

2 = 1:4

Ans60:(D)

Ans61: t1/t

2 = s

1/s

2= 1/2

Ans62: F = p/ t = 0.01x2x5/0.01= 10 N;

Ans63: (D)

Ans64: (C) x = kt2; v = dx/dt = 2kt; a=dv/dt = 2k

Ans65: (C)

Ans66:(B) t1 = v/g

1,; t

2 = 2v/g

2; g

1t

1 = g

2t

2

Ans67: t = 24/(5x0.6) = 8min

Ans68: v0 = 20 m/s at H=25; u2=900; u = 30 m/s

Ans69: (D);

Ans70: (B) S = 4 m

Ans71: t = 30/v’; v’ = vcos t = 2min

Ans72: (D) 60 m/min

Ans73: speed v/s time graph (C)

Ans74: vt –1/2(at2) = 125; t=5 s

Ans75: (A)

Ans76: Displacement component in x dir = 2sin 45 – 2cos 45 0=0; Displacement component

in y dir = 4 – 2cos 450 – 2sin 45= 4-2/2; Net displacement=4-2/2; Total distance

covered=8 m

Ans77: R = 2L/; Displacement = 2R = 22L/; Ratio = L/Displacement = /22

Ans78: (A) Distance always increases with time (C) Time doesn’t decrease

Ans79: Time = 4 x 8 + 5=37 sec

Ans80: (A) True: Nothing is in absolute rest because there is no object in the universe, which

is at rest and nothing is in absolute motion since there is no frame which is at absolute rest

and w.r.t that the body can have absolute motion

(B) False (C) True (C) True

Ans 81: No, since the speed of the object can’t be negative.

Ans82: (A) in a circular motion with constant speed.

Ans83: v=vx2+ v

y2=4a2t2+b2 (B)

Ans84: VA

:VB

=tan 30/tan 60=1:3

Ans85: S=Area under curve=55 m

Ans86: Vav

=2s/(s/40+s/60)=48 Km/hr

Ans87: Average velocity=Displacement/time=10/(30/3)=1m/s

Ans88: x = (t-3)2; dx/dt = 2(t-3) = 0; When t=3, x = 0

Ans89: Average speed= Distance traveled/time

Since dx/dt=8-6t and in the interval t=0 to t=1, dx/dt0

Distance traveled=5 m and Average speed=5 m/s

Ans90: v=dx/dt=k.exp(-bt)

Ans91: Instantaneous acceleration =dv/dt=d2x/dt2=2 a2

Ans92: v=202 m/s South East.

Ans93: (D)

Ans94: (a) A lives closer to the school than B since position of B on the curve is far than of

A .

(b) A starts earlier than B as evident from there initial positions x=0.

(c) B walks faster than A since slope of curve (dx/dt) is greater for B.

(d) A and B reach home at the same time.

Ans95: (D)

Ans96: 100=20t+202/8; t=2.5 sec

Ans97: (C)

Ans98: sn = u+(1/2)(a) (2n-1); a = 2

Ans99: (v1+v

2)t/2=4.50; v

1+v

2=19.6 m/s; v

22-v

12=98; v

1=7.3 m/s; s

0=v

12/2g=2.75 m

Ans100: sn =a/2(2n-1); H/g =2n–1; H=1/2gn2; n=(2 2)=3.41s; H=57 m

Ans101: sn=a/2(2n-1); 24.5=9.8/2(2n-1); n=3; s=1/2 gn= 44.1m

Ans102: ut-1/2gt2+1/2g(t-2)2=90; 2gH t-1/2gt2+1/2g(t-2)2=90; t=3.12s; s=83.69m

Ans103: 98t-1/2gt2=98(t-4)-1/2g(t-4)2; t = 12s

Multiple Choice Type Objective Questions

Ans1: (A)

Ans2: T= t1 (2+2)

Ans3: T=L/(L/60+L/90)=36 sec

Ans4: Comments: ( A) at t = 0, x = -1 m (D) The particle starts its motion at t=0 s; (B), (C)

Ans5: (A), (D)

Ans6:( B), (C), (D) Comments: (A) The acceleration at t=0 may or may not be zero.

Ans7: t1=2t/3; d =a (t

12/2)t

1+(at

12/2)t

1/2=at2/3

Ans8: (C), (D)

Ans9: am=(12-2)/10=1 m/s2

Ans10: (B)

Ans11: (A),(B),(C), (D) v = -u+gt1, v = u+gt

2 , t = 2u/g-t

l = 2s

Ans12: (A), (B), (D)

Ans13:(A), (C) a = -v; v = v0 e-t ; S

= v

0/

Ans14: ax+b = beat; x = b/a{eat-1}; accel=dv/dt = b {eat}(A),(B)

Ans15: v= -cos t-(t+1)-1+C1, S=-sint-ln(t+1)+C

1t+C

2; C

1=2 and C

2=0 and so S()=2-log

(+1)

Ans16: v2 = 108 – 9x2; a=dv/dt = -9x; At maximum distance v = 0; x = 12; (B),(C)

Ans17: (A),(B),(C)

Ans 18: (A), (C)

Ans19: x2/a2+y2/b2 = 1; x= a cos pt; dx/dt = - ap sinpt, d2x/dt2 = -ap2 cospt; y=b sin pt, dy/dt =

bp cos pt,

d2y/dt2 = bp2 sin pt; At t = /2p , v = -api^ , a= -b p2j^

Ans20: (A), (B), (C), (D)

Ans21: average speed=Distance moved/Elapsed time or average speed=2l/(l/v1+

l/v2)=2v

1v

2/(v

1+v

2)

Descriptive Type Questions

Ans1: since a = kv; v = ekt; x=1/k (ekt-1)

Ans2: No, As per the definition of average velocity <v> is given by the latter equation (rf -

ri)/(t

2-t

1)

Ans3: dv/dt means rate of change of magnitude of velocity vector irrespective of change

of direction while dv/dtmeans magnitude of rate of change of velocity vector or

acceleration vector and therefore are different quantities.

(i) When a particle describe circular motion dv/dt= 0, since magnitude of velocity vector

is constant whiledv/dt0 that is acceleration is non zero because of change in direction.

(ii) This case is not possible since change in magnitude of velocity vector imparts nonzero

acceleration, which suggests dv/dt0.

Ans4: Yes, movement in a circular path at constant speed still generates acceleration because

of change in

direction. It is possible to round a curve with constant and variable acceleration.

Ans5: VB =tan 600=3; V

A=tan30=1/3; Hence car B moves faster than A and will meet

when (3-1/3)t=1.0; t= 3/2s

Ans6: Since S=(u+v)/2t t=2S/(u+v)=60s and acceleration=(v-u)/t=2/9 m/s2

Ans7: v=5 m/s , a= -2 m/s2 , t=3s; u=v+at=11 m/s

2; S==(u+v)/2t =24 m

Ans8: (a) t1

= l/u+v; t2 = l-2v

l/u+v/(u+v) = lu/(u+v)2; t

1+t

2+t

3 ————+t

= l/u+v +

lu/(u+v)2+————-; or = l/u+v [1+u/u+v)+ ————] = (l/2u) =l/v=1 hour

(b) Distance traveled by the time =60 miles

Ans9: Since a=kt; v=kt2/2+c1; v=kt2/2; v

n=kn2/2; x=kt3/6+c

2; x=kt3/6

Ans10: a=9x; v dv/dx=9x; v dv=9x dx; v2/2=9/2x2+c1; at t=0,x=1,v=3m/s; or v2=9x2;

dx/dt=3x; x=e3t

Ans11: (i) 0t10; x = ½ at2; x10

= 210 m; v10

= 50 m/s; x20

= x10

+ v10t = 750 m

x30

= x20

+ v10t – ½ at2 = 1000 m

Ans12: v2 = u2+ 2as; Assumption made is that initial velocity of the arrow is zero; a = 10 4

m/s2

Ans13: Vx = 2-3cos60= 0.5i^; V

y = 3 sin60= 33/2j^;

Time to cross = Width of river (W)/Vertical component of velocity(Vy)=1/33 hour;

Direction of resultant velocity with vertical tan = Vx/V

y=1/33;

Distance from opposite bank (x0)= Vx t= 500/33m

Ans14: No, However if displacement and distance covered by any object under observation

are same in

magnitude that is to say motion in a straight line than the average velocity and average speed

will be the same.

Ans15: (i) Time of motion=5/2=2.5 s (ii) Distance traveled in the 2nd

second=u+1/2(2n-1)a=2

m

S2.5

=6.25 m , S2=6 m , S

3=6 m

Distance traveled in the third second=(S2.5

-S2)+(S

2.5-S

3)=0.5 m

Ans16: ta=2h/(g+r) and t

d=2h/(g-r) hence time of ascent is lesser than the time of descent.

Ans17: True

Ans18: Distance moved by the first ball in 3 sec (s1)=1/2x10x32=45m; Distance moved by the

second ball in 2 sec (s2)=1/2x10x22=20 m;

Separation of two balls s=25 m

Ans19: Instantaneous speed v=(ds/dt)=6t; At t=1 sec, v=6 m/s; At t=5 sec, v=30 m/s

Average speed=Distance traveled/time=72/4=18 m/s

Ans20: l2=x2+y2 or x dx/dt= -y dy/dt or dx/dt= K l2/x2-1.

Ans21: Length of skid marks = 19.2 ft; V2 = u2-2as; 0 = u2 – 2x32x19.2 or u = 35.05 ft/s =

23.96 mile/h < 30 mile/h

Ans22: 50 = - 5.0t + 1/2x10t2; t = -141/2; h = 63.50 m

Ans23: At t = 4.0s; vx = u

0 + a cos t=12.79 m/s; v

y = 3.61 m/s; v = 13 .28 m/s; x

t = u

xt+1/2

a t2= 41 .58m;

yt

= 7.22m; r = 42.20 m , = tan-1 7.22/4.58 = 9.85

Ans24: Displacement of train in a whole time 110 min D= 40 i^ + 14.14 i^ + 14.14 j^-50 i^ =

4.14 i^+ 14.14 j^

D = 4.142+ 14.142 = 14.73 mile; Average velocity = 0.13 mile/minute= 8.03 mile/h

At angle tan-1 (4.14/14.14) = 16.310 East of north

(B) (a) mean speed = r/t = 50.24 cm/s (b) Average velocity = 2R/t

Hence the driver was not speeding.

Ans25: t2=(t

1 + t

3); (20+v) t

1/2 = 200; v t

2=400; (v+20) t

3/2 = 600; Total time of motion=120

sec

Ans26: Time of motion = 2s; x = 1/2at2= 1/2x1x22= 2.0 m

Ans27: Distance from traffic light; X = L – [ut – ½ at2]= 40 m

Ans28: dx/dt=K/x ; K=0.04

x2=0.04 t+1.0;

so for x=2 m, t2=75 sec

Ans29: Let u be the initial velocity of the particle and A be the acceleration of motion

a=ut1+1/2At

12; a+b=u(t

1+t

2)+1/2A(t

1+t

2)2;

(A) = 2(bt1-at

2)/t

1t

2(t

1+t

2) if a=b then A=2b(t

1-t

2)/t

1t

2(t

1+t

2)

Ans30: (a) parabola (b) vertical straight line (c) parabola

Ans31: Vt1=v

1 i^+gt j^ and V

t2= -v

2 i^+gt j^

The velocities will be perpendicular to each other when Vt1.

Vt2=0

Time (t)= v1v

2/g, Distance= (v

1+v

2) .t=(v

1+v

2) v

1v

2/g

Ans32: Horizontal velocity component of flow at distance x from bank Vx=2 v

0x/c;

Drift=2

T/2

2 v0x/c dt

where T=time of reach up to the other bank= 2c/u and x=ut

=v0c/2u

Ans33: For the condition to satisfy the maximum velocity should be achieved at the middle

of it’s path and therefore Vm

2=2a L/2=La; or Vm= La

Ans34: V= t1=t

3 so t

1=V/ and t

3=V/; also (V/2)(V/)+V t

2+(V/2)(V/)= l;

t2=l/V-V/2(1/+1/)

Total time of motion= V/2(1/+1/)+l/V

Ans35: Let the distance of approach be x then x+1/2 (3/2)t2=60 t; 3t2-240 t+4x=0 for t to be

real x=1200 m

Ans36: For relative velocity of two particles to lie along AB the relative velocity component

perpendicular to AB should be zero and so 2v sin/2-(+)=v sin(/2-) or 2 cos (+)=v

cos

Also sin/r=sin(+)/2r or sin(+)=2 sin ; cot =2

Ans37: Let the person travel with velocity u at a direction East of North and wind is

blowing with velocity v at angle

East of south then as per condition u sin =v sin and cot =(2 u cos +v cos )/(2 u sin

-v sin )=2

Then =/2 and u= v ; Drift=2 2 v0x/c dt=4 v

0/u c xdx=v

0c/2u

Ans38: Let VA, VB be the velocities of blocks A and B relative to the earth frame. Hence

2(VA-VB)=VB 3 VB=2VA , after 3sec, VB=4a , VA=6a therefore acceleration of block

B(a)=0.03 m/s2 and acceleration of block a=0.045 m/s2

Ans39: For the condition to achieve the motorist will travel a distance of 800 m with the

decelearation a in the time interval of 15 sec and for that eq 800=(365/6)15-1/2 a(152) yields

a=1 m/s2 and speed of car as it passes the light=1021 m/s

Ans40: (a) Velocity of train relative to the automobile Vrel

=VA-V

B

Vrel=VA

2+VB

2+2 VA V

B cos= 1202+962+2x120x96 cos 600=187.5 km/h

(b) Change in position of the train relative to the automobile during four-second

interval=Vrel T=208 m

Ans41: Total time of motion (t ) = (AD – x)/v + l2+x2/nv;

Where x is the distance of the point of diversion from D.

