Question Bank Force - The Assam Valley School Today/WORKSH… · · 2013-07-24Class-X Question 1...

Class-X Question Bank 1

Question Bank

Force

1. What do you understand by the term momentum?

Or1

Define the term momentum.

Ans. The force possessed by a body, due to the combined effect of mass and

velocity of a body is called momentum.

2. State two factors which determine momentum of a body.

Ans. (i) Momentum is directly proportional to the mass of body.

(ii) Momentum is directly proportional to the velocity of body.

3. Is momentum a scalar or a vector quantity?

Ans. Momentum is a vector quantity.

4. State unit of momentum in the SI system.

Ans. Unit of momentum in the SI system is newton-second (Ns).

5. Define the term force.

Ans. Force is an external agent which changes or tends to change the state of rest

or uniform motion of a sbody in a straight line.


6. State the effects of a force applied on, (a) non-rigid body, (b) rigid body.

Ans. (a) The force acting on a non-rigid body, changes interspacing between its

constituent particles and, therefore, can change its dimensions and can

also produce motion.

(b) The force acting on a rigid body, does not change interspacing

between its constituent particles and, therefore, does not change its

dimensions. However, it can cause motion in a stationary body or can

speed up or stop a moving body.

7. State the effects a force can produce. Give one example of each effect.

Ans. (i) Force can set a body in motion. For example, a football is set in motion

on kicking.

(ii) Force can speed up a moving body. For example, a hand cart moves faster

when someone pushes it from behind.

(iii) Force can slow down or stop a moving body. For example, when brakes

are applied to a running bicycle, it slows down and then stops.

(iv) A force can change the direction of a moving body. For example, a

football player can change the direction of moving football by angling

his foot.

(v) A force can bring about change in dimensions. For example, on

compressing, the length of a coiled spring decreases and vice-versa.


8. Define Newton’s second law of motion.

Ans. The rate of change of momentum is directly proportional to the impressed

force and takes the direction in which force is applied.

9. How is the force related to the momentum of a body?

Ans. The force is directly proportional to the rate of change of momentum of

a body.

10. Prove mathematically, F = ma.

Or

Prove that, force = Rate of change of momentum

Ans. Consider a body of mass ‘m’ initially moving with a velocity ‘u’. Let a force

‘F’ act on the body for time ‘t’, such that it picks up a final velocity ‘v’, with

an acceleration ‘a’.

Change in momentum in time t = m(v – u)

Rate of change of momentum = m( – )ut

ν

But, ( – )u at

ν =

Rate of change of momentum = ma.

Also, by Newton’s second law, rate of change of momentum is ∝ applied

force.


∴ F ∝ ma

or F = k.ma

[where k is the constant of proportionality]

If there be a body of unit mass and unit acceleration, such that force possessed

by it is also unit, then :

1 = k × 1 × 1 or k = 1

∴ F = ma.

11. Under what condition, the Newton’s second law takes the form of F = ma?

Ans. F = ma is possible only, if we assume that a unit mass on moving with unit

acceleration possesses a unit force.

12. Define absolute unit of force in, (i) CGS system, (ii) SI system.

Or

Name and define the SI and CGS units of force.

Ans. (i) CGS unit of force is dyne.

When a body of mass 1 g is moving with a uniform acceleration of

1 cm s–2, the force possessed by it is called one dyne.

(ii) SI unit of force is newton (N).

When a body of mass 1 kg is moving with a uniform acceleration of

1 ms–2, the force possessed by it is called 1 newton.


13. Define gravitational unit of force in, (i) CGS system, (ii) SI system.

Ans. (i) In CGS system the gravitational unit of force is gramme-force (gf) : It is

defined as the force which produces an acceleration equal to the

acceleration due to gravity in a mass of 1 g.

1 gf = 1 g × 980 cms–2 = 980 dynes

(ii) In SI system, the gravitational unit of force is kilogram-force (kgf) : It is

defined is the force which produces an acceleration equal to the

acceleration due to gravity in a mass of 1 kg

1 kgf = 1 kg × 9.8 ms–2 = 9.8 newton.