For t to be minimum dt/dx = - 1/v+ 1/2nvl2+x2 . 2x = 0;

So x = ln /n2-1

Ans42: 0< t < 1; an = 1m/s2; S

1 = ½ a

n t2 = 0.5 m;

1<t<3; an = 0;

Displacement in the time interval (s2) = vt = 2m;

Total distance traveled at the last moment x3 = 0.5+2 = 2.5m;

3<t<6; an = - 1 m/s2;

Displacement in the time interval (s6)

= 1x3 – 1/2x1x32= - 1.50 m;

Total distance traveled at the last moment x6 = 2.50+1.50 = 3.5 m;

6< t <7; an = + 2.0 m/s2 ;

Displacement in the time interval=-2x1+1/2x2x1= -1 m

Total distance traveled at the last moment = 4.5 m

Ans43: Let be the direction of boat in the river frame and be the direction of boat in the

earth frame.

Horizontal component of boat velocity w,r,t earth frame Vx=u+v cos

and Vertical component of boat velocity w.r.t earth frame Vy= v sin

Total distance moved by the boat L= Vx2+Vy2. (T)

Where T=B/v sinis the width of river.

tan =v sin /(u+v cos)

L=(u+v cos)2+v2 sin2 (B/v sin)

For minimum L , dL/d=0 ; Cos =-u/v,-v/u

If (v>u) , cos u/vand so=/2

And if v<u then =tan-1(v/u2-v2)

Ans44: Let v be the velocity at impact to the first screen and v is the retardation produced by the

each screen

(v-v)t1=a ; (v-2v)t2=a ; v=2a/t1-a/t2=a(2/t1-1/t2)

Velocity at middle screen just before impact=a/t1

Ans45:(a) Velocity of A relative to B, VAB

==VA-V

B;

VAB

= VA

2+ VB

2+2 VA V

B cos(180-30) = 45 Km/hr

(b) Tangential acceleration of A relative to B,

aAB

= aA

2+ aB

2+2 aA a

B cos(30)=2.32 m/s2

Ans46: Let VA, V

B, V

C are velocities of blocks relative to earth frame and V

BP, V

CP are

velocities of blocks B, C relative to the pulley over which they hang. Then VB=V

BP-V

A,

VC=V

CP+V

A, V

A+V

C=0.32V

A+V

CP=0.3 and V

BP-2V

A= -0.2 ; V

A=0.125 m/s ; V

B=0.075 m/s

and VC=0.175 m/s

Ans47: s = ut – ½ gt2

; Height above window = v2/2g

Where v=u-gt so Height = 1/16 ft

Ans48:a = u+a/2(2x-1); b = u+a/2 (2y-1); c = u+a/2 (2z-1); a(y-z)+b(z-x)+c(x-y) =0

Ans49: Sn = u+a/2 (2n-1); Sn+1 = u+a/2 {2(n+1)-1}= u+a/2 (2n+1); Sn+2 = u +a/2 {2(n+2)-

1}= u+a/2 (2n+3);

Sn ,Sn+1, Sn+2 are in A.P. with difference of a

Ans50: a = ( v+v cos 600)dt = 3vT/2 or T=2a/3v

Ans51:

T

(u cos-v)dt=l; ;

T

v cos dt =uT ; T =lv/u2-v2

Ans52: Velocity of the train at moment when event 1 occur =wt

(1) The distance between two events in the reference frame attached to train would be equal

to L

(2) The distance between the two events in the reference frame attached to earth= L – wt T -

1/2 wT2

(3) Velocity of reference frame K so the two events occur at same point V.T +w t T+1/2 w

T2 = L

and so V =4.03 m/s

Ans53: t= t1 + t

2 + t

3 = s/20+s/12+s/v;

3s/t = (v1+v

4)/2;

20 = (v1+v

2)/2;

12 = (v3+v

4)/2;

v = (v2+v

3)/2;

Also (30+12-v)= (v1+v

4)/2=3s/t=3{1/( 1/20+1/12+1/v)}

2v2-4v-480 = 0;

On solving for v we get

v = 1+241

Ans54: t1 = 7/40h; t = ¼ h; (0+V

m/2) t

B1 + (V

m+0/2)t

B2 = S; V

m = 68.0 km/h

Ans55: Let x be the maximum distance for the just catching by the bus then

4.5 t=x+1/2(0.50) t2 or t2-18 t+4x=0

and for real value of t, x 20.25 m

At the shortest distance of meet the velocities will be equal at time t=9s and time taken by

train A to reach it’s maximum velocity=(20/3)(6)=40 sec

Time taken by train B to reach it’s maximum velocity=(40/3)(3)=40 sec

Let t be the time after 80 sec of motion since start of train A then at the point of overtaking

1200/3+20/3 t= 800/3+20/3 t t=20 sec ;Hence total time of motion =100 sec

Shortest distance=25+1/2(0.5)(92)-(4.5)(9)=4.75 m

Ans56: Maximum velocity of train A=24 Km/hr=20/3 m/s

Maximum velocity of train B=48 Km/hr=40/3 m/s

Ans57: Let u be the initial velocity of point at O and a the retardation; UA=U-aT; U

B = U-

2aT; Uc= U-4aT

(UA+ U

B)T/2=L and (U

B+ U

C)T=L; a = L/3T2; U = 3L/2T; OA = 4L/3 and CD=L/24

Ans58: (b-a) = un+1/2acn2; (c-a) = 2un+ 2acn2; d-a = 3[un+3/2acn2]; d-a = 3 (c-b); ac= (a+c-

2b)/n2

Ans59: v1+1/2at

1 = s/t

1; v

1+at

1 +1/2at

2 = s/t

2; v

1+at

1+at

2+at

3/2 = s/t

3 ; s/t

1-s/t

2+s/t

3 =

v1+1/2a(t

1+t

2+t

3) =3s/t

1+t

2+t

3; 1/t

1+1t

2 +1/t

3 = 3/t

1+t

2+t

3

Ans60: s1 = ½ ft

12 ; v = ft

1; t

2 = f/g t

1; s

2 = vt

2 – 1/2gt

22; a = 1/2ft

12+v(f/g)t

1 – 1/2f2/g.t

12; t

1 =

2ag/f(f+g); t2 = 2af/g(f+g); t = 2a(f+g)/fg

Ans61:(a) v1t=ft; v

2t =3ft; v

3t= 6ft; v

4t = 10ft; v

nt=n(n+1)ft/2

(b) s1t

= 1/2ft2; s2t

= ½ ft2+ft.t +1/2x2fxt2= 5/2ft2; s3t

= 7 ft2; s4t

= 15 ft2; snt = (0+1/2 ft2 )+

{(0+ft)t+1/2 (2ft) t 2 }+ { (ft+2ft)t+1/2 (3ft) t 2 }+……

={1+3+5+…..(n-1)terms} ft 2 + {1 2 + 2 2 + 3 2 +……n terms}= n(n+1)(2n+1)ft2/12

Ans62: s = ut+1/2pt2; s = vt+1/2qt2; t = 2(u-v)/(q-p); s = 2(u-v)(uq-pv)/(q-p)2

Ans63: a1/a

2 = m/n; t

1 = 2s/vm; t

2 = 2s/nv; t= s/v [1+1/m+1/n]; <v> = s/t = v/{1+1/m+1/n};

v/<v> = (1+1/m+1/n):1

Ans64: Let tr be the reaction time and a be the rate of deceleration then 30 = 20tr +200/a; 10

= 10tr +50/a

a = 10 m/s2

; tr = 0.58; s = 18.75 m

Ans65: AB = BC=CD; TAB

= 2s/g; TAC

= 4gs; TBC

= (4-2) s/g; TAD

= 6s/g; TCD

=(6-

4) s/g

TAB

: TBC

: TCD

:: 1: (21/2 -1) : (31/2 - 2 ½)

Ans66: Vau = u2-2gh; Vad = u2+2gh; Vad = 2Vau; u = 10/3 gh; hm

= u2/2g = 5/3 h

Ans67: h/n = 1/2gt2; t = 2h/ng; h(1-1/n) = u2h/ng – ½ g. (2h/ng); u = ngh/2; v2=u2–

2gh(1-1/n) = 2gh/n (n-2)/2; v1 = 2gh/n; v

2/v

1 = (n-2)/2

Ans68: The balls coincide in magnitude and direction after a lapse of time when ball 2 is

given a delay time of =nt1, where (n=0,1,2,3) and there after time interval of t1/2 and will

move for an interval of t1/2.

Ans69: v = dr/dt= a(1 –2t ) and w = -2 a ; r =0 at t=0, t=1/ and since v <0 after t=1/2

S=0

1/2

v dt +

v dt =a/2

Ans70: Since ft=2f(t2) t

2=t/2 sec; Vm=ft; Hence length of shaft=ft2/2+ft2/4=3/4ft2

Ans71: (a) 0<t<10, acceleration a=0 and displacement s=vt; 10<t<18 , acceleration a=0.5

m/s2 and

displacement s=20+2(t-10)+1/2(0.5)(t-10)2=2t+1/4(t-10)2

(b) 18<t<24, acceleration a=-1 m/s2 and displacement s=52+6(t-18)-1/2(1)(t-18)2=6t-56-

1/2(t-18)2 and max displacement at t=24 sec when v=0 equals to 70m or x=54 m

Ans72: Let ABCD be a square of side a, the direction of wind is

along the diagonal AC and plane moves along the path ABCD then

for path AB, vAB

=u cos 450+v cos=(u+2v2-u2)/2

Time taken to fly along AB=2 a/(u+2v2-u2), similarly

Time taken to fly along BC=2 a/(u+2v2-u2)

Time taken to fly along CD=2 a/(2v2-u2-u)

Time taken to fly along DA= 2 a/(2v2-u2-u)

Total time of fly=22a2v2-u2/(v2-u2)

Ans73: Resultant velocity of the man in flowing water when it rows in a direction

perpendicular to flow = u2 – v2

Time taken to row a distance L (t1) =L/( u2-v2) v

and t2= L/(u+v) then t

1/t

2=[(u+v)/(u-v)]1/2 u2-v2 u

Ans74: At any instant t, S=(l1-v

1t)2+(l

2-v

2t)2;

For S to be minimum dS/dt=0

At t=(v1l

1 + v

2l

2)/(v

12+v

22) and

Lmin

==(v2l

1 – v

1l

2)/(v

12+v

22)

Ans75: Velocity at the point of contact on water u= 2gh = 19.6 m/s; upward acceleration =

(/b-1) g = g.

Time the ball take to stop in water T1 = u/g

Time taken by ball to rise back to surface=2 T1 = 4 s

Ans76: Velocity of ball at the point of contact = gt1/2;

Upward acceleration =(dL/d-1)g

Total time of motion for ball to reach at the point of release =

t1+ 2 x g t

1/2(d

L/d-1)=t

1d

L/(d

L-d)

Ans77: let Ve be the velocity of the motorboat (m/s) relative to the flow and Vf be the flow

velocity (m/s)

Tt = L/V

f = T + (V

e +V

f)T -L/(V

e –V

f); V

f = L/2T

OR (Ve+Vf)T=Vf(T+t)+(Ve-Vf)t

Which implies that T=t

L=Vf(T+t)=2VfT

Vf=L/2T

Ans78: For collision to take place the unit vectors in the direction of vector joining initial

points and relative velocity vector should be equal

or (v1-v

2)/v

1-v

2 = r

2-r

1/ r

2-r

1

Ans79: Velocity vector of particle1 at any instant =v1 i^-gt j^

velocity vector of particle 2 at any instant =v2i^-gtj^

When these two vectors are perpendicular there dot product should zero .

(v1i^-gtj^).(-v

2i^-gtj^)=0

t=v1 v

2 /g

Since particles travel same vertical distance in the same time and therefore relative

displacement is only horizontal one and distance between particles at the instant=(v1+v

2)t

=(v1+v

2) v

1.v

2/g

Ans80: Velocity of heleicpter at the instant engine is switched off v = at1;

Displacement of the helicopter at that instant s = 1/2at12;

Time of last sound signal reaches to the point of take off = t2= t1+s/c;

OR t1 + 1/2at

12/c = t2;

3t12+640t

1 – 192000 = 0;

t1 = 26.67 s ; v =80 m/s

Ans81: Let v and vr be the velocities of the boat and river respectively than

L/(v+vr)=8;

L/(v-vr)=12;

Time required to travel boat in still water= L/v=9.6 Hours

Ans82: Position of third ball at the end of t=6 sec x3 = 1/2x10x32 = 45m;

Position of fourth ball at the end of t=6 sec x4 = 1/2x10x22 = 20m;

Position of fifth ball at the end of t=6 sec x5 = 1/2x10x12 = 5 m

Ans83: Velocity of the particle at the instant its acceleration changes sign v1 = a t

1;

Displacement at that instant S1 = 1/2at

12;

Now displacement at time t when the particle returns to its initial position from the instant

when acceleration changes sign or at time t=t1= -1/2at12 = at

1(t-t

1) – 1/2a(t-t

1)2;

t = t1 (22); since t = (2-2) t

1 < t

1 , not possible then t = (2+2)t

1

Ans84: resultant velocity of the plane VR = 1002+602 = 116.61km/h;

Inclination of plane from north = tan-1 (3/5) west of north;

Displacement at time t=20 sec, r(t) = VRt = 38.87 km

Ans85: dx/dt=x; 2x =t; Vav

=s/ts=s/2

Ans86: Velocity of elevator car at the instant bolt falls or after 2 sec of start V0 = 2.4 m/s

upward

Acceleration of bolt with respect to elevator floor = 9.8 + 1.20 = 11.0m/s2—downward

Time taken to travel a distance 2.7 m ;

2.7= ½ x11xt2;

t = 0.70 s

(b) In the reference frame attached to the shaft the displacement of the bolt during free fall =

- 2.4 x 0.7+ ½ x10 x 0.72= 0.77 m

(c) Distance covered during upward motion to=0.24s; s

1= 2.4 x 0.24-1/2 x 10 x 0.242=0.288 m

Distance arrival during downward and motion in next 0.46 s= 0x0.46+1/2 x 10 x 0.46 2

=1.058 m

Total distance covered in 0.70 s = 1.34 m

Ans87: Relative Acceleration of car w.r.t. truck = 0.5 m/s2

S = 0.1+1/2at2; t = 24.58 s; Sc = 1/2at2= 600 m; St = ½ x 1.5 x 600 = 450 m

Ans88: Let x be the distance between the two points x and t1, t2 be the time of motion in two

parts of journey each having measurement x/2

So t1=x/2v0

x/2 = v1t

2/2+v

2t

2/2; t

2 = x/(v

1+v

2) ; T=t1+t2= x/2v0 + x/(v

1+v

2)

Vav=x/T= 2v0(v1+v

2)/(2v0 + v

1+v

2 )

Ans89: Fig(a) (vel) (acc) Fig(b) (vel)

(acc)

OA + ve zero OA + ve -

ve

AB + ve - ve AB + ve

zero

BC zero zero BC + ve +

ve

CD - ve - ve CD + ve

zero

See here velocity is positive when dx/dt>0 and negative when converse is true. Similarly

acceleration is positive when dx/dt function is increasing or d2x/dt2>0 and negative when

converse is true.

Ans90: x = 1/2gt2 0 t t0

h-x= (2gh)(t-t0)-1/2g(t-t

0)2 t

0 t 2 t

0

x=h-(2gh)(t-t0)-1/2g(t-t

0)2 where h is the height of fall.

Ans91: Velocity of impact= 2gs1= 16 ft/s

Velocity of rebound = 2gs2 = 83 ft/s

Average acceleration=Rate of Change in velocity (v) = (16+83)/0.01= 2985.64 ft/s2

Ans92: S=33=(9-1)t + 1/2 (1-2) t2

t2 – 16t +66 = 0;

t = 16-8/2;

since t has imaginary number hence Q cannot catch B

Ans93: S=150=(72-18)x1000/3600 t + ½ (-1) t 2

t2 – 30t+300= 0;

t = 30-300/2;

Since solution is not a real number, hence A cannot catch B

Ans94: Case1: v2 = u2+2al;

a = v2-u2/2l;

vc2=u2- 2 a l/2

vc=v2+u2/2;

Velocity at center (vc ) = u+at

1;

t1= v

c-u/a ;

t2 = v-v

c/a;

since t1 = 2t

2, v

c-u = 2(v-v

c);

Substitute the value of vc =v2+u2/2;

7 u2 –8 uv +v2=0

v=7u

Ans95: T = 2t1+ t

2; where t1 is the time of acceleration and deceleration and t2 is the time of

uniform motion.

Distance traveled with acceleration s1 = 1/2at

1

2,

Distance traveled with uniform motion s2 = Vm t2= at

1t

2 ,

Distance traveled with deceleration s3 = at

1t

1-1/2at

12= 1/2at

1

2

;

½ at12+at

1t

2+1/2at

12 = <v> T;

t2 = T-2t

1= T1-4<v>/aT

Ans96: (a) Distance travelled in first 2s = Area of AOB = 1/2x2x10 = 10 m.

(b) Distance travelled in 2t4 = Area of BCD= 10 m.

(c) Displacement in 0t4 = 0

(d) acceleration at t=1/2s = slope of OA= 10 m/s2

(e) Acceleration at t =20s = -10 m/s2

Ans97: s = ½ Vm t1;

Time of acceleration t1 = 2s/Vm;

3s = Vm(t2)

Time of uniform motion t2 = 3s/Vm;

Vm/2 (t3 )= 5s;

Time of deceleration t3= 10s/Vm;

Vav (t1 + t2 + t3 )=9s

<v>/Vm = 3/5

Ans98: If h is the height of tower then

-h = ut1-1/2gt

12 , where t1 is the time of motion with velocity upward

h = ut2

+1/2gt22 where t2 is the time of motion with velocity downward

t1-t

2 = 2u/g;

t3 = 2h/g; where t3 is the time of motion with zero initial velocity

2h = g t1t

2;

t3 = 2h/g = t

1t

2

Ans99: ut –1/2gt2+1/2g(t-2)2 = 90;

ut+5{-t2+(t-2)2}=90;

u2-2gx90=0

u = 302m/s;

t =4.98 s

Ans100: Velocity attained at the end of t=1 min,

v= 0 + a t=600m/s;

Distance moved in time t= 1 min, h1 = 1/2x10x602 = 18 km

Let h2 be the distance moved further with acceleration g downward

0 = (600)2-2x10xh2;

h2 = 36x104/20 = 18 km;

Total max height attained = h1

+ h2

= 36 km; 0 = 600-10 x t ; t = 60 s

Ans101: a = t;

dv/dt = t;

v = 2/3t3/2

;

d= v dt = 4/15(t)5/2

= 4t;

t=(15)2/3

Distance d =4t = 4(15)2/3

Ans102: Distance moved by the passenger train before stopping

S = (40-20)x0.5+(40-20)2 /2=220 m <250 m

Ans103: Let the boat B moving in the direction perpendicular to current keeps its direction at

angle with the normal to the bank.

So sin = 1/2; = 300 ;

Time taken by boat to come back to original position after traveling distance of 500m

TB = 2x500/20cos= 100/3;

Time taken by boat A moving first in current direction and then returning back in the

direction opposite to current TA = 500/30+500/10= 200/3

So TA/T

B = 2/3

Ans104: Let Ve, Vwi and Vw be the velocities of ship, wind and stream water respectively as

observed on earth frame.

Now as per the observations Ve+Vwi = 4i^;

Ve+Vw = 18j^ ;

Ve+Vwi+Vw = 15.97i^+12.03j^;

Ve= - 12.0i^-20.0j^= 23.32 km/h

tan-1

(5/3) South of West

Ans105: Vw,0 = -(Vw cos60+Vs) i^+Vw sin60 j^ = -37.5i^+13 j^= 39.69 km/h , 19.100

Ans106: x = at2-bt3;

dx/dt = 2at- 3bt2 = 0; At t = 0, t = 2a/3b = 2s;

hence at t = 2s the particle reverses its direction of motion.

Distance traveled in t=2s, x1= 4 ft

Distance traveled in next 2s x2 =

(2at – 3bt2) dt = - 20;

Total distance traveled= 24 ft

Ans107: Case1: When the two trains move in opposite directions

2l/u1+u

2 = 15/2;

Case 2: When two trains move in the same directions

2l/v1-v

2 = 15;

Solving for v1and v2 , we get v1 = 36 m/s; v

2 = 12 m/s

Ans108: Velocity of the man on the escalator Vm = x/90;

Velocity of the escalator Ve = x/60;

( Vm+Ve)t = x;

(x/90+x/60) t = x;

t = 36 s

Ans109: Let L be the distance between the two particles when they can meet so

L = (v1+v

2)t –1/2 (a

1+a

2)t2;

(a1+a

2)t2 – 2(v

1+v

2)t +2L = 0;

Now for t to be real b2- 4 a c 0 or L (v1+v

2)2/ 2(a

1+a

2);

Lmax

= (v1+v

2)2/2 (a

1+a

2)

Chapter 5 Motion in a Plane

Motion in Two and Three Dimensions

In this chapter, we will now extend our thought of a motion to a more general case of motion

in two and three dimensions. So far, we have learnt the definition of displacement, velocity,

acceleration and relative motion with application for a most simplified case of motion in one

dimension. Now we are in the stage to extend it to the motion in the x-y plane and later to

the motion in three dimensions.

Location of particle during motion

Whenever we are concerned with a motion of a particle it may be one, two or three

dimensional, it is very first essential to locate the position of a particle at a particular instant

of observation. The location of a particle could be only established if we choose the correct

frame of reference and which is in itself a combination of coordinate axis system. However

in this chapter we are more concerned about motion in a plane therefore only two reference

axis are required to locate its instantaneous position. To locate the instantaneous position of

a particle, we might choose either the distances from Cartesian’s coordinate system that is

two perpendicular coordinate axes x & y axis or polar coordinate system that is distance

from a origin and angle made by the x axis. In terms of vector notation the position of

particle is defined by the position vector r whose head points to the origin O and tail points

to its position. Unlike the ordinary vector it is not a position independent vector. It is related

with the position of object whose location is under consideration and origin of reference

frame and establishes the relationship between the two. The change in position vector of the

object denotes the displacement of the object from its initial position.

If the particle is not confined to a plane but moves in three dimensions three numbers are

needed to specify its location. A simple approach is to describe its location with respect to

three perpendicular coordinate axes or an alternative approach is to use spherical

coordinates r, and . Spherical coordinates are useful when there is a spherical symmetry

that is radial distance from origin is constant and only and varies. The usual relationship

between Cartesian coordinates and spherical coordinate is shown here:

Displacement of particle during motion in a plane

Figure-4.1 shows a particle moving along a curved path in this plane. At any instant of

motion, say at time t1, it has occupied a position A as shown in the figure. As discussed

earlier its instantaneous position may be defined in terms of vector algebra by the position

vector r1. Now in a time interval moving along the curved path, it occupy another position

B as defined by the position vector r2. Now the displacement of particle from one position to

another is defined by the vector , which is defined as per vector notation as r2-r1.

Hence, it clearly indicates that displacement of particle is independent of the path the

particle traverses during its motion but depends only on initial and final positions during

motion and follows vector law in terms of its position vector. If the particle moves along

straight line path than displacement of particle equals the distance moved but if another path

is chosen than displacement will be always less than the distance moved during that time

interval.

Now the one question arises in one’s mind why the concept of vector displacement has

been generated although there is no practicable relationship with the physical movement of

the particle during its motion. The matter will be clear in the next stage while defining

concept of instantaneous velocity / average velocity/ instantaneous acceleration that how

important are these tools to analyze and describe the different types of motions observed in

the nature.

Average velocity of particle during motion in a plane

Now again consider the motion of particle along curved path in a plane as shown in fig 1. On

the same lines of the definition of average velocity defined earlier during the study of motion

along a straight-line path, it is again described as the rate of change of displacement and

written as . It is a vector quantity directed along the direction of vector and

physically denotes the rate of change of straight movement of particle from its initial position

to the next position in the said time interval. In our case the particle has moved from position

A to position B, whatever may be path than its average velocity is defined by

Where is the vector displacement r2-r1 in the time interval . However, the concept of

average velocity during the physical movement of a particle in a plane in any arbitrary path

has little physical meaning when time interval of motion is considered large enough. For a

large time interval of motion, there is a distinct difference between the actual distance

traversed and the distance considered for our purpose that is displacement. Therefore, the

average velocity does not attribute to the state of motion either at the start, middle or at end.

It is only an average value for general interpretation of motion.

Instantaneous velocity during motion in a plane

Now consider the motion of particle along curved path in a plane as shown in fig 4.2. The fig

shows the instantaneous positions of particle initiating motion at time t1, at different time

intervals , , , as P1, P2, P3, P4. As the time interval of motion under

consideration is now decreased slowly than in a limiting condition the physical displacement

is just equal to the distance traversed and the direction of vector displacement is the

direction of physical motion along tangent to the instantaneous position of particle.

The instantaneous velocity of the particle is now defined as limiting rate of displacement

when . In that case, the direction of velocity vector v is tangent to the path of the

particle.

The instantaneous velocity v(t) is defined as

As per definition, the instantaneous velocity vector denotes the instantaneous rate of

displacement along any path of movement of particle and in terms of direction, it is directed

along the tangent to the path of motion that should be. Therefore it clearly depicts the

instantaneous state of motion say at time t=t1 in terms of magnitude and direction.

The acceleration vector during motion in a plane

Now we shall define a term acceleration vector that is essentially the backbone of all

fundamental laws of Newtonian mechanics. The term has the same conceptual meaning as

has been defined earlier during the discussion of motion in one dimension. The acceleration

vector is a term defining change in velocity vector with respect to time. Since velocity vector

is defined as a vector quantity, therefore change in velocity vector is also predicted to be a

vector quantity. The acceleration vector is further divided into its two forms like as velocity

vector in terms of time interval of motion under consideration during study of motion in a

plane.