14. Define a relation between newton and dyne.

Ans. 1N = (1 kg) × (1 ms–2) = (1000 g)(100 Cms–2) = 100000 g Cms–2 = 105 dynes. 15. State the relationship between force, mass and acceleration. Draw graphs

showing the relationship between :

(1) acceleration and force for a constant mass

(2) acceleration and mass for a constant force.

Force = mass × acceleration.

Ans. (1) (2)


16. State the condition, when the change in the momentum of a body depends

only on the change in velocity.

Ans. The mass of body should be constant.

17. Complete the following statements :

(1) The mass of a body can be treated as constant, till the velocity of the body

is ...........................

(2) Mass × change in velocity = ................ × time interval.

Ans. (a) Far less than the velocity of light.

(b) Force.

18. The correct option in following formulae is :

(a) F = m2a (b) F = am

(c) F = ma (d) F = ma

Ans. (c), F = ma is the correct option.

19. A spaceship is moving at a constant speed in space by burning its fuel and

ejecting out the spent gases from its exhaust. Answer the following questions :

(1) Is there any change in the momentum of the rocket?

(2) Is there any force acting on the rocket?

Ans. (a) Yes, there is change in the momentum of rocket.

(b) Yes, there is a force acting on the rocket.


20. Choose the correct statement :

One newton is the force which produces :

(a) an acceleration of 1 ms–2 on a body of mass 1 g.

(b) an acceleration of 1 cms–2 on a body of mass 1 kg.

(c) an acceleration of 1 cms–2 on a body of mass 1 g.

(d) an acceleration of 1 ms–2 on a body of mass 1 kg.

Ans. (d) is the correct statement.

21. What do you understand by the term “force due to gravity”?

Ans. The force with which the earth attracts the body on or around its surface is

called force due to gravity or gravitational force.

22. Write the gravitational unit of force. How is it related to newton?

Ans. Gravitational unit of force is kilogram force (kgf).

1 kgf = 9.81 N = 10 N (approx.)

23. Explain, what do you understand by the statement :

1 kilogram force (kgf) = 9.8 newton.

Ans. From the above statement, it implies that when a mass of 1 kg is acted upon

by the acceleration due to gravity (9.8 ms–2), then the force acting on it is

1 kgf or 9.8 N.


24. The correct relationship between kgf and N is :

(a) 1 kgf = 1 N (b) 1 kgf = 9.8 N

(c) 9.8 kgf = 1 N (d) 1 kgf = 980 N

Ans. (b) 1 kgf = 9.8 N is the correct option.

25. Complete the following statements

(a) Force = mass × ..................

(b) 1 N = ..................... dynes

(c) 1 N = ..................... kgf (approx.)

(d) newton is the unit of .................. .

Ans. (a)acceleration (b) 105

(c) 0.1 (d) force

26. What do you understand by, (a) mass, (b) weight.

Ans. (a) The quantity of matter contained in a body, is called its mass.

(b) The force with which the earth attracts a body is called its weight on earth.

27. State SI units of, (a) mass, (b) weight

Ans. (a) SI unit of mass is kilogram (kg)

(b) SI unit of weight is newton (N).

28. How is the weight of body related to its mass?

Ans. Weight of body = Mass of body × Acceleration due to gravity.


29. The value of ‘g’ at the centre of earth is zero. What will be the weight of body

of mass m there?

Ans. Weight of body at the centre of earth

= mass × acceleration due to gravity.

= m × 0 = zero.

30. How much is the weight of a body of mass m kg?

Ans. Weight of body = m kg × g ms–2

= mg N

31. A man weighs himself in Moscow and then in Delhi. He finds that his weight

is less in Delhi. Give a reason.

Ans. Moscow being nearer to North pole has more acceleration due to gravity as

compared to Delhi. Now, weight of body is the product of mass and

acceleration due to gravity. Thus, the weight of man in Delhi is less compared

to that in Moscow.

32. A body of weight W on the surface of earth is taken to (a) a deep mine, (b) a

higher altitude. In each case state whether the body will weigh equal to, less

than or more than W? Give a reason.

Ans. The body will weigh less than W, both in (a) and (b).

It is because, the magnitude of acceleration due to gravity decreases in each

case.


33. Imagine a lift which is coming downward with an acceleration due to gravity

such that acceleration within the lift is zero. What will be the mass of a body

and its weight if the initial mass is x. Give a reason for your answer.

Ans. The body will have mass x as mass does not change with time or place.

The weight of body will be zero, because, the acceleration within the lift is

zero.

34. State whether the following statements are true or false.

(a) Mass or weight of a body can be measured by a physical balance

(b) 1 kgf = 9.8 N (approx)

Ans. (a) It is a false statement.

(b) It is a true statement.

35. Distinguish between mass and weight.

Ans.

Mass Weight 1. It is the measure of quantity of matter in a body. 2. It is constant quantity and does not change with time or place. 3. It is a scaler quantity. 4. It is measured by physical balance. 5. Its SI unit is kilogram.

1. It is the force with which a body is pulled towards the earth. 2. It is a variable quantity and changes from place to place, with the change in acceleration due to gravity. 3. It is a vector quantity. 4. It is measured by a spring balance. 5. Its SI unit is newton.


Numerical Problems.

1. A body having a velocity of 200 ms–1 has a momentum of 5 Ns. Find the mass

of the body.

Ans. m = ?

v = 200 ms–1

Momentum = 5 Ns.

Momentum = m × v.

5 Ns = m × 200 ms–1

m = 5

200 = 0.025 kg

2. Calculate the momentum of an electron of mass 9 × 10–31 kg, moving with

a velocity of 6 × 107 ms–1.

Ans. Momentum = ?

m = 9 × 10–31 kg

v = 6 × 107 ms–1

Momentum = m × v

= 9 × 10–31 kg × 6 × 107 ms–1

= 54 × 10–24 Ns.


3. A body of mass 1 kg is thrown vertically upward with an initial speed of

5 ms–1.What is the magnitude and direction of force due to gravity acting on

the body when it is at the highest point?

Ans. Magnitude of force = m × g

= 1 kg × 9.8 ms–2

= 9.8 N

The force will act in vertically downward direction.

4. Two balls of mass ratio 1 : 2 are dropped from the same height. Find :

(a) the ratio between their velocities, when they strike the ground.

(b) the ratio of forces acting on them during the motion.

Ans. (a) As the height is the same for both the balls so is the initial velocity (zero),

v2 = u2 + 2gh

v2 = (0)2 + 2gh

v = 2gh

In the above expression v = 2gh , velocity is independent of the mass,

therefore, both balls will have the same velocity.

(b) Force is given by the expression F = mg. As the mass of balls is in ratio

of 1 : 2, therefore, force acting on them must be in the ratio of 1 : 2.


5. What force will produce an acceleration of 3.5 ms–2, in a body of mass

12.5 kg?

Ans. F = ?

a = 3.5 ms–2

m = 12.5 kg.

F = ma

= 12.5 kg × 3.5 ms–2

= 43.75 N.

6. Calculate the acceleration produced in a body of mass 50 g, when acted

upon by a force of 20 N.

Ans. Mass of body (m) = 50 g

= 0.050 kg.

Force (F) = 20 N

Acceleration (a) = Fm

20N

0.050 kg

= 400 ms–2.


7. Calculate the mass of a body, when a force of 525 N, produces an

acceleration of 3.5 ms–2.

Ans. m = ?

F = 525 N

a = 3.5 ms–2.

m = F ÷ a

= 525 ÷ 3.5

= 150 kg.

8. A force of 50 kgf acts on a body of mass 1/2 tonne, find the acceleration

produced in the body. [Taking g = 10 Nkg–1]

Ans. F = 50 kgf

= 50 × 10

= 500 N;

m = 12

× 1000 = 500 kg;

a = ?

a = Fm

= 500500

= 1 ms–2.


9. What is the mass of a body, which is moving with an acceleration of 1.4 ms–2,

when a force of 50 kgf acts on it? [g = 9.8 ms–2]

Ans. F = 50 kgf

= 50 × 9.8 = 490 N;

m = ?;

a = 1.4 ms–2.

m = Fa

= 4901.4

= 350 kg.

10. A car of mass 600 kg is moving with a speed of 10 ms–1, while a scooter of

mass 80 kg is moving with a speed of 50 ms–1. Which vehicle will require

more force to stop in the same time interval and why?

Ans. Momentum of car (p) = mv= 600 kg × 10 ms–1

= 6000 Ns

Momentum of scooter (p) = mv= 80 kg × 50 ms–1

= 4000 Ns

The car will require a greater force to stop as it has more momentum.