Average acceleration

Now again consider the motion of a particle in a plane as shown in fig 4.3. The particle has

got as per definition instantaneous velocity vector at points P1

and P2

as v1, v2 at different

times of motion t1, t2 respectively. Now if the time interval of motion is considered large

enough than the average acceleration vector is defined as the ratio of the change in

instantaneous velocity vector Dv and time interval Dt,

Here is a vector difference of vectors v1 and v2 and therefore itself is a vector quantity..

Instantaneous acceleration

Now as shown in fig 4.3, if the time interval of motion is considered infinitesimal or tending to

zero than rate of change of velocity vector with respect to time is defined as instantaneous

acceleration. It is expressed as

Therefore, the instantaneous acceleration vector is the derivative of velocity vector with

respect to time. It is important to note here that if the particle during motion has velocity

changing in magnitude, direction or both the particle is said to be accelerating. We shall see

in the next chapter that a force is required to produce an acceleration of particle. Force is

required to be applied whether the acceleration is produced either due to change of

magnitude of velocity or change in direction of velocity vector or both. This is the reason why

the acceleration is defined in this way.

Motion with constant acceleration

Let us consider the motion of a particle in a plane with constant acceleration. In this case

when particle is moving with a constant acceleration therefore it implies that acceleration

vector is constant in magnitude and direction as well. Hence the components of vector a in

any particular reference frame will also remain constant. The situation is similar to the two

simultaneous constant acceleration motions occurring along two perpendicular directions.

Under the influence of that, the particle will traverse the curvilinear path however one of the

components may be zero.

An example of the above situation is the motion of a ball thrown into air, which follows a

curved path under the influence of gravity acceleration g acting downward. The horizontal

acceleration is zero.

Let us consider the general motion of a particle with constant acceleration:

ax= constant and ay=constant

Now under the following set of conditions, velocity vector at any instant

The above equation shows in a compact and more elaborate form that the velocity of

particle at any instant is the vector sum of initial velocity vector and at vector component

acquired in time t and also displacement vector r at any instant:

Projectile motion

An important case of curvilinear motion with constant acceleration is projectile motion. When

a body is thrown upward under the influence of gravity, then the body follows the path

known as projectile motion (fig 4.4). If the air resistance during follow up is neglected then

the body experiences the only acceleration due to gravity directed vertically downward.

As the acceleration vector is constant then instantaneous velocity vector, acceleration vector

and position vector lies in a plane. The motion is therefore a two-dimensional.

Let us choose the motion to be in the x, y plane and initial position of the particle be at origin

of the coordinate system. When the acceleration is constant then above equations can also

be applied considering motion along two axes separately. The x and y components of the

above equations are

The velocity components along each axis will be governed by the respective acceleration

component. Let us now apply these results to the motion of a projectile. The motion of

projectile is made complicated by the prevailing air resistance when body is drifted through

the air, rotation of earth on its own axis and variation in the acceleration due to gravity. For

employing lesser complexity, we neglect these complications. As said earlier the projectile

motion has a constant acceleration directed vertically downward with magnitude g=9.81

m/s2=32.2 ft/s2. If we take y-axis vertical with positive direction upward and x-axis horizontal

with positive direction in the direction of horizontal component of the projectile velocity at the

point of start, then we have

ay= -g and ax=0

Since there is no acceleration along the x-axis therefore the horizontal component, of

velocity vector remains constant and on the other hand, the motion along y-axis can be

considered to be with constant acceleration identical to that studied earlier. If the origin of

motion is considered from the origin of coordinate axis then the instantaneous velocity

components and instantaneous displacements are governed by the following equations,

where v0x and v0y are the initial velocity components or velocity components at origin of

axis ( see fig 4.4)

If the initial velocity vector makes an angle with the x-axis, the initial velocity

components will be

The general equation for the path y(x) can be obtained from equation by eliminating the

variable t between these two equations and is given by :

Which is the equation of a parabola therefore the trajectory of the particle follows a parabolic

path.

Now we shall try to find out the range (maximum horizontal distance) and time to reach the

highest point of the trajectory for the above motion:

Range of projectile

When the body reaches at the highest point P of the trajectory as shown in fig 4.4 the

vertical component of the velocity vector becomes zero therefore time of flight up to the

moment

and the range is the horizontal distance traveled in twice of this time and is given by

In terms of initial velocity and angle

Since the maximum value of is 1 when =90° or , then

maximum range is admitted when the body is projected at angle of 45° from the horizontal

and maximum range is given by vo2/g. It is to be noted also that for achieving maximum

range the two initial velocity components should be equal in magnitude. But if a body is

projected from a certain height then in this condition the body remains in air for a longer

duration therefore for maximum range the horizontal component is slightly higher than the

vertical component that is angle of projection is smaller than 45°. Studies have shown it to

be about 42°. It is also evident from the above equation that for a given range there are two

possible angles of projection q0 and 90-q0 that provide same range.

So far, in our discussion, we have not considered the effect of air resistance and the earth’s

orbital motion. As we would expect the air resistance reduces the range for a given angle of

projection and reduces slightly the optimum angle of projection. Due to earth’s orbital motion

the projectile motion doesn’t remain in the plane formed by the initial horizontal and vertical

components but gets slightly drifted to right in northern hemisphere and to left in the

southern hemisphere. It is due to the Coriolis effect, which arises due to the surface of earth,

is accelerating radially because of the earth rotation on its own axis.

Maximum height of projectile

At maximum height of projectile

So

Projectile thrown parallel to horizontal

Consider a projectile thrown from a point O at some height h from the ground with a velocity

. Now the equation of its motion in two directions may be depicted here as

This is the equation of trajectory.

Time of flight

Horizontal range=

Now if particles are thrown at an angle with horizontal then all the particles will arrive at

ground with equal velocities whatever is the time of arrival. The particle thrown in the

downward direction will take least time while thrown upward will take maximum time.

4.6.1.4 Projectile thrown from an inclined plane

Consider a particle thrown from the base of an inclined plane with a velocity v at an angle

from the horizontal. The angle of inclination of plane from horizontal is .

For the motion perpendicular to the plane

For the motion along the plane

Range is maximum when is and projectile hits the plane at right angles.

Projectile thrown with variable acceleration

Now consider the motion of a particle in the x–y plane with instantaneous velocity

where a and b are constants and particle is situated at x=0, y=0 at t=0

So

Eliminating t from these two equations

The radius of curvature of trajectory at any instant t is given by

Exercise 1

Q1. If T were the total time of flight of a current of water and H be the maximum height

attained by it from the point of projection, then H/T will be-

(u=projection velocity, =projection angle)

(A) (1/2 )u sin (B) (1/4) u sin (C) u sin (D) 2u sin

Ans1: Hmax= u2 sin2 /2g; T = 2u sin/g; H/T = u sin/4

Q2. A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant

bullet leaves the gun, monkey drops, and the bullet.

(A) hits the monkey (B) misses to hit the monkey (C) cannot be said (D) None of these

Ans2: (A)

Q3. A projectile can have the same range r for two angles of projections. If t1 and t2 be the

times of flight in two cases, then the product of times of flight will be-

(A) t1t2 R (B) t1t2 R2 (C) t1t2 1/R (D) t1t2 1/R2

Ans3: t1t2=4 u2sincos/g2=2 u2sin2/g2 R

Q4. A body is thrown with a velocity of 9.8 m/s making an angle of 300 with the horizontal. It

will hit the ground after a time.

(A) 3 s (B) 2 s (C) 1.5 s (D) 1 s

Ans4: Time of flight = 2u sin/g = 1 s

Q5. A body A is dropped from a height h above the ground. At the same, time another body B at a distance d from the projection of A from the ground is fired at an angle a to the horizontal. If the two collide at the point of the maximum height of trajectory of B the angle of projection is given by :

(A) = tan–1[d/h] (B) = sin–1[h/d] (C) = tan–1 [h/d] (D) =cos–1 [h/d]

Ans5: u t=d; v t=h; tan= v/u=h/d

Q.6 The trajectory of the projectile is shown in fig. t and T refer to times to reach the points P

and B respectively. h and H refer to heights of points above horizontal plane. Then H/h will

be equal to

(A) (T/t)2 (B) (T/t) (C) T/t/(2-(t/T)) (D) (T/t)2/{2(T/t)-1}

Ans6: (C) h= v t- ½ g t2; H=v T-1/2 g T2 where T= v/g; H/h=( T/t)/(2-(t/T))

Q7. An airplane is flying horizontally with a velocity of 720 km/h at an altitude of 490 m.

When it is just vertically above the target a bomb is dropped from it. How far horizontally it

missed the target?

(A) 1000 m (B) 2000 m (C) 100 m (D) 200 m

Ans7: ½ gt2 = 480 or t=10s ,X = ut = 2000 m

Q8. An airplane is moving with a horizontal velocity u at a height h above the ground, if a

packet is dropped from it, the speed of the packet when it reaches the ground will be-

(A) (u2 +2gh) (B) 2gh (C) (u2-2gh) (D) 2gh

Ans8: Vv = 2gh; Vh = u; V = u2+ 2gh

Q9. From the top of a tower of height h a body of mass m is projected in the horizontal

direction with a velocity v, it falls on the ground at a distance x from the tower. If a body of

mass 2m is projected from the top of another tower of height 2h in the horizontal direction so

that it falls on the ground at a distance 2x from the tower, the horizontal velocity of the

second body is-

(A) 2v (B) 2v (C) v/2 (D) v/2

Ans9: v= 2hg + u2

Q10. A stone is thrown from a bridge at an angle of 300 down with the horizontal with a

velocity of 25 m/s. If the stone strikes the water after 2.5 sec then calculate the height of the

bridge from the water surface-

(A) 61.9 m (B) 35 m (C) 70 m (D) None

Q11. A projectile is thrown with a velocity of 20 m/s. at an angle of 600 with the horizontal.

After how much time the velocity vector will make an angle of 450 with the horizontal (Take

g=10 m/s2 )

(A) 3 sec (B) 1/3 sec (C) (3+1) sec (D) (3-1) sec.

Q12. An object of mass m is projected with velocity v at an angle 450 with the horizontal. The

angular momentum of the projectile at the maximum height h about point of projection will

be-

(A) Zero (B) mv3/42g (C) mv3/22g (D) None of these

Q13. An object is thrown at an angle a to the horizontal (00<<900) with a velocity u then

during ascent (ignoring air drag) the acceleration-

(A) With which the object moves is g at all points

(B) Tangential to the path decreases

(C) Normal to the path increases, becoming equal to g at the highest point

(D) All of the above

Q14. A bomber is moving with a velocity v (m/s) above H meter from the ground. The

bomber releases a bomb to hit a target T when the sighting angle is . Then the relation

between , H and v is-

(A) = tan-1 v 2Hg (B) = tan-1 v 2/gH (C) = tan-1 v H/2g (D) None of the above

Q15. Three projectile A, B and C are thrown from the same point in the same plane. Their

trajectories are shown in the figure. Then which of the following statement is true-

(A) The time of flight is the same for all the three

(B) The launch speed is greatest for particle C

(C) The horizontal velocity component is greatest for particle C

(D) All of the above.

Q16. A ball rolls off top of a stairway with a horizontal velocity u m/s. If the steps are h m

high and b m wide, the ball will just hit the edge of n th step if n equals to-

(A) hu2/gb2 (B) u2g/gb2 (C) 2hu2/gb2(D) 2u2/hb2

Q17. A boy throws a ball with a velocity v0 at an angle to the horizontal. At the same

instant he starts running with uniform velocity to catch the ball before it hits the ground. To

achieve this, he should run with a velocity of-

(A) v0 cos (B) v0 sin (C) v0 tan (D) v02 tan

Q18. A hosepipe lying on the ground shoots a stream of water upward at an angle of 600 to

the horizontal. The speed of the water is 20 m/sec as it leaves the hose. It will strike a wall

10 m away at a height of-

(A) 10.5 m (B) 12.32 m (C) 10 m (D) 20 m

Q19. A canon on a level plain is aimed at an angle above the horizontal and a shell is fired

with a muzzle velocity v0 towards a vertical cliff a distance R away. The height from the

bottom at which the shell strikes the sidewalls of the cliff is-

(A) R tan-1/2gR2/v02cos2 (B) R tan -1/2gR2/v0

2 (C) R sin-1/2gR2/v02sin2

(D) R tan +1/2gR2/v02

Q20. A ball is thrown at an angle of 450 with the horizontal with kinetic energy E. The kinetic

energy at the highest point during the flight is-

(A) Zero (B) E/2 (C) E (D) (B) ½ E

Q21. A projectile of mass m is fired with velocity v from the point P at an angle 450 with the

horizon. The magnitude of change in momentum when it passes through the point Q on the

same horizontal line on which P lies is-

(A) mv/2 (B) 1/2mv (C) zero (D) 2mv

Q22.A gun has a maximum range of 5 km on a horizontal plane. If a shell is fired vertically

upwards. Then the maximum height attained is-

(A) 5/2 km (B) 52 km (C) 5/2 km (D) 50/49 km.

Q23. If a particle follows the trajectory y= x-1/2x2, then the time of flight is-

(A) 1/ g (B) 2/g (C) 3/g (D) 4/g

Q24. For angle of projection of a projectile at angles (45+) and (45-), the horizontal ranges

described by the projectile are in the ratio of (if 450)-

(A) 2:1 (B) 1:2 (C) 1:1 (D) 2:3

Q25. Two particles are projected simultaneously in the same plane from the same point on

Earth’s surface. The particles are given different initial velocities and projected in different

direction. The path of one projectile as seen from the other projectile, is a-

(A) Circle (B) parabola (C) Hyperbola (D) straight line.

Q26. An artillery piece, which consistently shoots its shell with the same muzzle speed, has

a maximum range of R. To hit a target, which is R/2 from the gun and on the same level, at

what elevation angle should the gun be pointed-

(A) 300 (B) 450 (C) 600 (D) 750

Q27. For a projectile its range is equal to maximum height attained then the value of angle

of projection is given by –

(A) sin =1/2 (B) tan =4 (C) cos =3/4 (D) cos = ¼

Q28. Two bullets are fired at angles and (90-) to the horizontal with the same speed. The

ratio of their times of flight is-

(A) 1:1 (B) tan : 1 (C) 1 : tan (D) tan/2 : 1

Q29. A ball is projected at an angle of 600 from the horizontal with a velocity of 6 m/s. A

person observes the projectile from a vehicle moving horizontally with uniform velocity of 3

m/s, then the ball appears to move (neglect air resistance)-

(A) In a parabolic path (B) vertically upwards and then fall vertically downward (C) In a

hyperbolic path (D) none of these

Q30. A projectile has the maximum range of 500 m. If the projectile is now thrown up an

inclined plane of 300 with the same speed, the distance covered by it along the inclined

plane will be -

(A) 250 m (B) 500 m (C) 750 m (D) 1000 m

Q31. It was calculated that a shell when fired from a gun with a certain velocity and at angle

of elevation 5/36 radian would strike a given target. In actual practice it was found that a hill

just prevented in the trajectory. At what angle of elevation the gun is fired to hit the target-

(A) 5/36 rad (B) 11/36 rad (C) 7/36 rad (D) 13/36 rad

Q32. A large number of bullets are fired in all direction with the same speed v. What is the

maximum area on the ground on which these bullets will spread-

(A) v2/g (B) v4/g2 (C) 2v4/g2 (D) 2v2/g2

Q33. A horizontally flying aeroplane releases a bomb. The trajectory of the bomb is a –

(A) Straight line (B) parabola (C) Hyperbola (D) circle

Ans:33. (B) parabola

Q34. At the top of the trajectory of a projectile the direction of its velocity and acceleration

are-

(A) Parallel to each other (B) Inclined at an angle of 450 to the horizontal

(C) Perpendicular to each other (D) none of the above statement is correct.

Ans:34.(C) V cos 450 = 20; V = 202; Vv =20 m/s

Q35. A hunter aims the gun and fires a bullet directly towards a monkey sitting at a distance

tree. To save itself monkey can-

(A) Either sit at the position or drop the downward (B) Either sit at the position or jump

upward (C) Either jump upward or drop downward (D) Nothing can be said

Ans35: (B)

Q36. A bullet is fired in a horizontal direction from a tower while a stone is simultaneously

dropped from the same point then-

(A) The bullet and the stone will reach the ground simultaneously (B) The stone will reach

earlier C) The bullet will reach earlier (D) Nothing can be predicted.

Ans36: (A)

Q37. A man in the parachute falls from an airplane moving with uniform horizontal velocity

(v). At the moment he just touches ground -

(A) He would be just below the airplane at that moment. (B) He would be left behind the

plane. (C) He would go ahead of the plane (D) Nothing could be said with certainly.

Ans37: (A)

Q38. Two bullets are fired simultaneously, horizontally and with different speeds from the

same place. Which bullet will hit the ground first-

(A) The faster one (B) The slower one (C) Both will reach simultaneously (D) Depends on

the masses.

Ans38: (C)

Q39. If it is possible to project a particle with a given velocity in two possible way so as two

make it pass through a point at a distance r from the point of projection. The product of the

time taken to reach these points in the two possible ways is then proportional to-

(A) r3 (B) r2 (C) r (D) 1/r.

An39: T = 2u sin/g; T1 = 2u sin/g; T2 = 2u cos/g; T1T2 =2r/g; T1T2 r

Q40 A body is thrown a velocity v0 at an angle to the horizon. Determine v0 and , if the

maximum height to which the body reaches is 3 metres and the radius of curvature of the

upper point of trajectory is 3 meters.

(A) v0= 6.4 m/sec, =450 (B) v0= 9.4 m/sec, = 540 44’(C) v0=12 m/sec, =300

(D) v0= 15 m/sec, =150

Ans40: (v0 cos )2/R=g & v02sin2/2 g=H

=54044’, v0=9.4 m/s

Q41. At any instant a projectile is moving with a velocity u in a direction making an angle

with horizon. After what time the direction of motion turns through an angle ?

(A) u cos/gsin(-) (B) u sin /gcos(-) (C) u/g sin(-) (D) u/g cos (-)

Ans41: u cos/cos(-) sin(-)= usin - gt; t= u sin/g cos(-)

Q42. A particle is thrown over a triangle from one and of a horizontal base and grazing the

vertex fall on the other end of the base. If and be the base angles and the angle of

projection then correct relation between (q), (a) and (b) is-

(A) tan =tan+tan (B) tan=tan+tan (C) tan= tan-tan (D) tan= tan+tan

Ans42: R=h (cot+cot)= u2 sin 2/g; u2 sin 2/g = h(cot+cot); h=u2sin2/2g; tan = tan+tan

Exercise 2There are some statements at the end of each paragraph. Ascertain true and false

statements for each one and give reasons for the assertion:

Q1. In a projectile motion –

(A) Change in linear momentum between the initial and final point is mg T, downwards.

(True/False)

(B) Angular momentum with the respect to the point of projection remains constant.

(True/False)

Q2. A cart moves with a constant speed along a horizontal circular path. From the cart a

particle is thrown up vertically with respect to the cart-

(A) The particle will land somewhere on the circular path. (True/False)

(B) The particle will follow a parabolic path. (True/False)

Q3. (A) Two particles thrown with same speed from the same point at the same instant but

at different angles, can not collide in mid air

(B) A body projected in a uniform gravitational field follows a parabolic path. (True/False)

(C) In projectile motion, velocity is never perpendicular to the acceleration. (True/False)

(D) A particle dropped from rest and blown over by a horizontal wind constant velocity traces

a parabolic path. (True/False)

Q4. A particle is projected from the end B of a horizontal track BC at a given angle a to the

horizontal after just grazing the vertices A, A’, A,”……. of the triangles BAC, BA’C,

BA’’C………..on its way. It lands up at the point C on the other end of the horizontal line; the

sum of the tangents of the base angles of any of these triangles is a constant. (True/False).

Q5. A shell bursts on contact with the ground and the piece from it fly off in all directions with

all speeds up to 30 m/s. A man standing 30 m away is in danger for nearly 6 seconds.

(True/False)

Q6. Two projectiles are thrown with different velocities and at different angles so as to cover

the same maximum height. Then the sum of the times taken by each to reach the highest

point is equal to the total time taken by either of the projectiles. (True/false)

Q7. A batsman hits a pitched ball at a height 4.0 ft above ground so that its angle of

projection is 450 and its horizontal range is 350 ft. The ball fells down the left field line where

a 24 ft high fence is located 320 ft from home place. The ball clears the fence. (True/False)

Q8. The maximum horizontal range is four times the maximum height attained by the

projectile, when fired at an inclination so as to have maximum horizontal range.

(True/False).

Q9. A particle is projected vertically upwards in a vacuum with a speed u-

(A) When it rises to half its maximum height, its speed becomes u/2

(B) The time taken to rise to three –fourths of its maximum height is half the time taken to

reach its maximum height.

Q10. A pilot is supposed to fly due east from A to B and then back again to A due west. The

velocity of the plane in air is v and the velocity of the air with respect to the ground is u. The

distance between A and B is l and the plane’s air speed v’ is constant.

(A) If u=0 (still air), the time for a round trip is then TE= t0/1+u2/(v’)2 (True/False)

(B) If the air velocity is due north (or south). The time for a round trip is then

tN = t0/1-u2/(v’)2 (True/False)

Q11. Two particles are projected from the same point on the ground, with same speed

simultaneously at angles and with the horizontal. They strike the ground at same point

after times t1 and t2 respectively then-

(A) += 900 (True/False)

(B) t1/t2= tan (True/False)

Q12. In case of projectile motion if two projectiles A and B are projected with same speed at

angles 150 and 750 respectively to the horizontal then-

(A) HA>HB (True/False)

(B) TA>TB (True/False)

Q13. Three stones are projected simultaneously in the same vertical plane with velocities u,

v, w in directions making angles , and with the horizontal. The area of the triangle

formed by the stones at any instant is proportional to the square of the time elapsed from

the instant of projection. (True/False)

Q14. A particle is projected from a point on the inclined plane and at the rth impact strikes the

plane perpendicularly and at the nth impact it is at the point of projection. If e is the coefficient

of restitution then en-2 er+1=0 (True/False)

Q15. The motion of one projectile as seen from another projectile will always be a straight-

line motion. (True/False)

Q16. There are two values of time for which a projectile is at the same height and the sum of

these two times is equal to the time of flight. (True/False)

Exercise 3

Q1. If a baseball player can throw a ball at maximum distance d over a ground, then find out

the maximum vertical height to which he can throw the ball. (Balls have same initial speed in

each case)

Q2. The height y and the distance x along the horizontal plane of a projectile on a certain

planet (with no surrounding atmosphere) are given by y=(8t-5t2) meter and x=6t meter where

t is time in seconds. Find out the velocity of projection.

Q3. A particle is projected from O at an elevation and after t second it has an elevation

as seen from the point of projection. Find its initial velocity of projection.

Q4. Two stone are projected with the same speed but making different angles with the

horizontal the ranges are equal. If the angle of projection of one is /3 and its maximum

height is y1 then find out the maximum height of ascent of the other stone.

Q5. A ball is projected upwards from the top of a tower with a velocity 50 m/s, making an

angle 300 with the horizontal. The height of the tower is 70 m. Find out after how many

seconds from the instant of throwing the ball reach the ground.

Q6. Two tall buildings face each other and are at a distance of 180 m from each other. With

what velocity must a ball be thrown horizontally from a window 55 m above the ground in

one building, so that enters a window 10.9 m above the ground in the second building.

Q7. What is the average velocity of a projectile between the instants when it crosses half the

maximum height it is projected with a speed u at an angle with the horizontal?

Q8. A canon ball has a range R on a horizontal plane. If h and h’ are the greatest heights in

the two paths for which this is possible, then find out the expression between R, h and h’.

Q9. The trajectory of a projectile in a vertical plane is y=ax-bx2, where, a and b are

constants, and x, and y are respectively the horizontal and vertical distance of the projectile

from the point of projection. Find out the maximum height attained and the angle of

projection from the horizontal.

Q10. An airplane is flying at a height of 1960 m in horizontal direction with a velocity of 360

km/hr, when it is vertically above the point A on the ground drops a bomb. The bomb strikes

a point B on the ground. Then find out the time taken by the bomb to reach the ground.

Q11. A particle moves in the plane x-y with a velocity , where and are

the unit vectors along x and y axis, and k1 and k2 are constants. At the initial moment of time

the particle was located at the point x=y=0 then find out the equation of the particle’s

trajectory y(x).

Q12. Three particles A, B and C are thrown with speeds VA, VB, and VC, with A horizontally,

B at an angle of 300 with the horizontal and C, vertically in such a manner that they collide

simultaneously at H, the highest point of the parabolic path of B. If the acceleration due to

gravity is g, then find out the ratio of the speeds VA : VB : VC.

Q13. A body is thrown horizontally with velocity [2gh] from the top of a tower of height h. It

strikes the level ground through the foot of the tower at a distance x from the tower. Find out

the value of x.

Q14. A stunt performer is to run and dive off a tall platform and land in a net in the back of a

truck below. Originally the truck is directly under the platform. It starts forward with a

constant acceleration a at the same instant the performer leaves the platform. If the platform

is H above the net in the truck, then find out the horizontal velocity u that the performer must

have as he leaves the platform.

Q15. From points A and B at the respective heights of 2 m and 6 m, two bodies are thrown

simultaneously towards each other, one is thrown horizontally with a velocity of 8 m/s. and

the other, downward at an angle of 450 to the horizontal at an initial velocity such that the

bodies collide in flight. The horizontal distance between points A and B equals 8 m.

Calculate the initial velocity v0 of the body thrown at an angle 450.

Q16. Two inclined planes OA and OB having inclinations (with horizontal) 300 and 600

respectively, intersect each other at O as shown in fig. A particle is projected from point P

with velocity u=103 m/s along a direction perpendicular to plane OA. If the particle strikes

plane OB perpendicularly at Q, calculate,

(A) Velocity with which particle strikes the plane OB

(B) Time of flight.

(C) Distance PQ

(D) Vertical height h of P from O

(E) Maximum height from O, attained by the particle and

Q17. A ball A is projected from origin with an initial velocity v0= 700 cm/sec, in a direction

370 above the horizontal. Another ball B 300 cm from origin on a line 370 above horizontal is

released from rest at the instant A starts. Then find out how far will B have fallen when A hits

it.

Q18. A golfer standing on level ground hits a ball with a velocity of u= 52 m/s. at an angle

above the horizontal. If tan = 5/12, then find out the time for which the ball is at least 15 m

above the ground (i.e. between A and B). (Take g = 10 m/s2)

Q19. A particle is projected with a speed V from a point O making an angle of 300 with the

vertical. At the same instant, a second particle is thrown vertically upwards from a point A

with velocity v. The two particles reach H, the highest point of the parabolic path of particle

simultaneously. Then find out the ratio V/v.

Q20. If R is the range of a projectile on a horizontal plane and h is its maximum height, then

find out the maximum horizontal range with the same velocity of projection.

Q21. A body projected from the top of a tower horizontally with an initial velocity 20 m/s hits

the ground at an angle of 450. Then find out the vertical component of velocity at the time of

hitting.

Q22. A particle is projected from a point O with a velocity u in a direction making an angle a

upward with the horizontal. After some time at point P it is moving at right angle to its initial

direction of projection. Find out the time of flight from O to P.

Q23. An aircraft drives towards a stationary target, which is at sea level, and when it is at a

height of 1390 m above sea level it launches a missile towards the target. The initial velocity

of the missile is 410 m/s. in a direction making an angle below the horizontal where tan=

9/40. Then find out the time of flight of the missile from the instant it was launched until it

reaches sea level.

Q24. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill

is = 300, and the angle of the barrel to the horizontal =600. The initial velocity v of the shell

is 21 m/sec. Then find out the distance of point from the gun where shell will fall.

Q25. Two guns, situated on the top of a hill of height 10 m, fire on shot each with the same

speed 53 m/s at some interval of time. One gun fires horizontally and other fires upwards at

an angle of 600 with the horizontal. The shots collide in air at a point P. Find (A) the time

interval between the firings, and

(B) the co-ordinate of the point P.

Take origin of the coordinate system at the foot of the hill right below the muzzle and

trajectories in x-y plane.

Q26. A boy throws a ball horizontally with a speed of v0=12 m/s from a bridge. In an effort to

hit the top surface AB of a truck traveling directly underneath the boy on the bridge. If the

truck maintains a constant speed u=15 m/s, and the ball is projected at the instant B on the

top of the truck appears at point C, determine the positions when the ball strikes the top of

the truck.

Q27. (a) A particle is projected with a velocity of 29.4 m/s at an angle of 600 to the

horizontal. Find the range on a plane inclined at 300 to the horizontal when projected from a

point of the plane up the plane.

(b) Determine the velocity with which a stone must be projected horizontally from a point P,

so that it may hit the inclined plane perpendicularly. The inclination of the plane with the

horizontal is and P is h meter above the foot of the incline as shown in the figure.

Q28. A cricket ball thrown from a height of 1.8 m at an angle of 300 with the horizontal at a

speed of 18 m/s is caught by another field’s man at a height of 0.6 m from the ground. How

far were the two men apart?

Q29. A projectile is projected from the base of a hill whose slope is that of right circular

cone, whose axis is vertical. The projectile grazes the vertex and strikes the hill again at a

point of the base. If be the semi-vertical angle of the cone, H its height, u the initial velocity

of projection and the angle of projection, then find out

(i) Relationship between and

(ii) Velocity of projection in terms of h and .

Q30. Two particles move in a uniform gravitational field with an acceleration g. At the initial

moment the particles were located at one point and moved with velocities 3 m/s and 4 m/s

horizontally in opposite direction. Find the distance between the particles at the moment

when the velocity vector become mutually perpendicular.

Q31. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a

horizontal road. The man throws a ball in such a way that it returns to the truck after the

truck has moved the distance of 58.8 m.

Find the speed and the angle of projection

(a) as seen from the truck.

(b) as seen from the road.

Q32. A body falling freely from a given height ‘H’ hits an inclined plane in its path at a height

‘h’ As a result of this impact the direction of the velocity of the body becomes horizontal. For

what value of (h/H) the body will take maximum time to reach the ground?

Q33. A spherical bowl with innumerable holes is placed in a lawn at a distance ‘d’ from the wall of a

building. If R is the maximum range of the jet that is produced when the bowl is connected to the nose of

a fire engine, that the portion of the wall that is hit by the jet water is bounded by a parabola. Find the

height and breadth of the bounded parabola.

Q34. The slopes of the windscreen of two motorcar are 1=300 and 2=150 respectively. At

what ratio v1/v2 of the velocities of the cars will their drivers see the hailstones bounced by

the windscreen of their cars in the vertical direction? Assume that hailstones fall vertically.

Q35. Figure shows an 11.7 ft wide ditch with the approach road at an angle of 150 with the

horizontal, with what minimum speed should a motorbike be moving on the road so that it

safely crosses the ditch? Assume that the length of the bike is 5 ft; and it leaves the road

when the motorbike runs out of the approach road.

Q36.(a) If the maximum height of projectile above a horizontal through the point of projection

were h and a be the angle of projection, find the interval between the instants at which the

height of the projectile is h sin2a.

(b) Two particle are projected simultaneously from the same point at angles of projection and

respectively. If they simultaneously strike the top and the bottom of a vertical pole subtending an angle

at the point of projection, then find out the relationship between , and .

Q37. Consider a projectile at the top of its trajectory.

(a) What is its speed in terms of v0 and 0?

(b) What is its acceleration?

(c) How is the direction of its acceleration related to that of its velocity?

(d) Over a short distance a circular are is a good approximation to a parabola. What then is

the radius of the circular arc approximating the projectile’s motion near the top of its path?

Q38. A ball A is projected from O with an initial velocity 8 m/s in a direction 370 above the

horizontal. A ball B, 4 m away from O on the line of initial velocity of A, is released from rest

at the instant A is projected.

Find (i) the time when B is hit by A

(ii) the height through which B falls, before it is hit by A

(iii) the direction and magnitude of velocity of A at the time of impact.

[Take g=10 m/s2, sin 370= 0.6 & cos 370 = 0.8]

Q39. Two identical shells are fired from a point on the ground will same muzzle velocity at

angles of elevation =450 and =tan-13 towards top of the cliff, 20 m away from point of

firing. If both the shells reach the top simultaneously, calculate 

(a) Muzzle velocity (b) Height of the cliff and (c) Time interval between two firings.

Q40. A target is fixed on the top of a pole 13 m high. A person standing at a distance of 50

m from the pole is capable of projecting a stone with a velocity 10g m/s. If his aim is to

strike the target in the quickest time possible, at what angle of elevation should he throw the

stone?

Q41. A particle is moving along a vertical circle of radius R=20 m with a constant speed

v=31.4 ms-1 as shown in figure. Straight line ABC is horizontal and passes through the

center of the circle. A shell is fired from point A at the instant when particle is at C. If

distance AB is 203 m and shell collides with the particle at B, calculate

(a) Smallest possible value of the angle of projection

(b) Corresponding velocity u of projection.

(Take =3.14 and g=10 ms-2)

Q42. A projectile aimed at a mark which is in a horizontal plane through the point of

projection fall a meter short of it when the elevation is and goes b meter too far when the

elevation is . If the velocity of projection be the same in all cases, then find out the proper

elevation for projection.

Q43. A stone is projected from a point on the ground in such a direction so as to hit a bird on

the top of the telegraph post of height h and then attain the maximum height 2h above the

ground. If, at the instant of projection, the bird were to fly away horizontally with uniform

speed, find the ratio between the horizontal velocities of the bird and stone, if the stone still

hits the bird while descending.

Q44. Two persons simultaneously aim their guns at a bird sitting on a tree. The first person

fires his shot with a speed of 100m/s at an angle of projection of 300. The second person is

ahead of the first by a distance of 50 m and fire his shot with a speed of 80 m/s. How must

he aim his gun so that both the shots hit the bird simultaneously. Calculate

(A) the distance of the foot of the tree from two persons and the height of the tree.

(B) With what velocities and when do the two shots hit the bird.

Q45. A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4

m and falls at a distance of 14 m from the wall. Find the magnitude and direction of the ball.

Q46. Two particles are projected from a point simultaneously with velocities whose

horizontal and vertical components are u1, v1 and u2, v2 respectively. Find out the interval

between their passings through the other common point of their path.

Q47. Two particles are simultaneously thrown from roofs of two high building, as shown in

fig. Their velocities of projection are 2 ms-1 and 14 ms-1 respectively. Horizontal and vertical

separations between points A and B are 22 m and 9 m respectively. Calculate the minimum

separation between the particles in the process of their motion.

Q48. A gun of muzzle speed v0 is situated at height h above a horizontal plane. Find out the

angle at which it must be fired so as to achieve the greatest range on the plane.

Q49. A projectile is fired from a point on a cliff to hit a mark 10 m horizontally from the point

and 10 m vertically below it. The velocity of projection is equal to that due to falling freely

under gravity through 5 m from rest. Find out the two possible directions with angles of

projections and time of flights respectively. (g=10 m/s2)

Q50. From points A and B, at the respective height of 2 m and 6 m, two bodies are thrown

simultaneously towards each other, one is thrown horizontally with a velocity of 8 m/s and

the other, downward at an angle of 450 to the horizontal at an initial velocity such that the

bodies collide in flight. The horizontal distance between points A and B equals 8 m.

Calculate

(A) the initial velocity v of the body thrown at an angle 450.

(B) the coordinates (x, y) of the point of collision.

(C) the time of the flight of the bodies before colliding.

(D) the velocity of two bodies at the instant of collision.

Assume that the trajectory lies in a single plane.

Q51. Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B,

knowing that the conveyor belt moves at the constant speed v0. Determine the smallest

value of v0 for which sand can be deposited on the stoke pile at B and the corresponding

value of a.

Q52. A particle is projected under gravity v in a direction making an angle with respect to

an inclined plane and having gradient , the point of projection being on the plane. If the

motion is in the vertical plane and the particle strikes the plane at an angle to the

horizontal. Find out

(a) Value of

(b) Velocity with which the particle strikes the plane

Q53. If the horizontal range of a projectile were a and the maximum height attained by it was

b, then find out the velocity of projection in terms of a and b.

Q54. Particle P and Q of mass 20 gm and 40 gm respectively are simultaneously projected

from points A and B on the ground. The initial velocities of P and Q make 450 and 1350

angles respectively with the horizontal AB, as shown in the figure. Each particle has an

initial speed of 49 m/s. The separation AB is 245 m. Both particles travel in the same vertical

plane and undergo a collision. After collision P retraces its path. Determine (A) the position

of Q when it hits the ground.

(B) How much time after the collision does the particle Q takes to reach the ground.

Take (g = 9.8 m/s2)

Q55. A particle is projected from point O on the ground with velocity u= 55 ms-1 at angle

= tan-1(0.5). It strikes at a point C on a fixed smooth plane AB having inclination of 370 with

horizontal. If the particle does not rebound, calculate

(a) Coordinates of point C in reference to coordinate system shown in figure.

(b) Maximum height from the ground to which the particle rises.

Take (g=10 m/s2)

Q56. The range of a rifle bullet is 1000 m, where is the angle of projection. If the bullet is

fired with the same angle from a car traveling at 36 km/h towards the target, then find out

the increase in the range of the projectile.

Q57. A cart is moving along + x direction with a velocity of 4 m/s. A person on the cart

throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the

cart, the stone is thrown in y-z plane making an angle of 300 with vertical z-axis. At the

highest point of its trajectory, the stone hits an object of equal mass hung vertically from the

branch of tree by means of a strings of length L. A completely inelastic collision occurs, in

which the stone gets embedded in the object. Determine:

(A) The speed of the combined mass immediately after the collision with respect to an

observer on the ground.

(B) The length l of the string such that the tension in the string becomes zero when the

string becomes horizontal during the subsequent motion of the combined mass.

Q58. A cyclist moves along a straight line with a velocity of 9 km/h. In what direction should

he throw a stone with a velocity of 16 m/s and parallel to the ground so that the resultant

motion of the stone may be at right angles to the direction of motion of the cyclist.

Q59. A boat is moving directly away from a gun on the shore with speed v1. The gun fires a

shell with speed v2 at an angle of elevation and hits the boat. Find out the distance of the

boat from the gun at the moment it is fired.

Q60. A gun is fired from a moving platform and range of the shot are observed to be R1 and

R2 when the platform is moving forwards and backwards respectively with velocity v. Find

the elevation of the gun in terms of the given quantities.

Q61. Two towers AB and CD are situated a distance ‘d’ apart as shown in figure. AB is 20 m

high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB

horizontally with the velocity of 10 ms-1 towards CD. Simultaneously another object of mass

2m is thrown from the top of CD at an angle of 600 to the horizontal towards AB with same

magnitude of the first object. The two object move in the same vertical plane, collide in mid

air and stick to each other.

(a) Calculate the distance between the tower and

(b) Find the position where the objects hit the ground.

Q62. An object A is kept fixed at the point x=3 m and y= 1.25 m on a plank P raised above

the ground. At time t=0 the plank starts moving along the +x direction with an acceleration

1.5 m/s2. At the same instant a stone is projected from the origin with a velocity u as shown.

A stationary person on the ground observes the stone hitting the object during its downward

motion at an angle of 450 to the horizontal. All the motions are in x-y plane. Find u and the

time after which the stone hits the object.

Take g=10 m/s2.

Q63. A small ball thrown at an initial velocity v0 to an angle a to the horizontal strikes a

vertical wall moving towards it with a horizontal velocity v and is bounce to the point from

which it was thrown. Determine the time t from the beginning of motion to the moment of

impact, neglecting friction losses.

Q64. Rain appears to fall vertically downward to a man moving with a velocity of 10 m/s.

When he doubles his speed then the rain appears to strike him at an angle of 600 from

horizontal, then what is actual velocity of the rain?

Q65. At what angle to the horizon should a stone be thrown from the steep bank of a river

so that it may fall into the water as far as possible from the bank? The height of the bank is h

and the velocity of projection is v.

Q66. A particle is projected from the ground to graze the four upper vertices of a regular

hexagon whose side is 2a and which is placed vertically with one side on the ground. Find

(a) the time of flight,

(b) range on the ground.

Q67. A particle is projected with a velocity u and it strikes at right angles plane through the

point of projection inclined at an angle to the horizontal. Find the height of the point strike

above the horizontal plane through the point of projection and the time of flight up to that

instant.

Q68. A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5

m away from the center of the boat. He wishes to throw an apple into the boat. If he can

throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of

projection for successful shot. Assume that the point of projection and the edge of the boat

are in the same horizontal level.

Q69. The man stands 18 m from the wall and throws a ball at a speed of v0=15 m/s.

Determine the angle at which he should release the ball so that it strikes the wall at the

highest point possible. What is the height? The room has a ceiling height of 5 m.

Q70. A point moves in the plane xy according to the law

x = at,

y = at(1-t) , where a and are positive constants , and t is time.

Find

(a) The equation of the point‘s trajectory y (x); plot this function;

(b) The velocity v and the acceleration w of the point as functions of time; the moment t0 at

which the velocity vectors forms an angle /4 with the acceleration vector.

Q71. Consider a balloon that rises from the surface of the earth. The ascension rate is

constant and is given by v0. Now the blowing action of wind causes the balloon to gather

horizontal velocity component vx= ay where y is height of ascent. Find out the horizontal drift

of the balloon as well as total, tangential and normal acceleration of the balloon.

Q72. A particle moves in the plane xy with constant acceleration w directed along the

negative y-axis. The equation of motion of the particle has the form y = ax – bx2, where a

and b are positive constant. Find the velocity of the particle at the origin of coordinates.

Q73. A ball starts falling with zero initial velocity on a smooth in inclined plane forming an

angle with the horizontal. After having fallen the distance h, the ball rebounds elastically

off the inclined plane. At what distance from the impact point will the ball rebound for the

second time?

Q 74. A cannon and a target are 5.10 km apart and located at the same level. How soon will

the shell launched with the initial velocity 240 m/s reach the target in the absence of air

drag?

Q75. A particle A moves in one direction along a given trajectory with a tangential

acceleration wt = a t where a is a constant vector coinciding in direction with the x axis as

shown in fig, and t is a unit vector coinciding in direction with the velocity vector at a given

point. Find how the velocity of the particle depends on x ‘provided that its velocity is

negligible at the point x = 0

Q76. A point moves in the plane so that its tangential acceleration wt = a, and its normal

acceleration wn= bt4, where a and b are positive constant, and t is time.

At the moment t = 0, the point was at rest. Find how the curvature radius R of the points

trajectory and the total acceleration w depend on the distance covered s.

Q77.A particle moves along the plane trajectory y (x) with velocity v whose modulus is

constant. Find the acceleration of the particle at the point x= 0 and the curvature radius of

the trajectory at the point x = 0 and the curvature radius of the trajectory has the from

(a) of a parabola y = ax2;

(b) of an ellipse (x/a)2 + (y/b)2 = 1; a and b are constants here.

Q78. Projectiles are hurled at a horizontal distance R from the edge of a cliff of height h in

such a way as to land a horizontal distance x from the bottom of the cliff. If you want x to be

as small as possible, how would you adjust 0 and v0, assuming that v0 can be varied from

zero to some maximum finite value and that 0 can be varied continuously? Only one

collision with the ground is allowed (see figure).

Q79. A projectile is fired with a speed of 40 m/s at an angle of 600 with the horizontal

towards the standing roof of a building. The roof is standing at an angle of 450 to the

horizontal as shown in figure.

(a) At what point the projectile hits the roof.

(b) What is the velocity of projectile when it hits the roof?

(c) What should be the angle of projection of the particle, so as to reach the roof in minimum

time and what is the value of this minimum time?

Q80. A canon fires from under a shelter inclined at an angle to the horizontal (fig). The

canon is at point A at a distance l from the base of the shelter (point B). The initial velocity of

the shell is v0, and its trajectory lies in the plane of the figure. Determine the maximum range

Lmax of the shell.

Exercise 1:

Ans10: H = (u sin ) t +1/2 g t2; H= 61.9 m

Ans11: Vv = 103 & Vh= 10; At time t, 10 = 10 3 – 10t; t = (3-1)s (D)

Ans12: Angular momentum = m (v cos 450)= M v3/42g (B)

Ans13: (A)

Ans14: R = H tan H = ½ gt2; t = 2H/g; R = u 2H/g = H tan or = tan-1 2u2/gh

Ans 15: (D)

Ans16: R = nb; H = nh= ½ gt2; ut = nb; t = nb/u; nh = ½ gxn2 b2/u2 or n = 2u2 h/gb2

Ans17: v0 cos

Ans18: Time of motion =10 / 20 cos60= 1s; Y = 12.32 m

Ans19: T = R/ V0 cos; Y = V0 sin(r/ V0 cos ) – 1/2g (R2/ V0 2 cos2)= R tan - ½ gR2/ V0 2

cos2 (A)

Ans20: At the highest point v = u cos 450= u/2; KE = 1/2mv2= 1/4mu2 = 1/2E

Ans21: Change in momentum p = 2mv sin450 = 2 mv

Ans22: Rm = 5 = u2/g; u = 5x103g; Hm = u2/2g = 2.5 km

Ans23: y = x-1/2x2; y = x tan - gx2/2u2 cos2 tan=1 or =450; g/u2 cos2 = 1; u = 2g; tf = 2u

sin/g = 2/g (B)

Ans24: R = u2 sin2 /g; R1/R2 =1:1

Ans25: (D)

Ans26: R= u2/g; u = Rg; R/2 = (Rg)2 sin2/g = Rsin2 sin = ½; = 150 & 750

Ans27: Since R=Hmax ; u2 sin 2/g=u2 sin2/2g; tan = 4

Ans28:Tf = 2u sin/g; Tf1 = 2u sin /g; Tf2 = 2u cos/g; Tf1/Tf2 = tan

Ans29: (B)

Ans30: Rm = u2 /g; u = 502 m/s; y = u sint-1/2gt2= 50 t – 5t2; x = 50t; tan30 = 50t-5t2/50t; t=

4.2s; y = 121.8; x = 210m; s = 242m

Ans31: = /2-5/36=13/36 rad

Ans32: Rm = v2/g ; A = v4/g2

Exercise 2:

Ans1: p = 2mv sin V sin = gT/2; p = mgT ; L = mv.r = mu.y – mv.x; L continuously changes

Ans2: False, True

Ans3: (A) True, (B) True, (C) False (D) True

Ans4: Let and be the base angles of the triangle and be the angle of projection of the

projectile.

y=x tan -g x2/2u2cos2 h= h cot tan –g h2cot 2/2 u2 cos21= cot tan –g h cot 2/2 u2

cos2

also h(cot + cot )=u2 sin 2/g; tan =tan + tan

Ans5: R = u2 sin/g; 30 = (30)2 sin2/g; sin2= 1/3; = 9.730, 80.260; tf = 2u sin/g; tf = 5.91s

say 6.0s

Ans6: u12 sin21 = u2

2 sin22; u1 sin 1 = u2 sin2; t1h+t2h = u1 sin1/g + u2sin2/g= 2u1

sin1/g = 2u2 sin2/g

= time of flight or either one.

Ans7: Consider the vertical motion of projectile -4.0 = 4 sin45t-1/2gt2 ;u cos45t = 350; t = 4.70s; u =

105.31 ft/d; t2 = 320/u cos45 = 4.29s; y2 = 29’; Ball clears the fence by 5’

Ans8: Rmax = u2/g; Hmax = u2/4g; Hmax = Rmax/4

Ans9: H = u2/2g; ½ mv2 = ½ mu2/2; v = u/2; v2 = u2 – 2g. (3/4 u2/2g)= ¼ u2; v = u/2; t’ = u/2g

=T/2

Ans10: (a) t0 = 2l/v; T = l/v-u+l/v+u= 2lv/v2-u2 = 2l/v{1-u2/v2}= t0/1-u2/v2

(b) sin = u/v; t1 = l/v cos=l/v1-u2/v2; t = 2t1=t0/1-u2/v2

Ans11: T1 = 2u sin /g , T2 = 2u sin/g; T1/T2 = sin/sin =tan=cot

(A) True (B) True

Ans12: (A) False, (B) False

Ans13: (ab)h = (u cos-vcos)t=k1 t; (ab)v = (u sin-v sin)t=k

2 t; (bc)h=( w cos-vcos)t=k

3 t

(bc)v=( w sin-v sin)t=k4 t; (ca)h==( w cos-u cos)t=k5 t; (ca)v==( w sin-u sin)t=k6 t

a=(ab)h 2+ (ab)v

2; b=(bc)h 2+ (ca)v

2; a=(ab)h 2+ (ca)v

2; s=(a+b+c)/2; Area =1/2s(s-a)(s-b)(s-

c)=K t2

Ans14: Tr=2 v/g cos(1+e+e2+e3+….+e(r-1))= 2 v (er-1) /g cos

Tn=2 v (en-1) /g cos; u= g sin Tr & 2u = g sin Tn; en-2 er+1=0

Ans15: Let u1, 1, be the velocity and angle of projection of first particle and u2, 2 be the values

for other one then position of particles at any instant t; x1 = u1 cos 1 t ; x2 = u2 cos 2 t ; y1 = u1 sin

1 t – ½ gt2

y2 = u2 sin2t – ½ gt2; Relative position of particle 2 w.r.t one x = (u2 cos2 – u1 cos1)t; y =

(u2 sin2-u1 sin1)t ; y/x = (u2 sin2-u1 sin 1/u2 cos2-u1 cos 1) = k

Ans16: h= u sin. t-1/2gt2; gt2-2u sin t + 2h = 0 ; t=2u sin4u2 sin2-5gh/2g; t=u sin/g u2

sin2 – 2gh/g; t1+t2 = 2u sin/g = time of flight

Exercise 3:

Ans1: R = u2 sin2/g ; d = u2/g or u = gd; Hmax= u2/2g = d/2

Ans2: dy/dt = 8-10t ; At t = 0 ; Vy = 8; Vx= dx/dt = 6 so V=10 m/s

Ans3: tan = u sin-gt/ u cos ; u = gt cos/sin(-)

Ans4: y1 = u2 sin2/2g= 3u2/8g; u = 8yg/3; y2 = u2 sin2 (/6)/2g = u2/8g= y1/3

Ans5: 70= - 25t+1/2x10t2 =5t2-25t; t = 7, t= -2

Ans6: t = 180/u; H = ½ gt2 ; (55-10.9) = 1/2x10x(180/u)2; u = 60.60 m/s

Ans7: Average velocity = distance between points / time= u cos

Ans8: R = u2 sin2 /g; h= u2 sin2/2g , h’ = u2 cos2/2g; hh’ = u4/16 g2 (4 sin2 cos2) =

u4/16g2, sin22

hh’ = 1/16 R2 or R = 4hh’

Ans9: y = ax – bx2 = x tan - gx2/2u2 cos2; tan = a, g/2u2 cos2 = b; u2 = g/2b (1+a2);

Hm= u2sin2/2g=a2/4b

Ans10: H = 1/2gt2; t = 2h/g = 20 s;

Ans11: v = k1 i^+k2xj^; dx/dt = k1 or x = k1t; dy/dt = k2x=k1 k2t=>y = 1/2k1k2t2; y = k2/2k1 x2

Ans12: VA = VB cos VA = 3/2 VB; VB sin 30 = Vc; VB = 2Vc; VA: VB: VC = 3:2:1;

Ans13: Time of flight = 2h/g; X = 2ghx2h/g=2h

Ans14: t = 2H/g; X = aH/g = u2H/g; u =aH/2g

Ans15: (8+v0/2)t = 8 ; v0 sin 45t = 4; v0/2t=4; t=1/2s; v0 = 82 = 11.31 m/s

Ans16: tf = u/gcos30 = 2u/3g = 2s; Vv = gsin30t= 10m/s; Distance PO=1/2 g sin 30 tf2 =10 m

Hence h= PO sin 30=5m; Maximum height attended = h +(usin60) 2/2g= 16.25 m;

Applying principal of energy conservation from P to Q

½ m u2 =1/2 m Vv

2 + mgh; h =10m; PQ = 17.322+102= 20 m

Ans17: BA = 300 sin370-7002 sin2 37/2x980; BA= 90 cm

Ans18: Time of said motion =Time of flight-2(t15)=2 sec

Ans19: Vcos 300=v; V/v=2/3

Ans20: R = u2 sin2/g = 2u2 sin cos/g; H = u2 sin2/2g; R2 = 4u4/g2 sin2 (1-sin2) = 48h/g

(u2-2gh)

Rm = u2/g= R2/8h+2h

Ans21: 20 m/s

Ans22: u=g sin.t; t = u/g sin= u/g cosec

Ans23:1390 = 90t + 5t2; t= 10.0 s

Ans24: T = 2v sin30/g sin60 = 2v/3g; S = v cos30 (2v/3g )-1/2g cos60 (4v2/3g2)=30 m

Ans25: 53t = 53cos600(t+t); t = t; -5t sin60(t+t)+1/2g(t+t)2= 1/2gt2; t = 1s; coordinate of P,

x = 53, y = 5m

Ans26: Time of flight=2h/g=1.27 s; Position of ball=(u -v0)T=3.81 m

Ans27: (a) 0 = 29.4 sin30t – 1/2g sin600 t2; t = 5.88 s; R = 29.4 cos30t – 1/2g cos600 t2= 58.8 m

(b)t = u/g cot; h cos = u sint +1/2 g cos.t2; u = 2gh/(2+cot2)

Ans28: In a vertical motion 1.20 = 18 sin30t – 1/2gt2; 5t2 – 9t – 1.20 = 0; t = 1.92 s;

x = 18 cos30x 1.92 = 29.92 m

Ans29. (i) R/h= 4 cot ; R/2h= tan tan = 2 cot

(ii) R = u2sin2/g = 2h tan 2u2 tan/(1+tan2) = 2gh tan u2=gh/2 (4+tan2)

Ans30: v1 = - 3 i^+gt j^; v2 = 4i^ + gtj^; v1 . v2 = - 12 +g2t2 = 0; t = 12/g; x = 2.42 m

Ans31: Time of flight (T)= 58.8/14.7= 4s; 2 u sin /g= 4 u sin =2g; u cos=0 =/2 and

u=19.6 m/s

and relative to earth v=14.72+19.62 =24.5 m/s at angle of 530 from horizontal.

Ans32: Time of flight (T)= 2(H-h)/g + 2h/g; dT/dh= 2/g1/2h-1/2H-h=0; H=2h or h/H=1:2

Ans 33: Y = x tan - gx2/2u2 cos2 Y = d tan - gd2/2u2 sec2 dy/d = d sec2 - gd2/2u2 (2

sec2 tan)

tan = u2/gd=R/d; Ym = (R2-d2/2R); B = 2R2-D2

Ans34: Let v0 be the velocity of fall of hailstones falling vertically with respect to the Earth

frame.Then 2v1 cot21=v0 and 2v2 cot22=v0; v1/v2=tan 21/tan 22 =3

Ans35: y = x tan - gx2/2u2cos2 (dy/dx) = tan - gx/u2 cos2 At x =(11.7+5 cos 150)= 16.52’,

dy/dx=- tan

2 tan = gx/u2 cos2 or u = (gx)/sin u = 32.49 ft/s

Ans36: (a) h = u2 sin2/2g; u = 2gh/sin2 = 2gh/sin Vf2 = Vi2 – 2gh sin2 Vf = 2gh cos.

(b) u22 = gx/2 sin cos ; o = x tan - gx2/2u2

2cos2 x tan = x tan - gx2/2u12 cos2 gx/2u1

2

cos2 = tan - tan u22 sin cos/u1

2 cos2 = tan tan since u1 cos = u2 cos tan = tan-

tan

Ans37: V = V0 cos0; a = gav; V2/R = g; R = V2/g = u02 cos20/g

Ans38: t = 4/u, t =1/2s; BF = 1/2gt2 = 1.25m; VvA = u sin - gt =0.1m/s; VH

A = 6.3 m/s; VA = 6.38

m/s, tan =0.01

Ans39: H = 20 tan - gx102/2u2cos2 = 20 tan - gx102/2u2cos2 u = 25(sec2-sec2)/(tan-

tan)

u = 20 m/s; H=10m & t=1.72 s

Ans40: y = x tan - gx2/2u2 cos2 12.5 tan2 - 50tan+25.5 =0; tan= 3.4, 0.6; = 73.60, 30.960

Angle of projection for quickest time of throw is with least sin that is= 30.960

Ans41: Tf = 2u sin/g=2s; R=203 = u2sin2/g ; tan=1/3 =>=300; u = 20 m/s

Ans42: g(R-a) = u2 sin2 ; g(R+b) = u2 sin2 R =u2sin2/g= u2/g(asin2+bsin2)/(a+b);

= ½ sin-1(a sin2+bsin2)/(a+b)

Ans43: h =(u sin.)t – 1/2gt2; t= (2+2)h/g; t2 = (2-2) h/g; u cost2- t= v t2;

u cos/v= t2 /(t2- t)=(2/2+1)

Ans44: 100 cos30.t = 80 cos t + 50; 100 sin30.t – ½ gt2 = 80 sin t –1/2 gt2; sin = 5/8; t = 2.07 s.

Distance of foot of tree from first person x1 = 80 cost + 50 = 179.27

Distance of foot of tree from second person x2= 80 cost = 129.27 m

Height of tree = 100 sin30 t – ½ gt2 = 82.50 m

Ans45: u cos t = 4; u sin t – ½ gt2 = 4; u2 sin2/g = 18 ; u sin /g t = 9/4; 9/4 gt2 – ½ gt2 = 4; t =

0.47 s

u = 13.71 m/s; = 51.630

Ans46: u1t1 = u2 t2 ; v1 t1 – ½ gt12 = v2t2 – ½ gt2

2 ; v1/t1 – 1/2g = u2 (t2/t1)1/t1 – ½ g(t2/t1)2;

Since t2/t1 = u1/u2 so t1 = 2u2(u2v1-u1v2)/g(u22-u1

2)’; t2 = 2u1 (u2u1-u1v2)/g(u22-u1

2);

t = t1-t2 = 2(u2v1-u1v2)/g(u1+u2)

Ans47. At any instant Horizontal Separation = 22-(16/2) t; Vertical Separation =(9-12/2t);

Z=(22-16/2t)2 +(9-12/2t)2; dz/dt =0; x = 6 m

Ans48: gt2-2v0sint-2h = 0; t = 2v0 sin4v02sin2+2gh/2g; R =

v02sincos/g+v0cos/gv0

2sin2+2gh

dR/d = 0; cos2 = gh/v02+gh; = ½ cos-1 {gh/v0

2+gh}

Ans49: u=2gh=10m/s; -10 = 10sin t-5 t2; t = 1/cos; tan =(12)/2; 1 = 67.50, 2 = -22.50

Ans50: 8t+vcos45t = 8; v sin45t = 4; vt = 42; t=1/2s; v = 11.31 m/s;

Ans51: V0 cos t = 8; V0 sint= 5t2-6; V02t2 = 25t4-60t2+100; Z = V0

2=25t2+100/t2-60; dz/dt =

50t+100(-2/t3) = 0; t = 1.41s; V0 = 6.30m/s; cos = 0.89; = 26.940

Ans52: t = 2v sin/g cos Vh = v cos- gsin.(2vsin/gcos);

Vv = -vsin+ g cos t; tan2 = Vv/Vh

(1+3tan2) tan = 2tan

Ans53: 2u2 sin cos/g =a ; u2 sin2 /2g = b; tan = 4b/a; sin=4b/a2+16b2; u = 2g(b+a2/16b)

Ans54: At the point of collision UAX = u cos 450 = u/2; UAY = u/2 – gt=0 ; UBX = -u/2; UBY =

u/2-gt=0

mu/2-2mu/2 = -mu/2+2mu’x; u’x = 0; Hence body Q falls freely under gravity

Ans55: 10/3 sin 37=55 cos t+1/2 g cos 370 t2 where =79.560; t = 0.50s; MC=55 sin t-1/2 g

sin 370 t2=4.75m

AC=4.75-10/3 cos 370= 2.08 m; X=5.0m & Y=1.25 m; Vp=7.8 m/s; ½ m Vp2=m g h; hm =

h+Y=4.45 m

Ans56:103 =u2sin2/10; u=104/sin2 t = 2u sin/g; Rn = (104/sin cos+10)

(200/sin2)sin/10-103

= 1000/7 tan

Ans57: At highest point Vh = Vn2+vz2 = 5 m/s; Mv+0= 2m.v’; V’ = 2.5 m/s; ½ 2m(v’)2 = 2mgl;

l=0.32 m

Ans58: 16cos = 2.5; cos = 0.15; = 810

Ans59: X = (v2 cos-v1)t1= 2v2sin/g(v2cos-v1)

Ans60: R1 = (u+v)tf ; R2 = (u-v)tf; R1/R2 =(u+v)/(u-v); u=(R1+R2)v/ (R1-R2); tan = g(R1-

R2)2/4v2(R1+R2)

Ans61. 15t = d; (10sin60t+1/2gt2)-(1/2gt2)= 10 m; t = 2/3 ;d = 17.32 m; At the instant of collision

VnA = 10 , VvA = 1/2gt2 = 20/3; VnB = 5, VvB = 10/3 (3+2); From principal of conservation of

momentum

m(10) – 2m(5) = 3m(Vn); Vn = 0

Hence particle after collision falls vertically at distance=10t = 20/3m from AB

Ans62: u cost = 3+0.75t2 ; u sint = 1.25+5t2; u cos = -usin+gt; t = 1s; u = 7.28 m/s

Ans63: X = v0cost; Y = v0sint-1/2gt2; Y = -(v0sin-gt)t1+1/2gt12; X = (v+v0cos)t1;

T=v0sin(v0 cos+2 v)/g (v0 cos +v)

Ans64: Vsin = 10; tan600 = Vcos/20-Vsin V cos = 103; V = 20 m/s

Ans65: h=-R tan +1/2g/v2 R2(1+tan2); Differentiating both sides with respect to , R=v2/g tan ;

tan=v/v2+2 gh

Ans66: Let the origin be at the point of projection and b is the abscissa of the first vertex of the

Hexagon above plane 3 a= b tan - g b2/2 u2 cos2 23 a= (b+a) tan - g (b+a)2/2 u2 cos2

23 a= (b+3a) tan - g (b+3a)2/2 u2 cos2 3 a= (b+4a) tan - g (b+4a)2/2 u2 cos2

On solving (u cos)2=(3 a g/2) And b= (7-2)a; So Range R=4a +2b=27 a

Ans67: t= 2un/g cos up= 2 un tan so un=u /1+4 tan2 h=2 u2 sin2 /g(1+3 sin2);

Ans68. Possible range of shot=5.50 – 0.50 = 5 m; and 5.50+0.50 = 6 m; R = u2 sin2/g; = 150, 750

(Sin2)m = 3/5; m = 18.430 ,71.570; Hence can have value between 150 and 18.430 or 71.570

and 750

Ans69: Let hmax=v0 sin2 /2g 5sin2/3

Also h=18 tan -36/5 (1+ tan2); For dh/d=0 tan=5/42/3 hence take =sin-1(2/3) ; h=3.14

Ans 70: Equation of the motion in the plane x= at; y = at(1-t); y = x (1-x/a); v = ai + (a-

2at)j

w = (-2a)j

angle the velocity vector makes with the y axis tan = a/(2at - a) = 1; t = 1/

Ans71: dy/dt = V0; dx/dt = ay ; tan =dy/dx= v0/ay; d2y/dt2 = 0; d2x/dt2 = av0;total = av0; t =

total cos = total

/1+ tan2; t = a2 v0 y/v02+a2.y2

Ans72. w = constant = -k j where k is positive constant; dy/dx = a – 2bx; and d2y / dx2 = - 2b;

vy =dy/dt= vx(a-2b x); -w= dvy/dt=-2 bvx2; vx = w/2b; v = w/2b i + (a – 2bx)j

Ans73: Time to reach peak tp = v cos / g cos = v/g

Total time of motion up to second collision = 2v/g

Distance traveled along plane (AB) = 4v2/g sin=8 h sin

Ans74 : let the be the angle projection for minimum time of flight

so v0 cos tm

= L; tm = 2v0 sin /g; v0 cos (2v sin)/g = L

v02 sin2/g = L; sin2 gl/v0

2 = 0.86; = 30.140 or 59.850; tm = 2 x 240 sin 30.140/10 = 0.40

min and 0.69 min

Ans75: dv/dt = a= a.v/v; vdv/dx = a ; v = 2a x

Ans76: Tangential acceleration wt = a; Normal acceleration wn = bt4; s = at2/2; Hence wn =b(2s/a)2

total acceleration = a2 +b2 (2s/a)4 = a2 +16b2s4/a4

Ans77: (a) equation of motion y = ax2; Since v = constant implies tangential acceleration is zero at

every point and acceleration is directed normal to path; Also R= {1+(dy/dx)2 }3/2 / d2y/dx2 = a/2;

atotal= 2v 2a

(b) ellipse (x/a)2+(y/b)2 = 1; radius of curvature = a2v2/bv2 = a2/b

a=an= -[bav 2/a3]= - bv2/a

Ans78: Time of fall of height h(t)= v2sin2 +2 g h /g –v sin/g

x=v cos v2sin2 +2 g h /g –v sin/g =(R2+2h/g v2 cos2) –R/2

For x to be minimum v cos should be minimum or v sin should be maximum which is for v to be

vmax and be the corresponding max value given by ½ sin –1(R g/vmax2)

Ans79: 8sin t-t2 = 4-8cost

{cos t+sin dt/d}-2tdt/d = -{-sin t +cos dt/d}; = /4

Ans80: tan=V0 sin - g t/ V0 cos t= V0 /g(1- tan) since for Lmax,=450 and path is tangent to

shelter

Also l +V0/2 t tan = V0/2 -1/2 g t2 ; Sin2 = V02/ (V0

2+ 2gl) or ½ sin-1 V02/ (V0

2+

2gl)

Then Lmax = V02/g ; If ½ sin-1 V0

2/ (V02+ 2gl) , Lmax=V0

2/gsin 2(+ sin-1 gl sin 2/

V0