11. How much acceleration will be produced in a body of mass 10 kg, acted

upon by a force of 2 kgf? [g = 9.8 ms–2]

Ans. Force (F) = 2 kgf = 2 × 9.8 = 19.6 N

Mass (m) = 10 kg

Acceleration (a) = Fm

= 19.6 N10 kg

= 1.96 ms–2.

12. Two bodies have masses in the ratio 3 : 4. When a force is applied on smaller

mass, it moves with an acceleration of 6 ms–2. How much acceleration will

the same force produce in the bigger body?

Ans. Let the masses of the bodies be 3 x and 4 x.

Case (i) : Force produced in the body of mass 3 x

(F) = 3 x × 6 ms–2

Case (ii) : Force acting on the body of mass 4 x

(F) = 4 x × a ms–2.

From the question, 3x × 6 = 4x × a

∴ a = –23 6 ms

4x

x×

= 4.5 ms–2


13. A car initially at rest, picks up a velocity of 72 kmh–1 in 20 s. If the mass of

the car is 1000 kg. calculate, (i) Force developed by its engine

(ii) Distance covered by the car.

Ans. u = 0;

v = 72 kmh–1 = 20 ms–1;

t = 20 s;

m = 1000 kg;

F = ?;

S = ?

Applying v = u + at

20 = 0 + a × 20 ∴ a = 1 ms–2

(i) Force developed, F = ma = 1000 kg × 1 ms–2

= 1000 N.

(ii) Applying v2 – u2 = 2aS

(20)2 – (0)2 = 2 × 1 × S.

S = 4002

= 200 m.

14. A golfer hits a golf ball at rest, such that the contact between the ball and

golf stick is for 0.1 s. If the golf ball covers a linear distance of 400 m in 2 s,

calculate the magnitude of force applied. Mass of golf ball is 50 g.


Ans. u = 0;

t = 0.1 s;

S = 400 m;

time for linear distance = 2 s;

m = 50 g = 0.05 kg;

F = ?

Distance covered by ball in 2 s = 400 m

∴ Distance covered by ball in 1 s = 400 m ÷ 2

= 200 m

∴ Final velocity of ball = 200 ms–1

Applying, v = u + at

200 = 0 + a × 0.1

∴ a = 2000 ms–2.

∴ Force acting on ball F = ma

= 0.05 × 2000

= 100 N.

15. A car of mass 800 kg moving at 54 kmh–1 is brought to rest over a distance

of 15 m. Find the retarding force developed by its brakes.

Ans. m = 800 kg;

u = 54 km h–1 = 15 ms–1;


v = 0; S = 15 m; a = ? R.F. = ? Applying, v2 – u2 = 2aS (0)2 – (15)2 = 2a × 15 ∴ a = –7.5 ms–2 ∴ Retardation = 7.5 ms–2 ∴ Retarding force = mass × retardation = 800 kg × 7.5 ms–2 = 6000 N. 16. A cricket player holds a cricket ball of mass 100 g by moving his hands

backward by 0.75 m. If the initial velocity of ball is 108 kmh–1, calculate the

retarding force applied by the player.

Ans. m = 100 g = 0.1 kg; S = 0.75 m; u = 108 kmh–1 = 30 ms–1; Retarding force = ? Applying, v2 – u2 = 2aS (0)2 – (30)2 = 2a × 0.75 ∴ a = – 600 ms–2 ∴ Retardation = 600 ms–2 ∴ Retarding force = mass × retardation

= 0.1 kg × 600 ms–2 = 60 N.


17. A force of 0.6 gf acts on a glass ball of mass 200 g for 12 s. If initially the

ball is at rest, calculate

(i) final velocity, (ii) distance covered. [g = 1000 cms–2]

Ans. F = 0.6 gf = 0.6 × 1000 = 600 dynes;

u = 0;

m = 200 g;

t = 12 s;

v = ?;

S = ?

Applying, F = ma

600 = 200 × a or a = 3 cms–2.


v = 0 + 3 × 12 = 36 cms–1.

Applying, S = ut + 12 at2

= 0 × 12 + 12 × 3 × (12)2

= 216 cm.


18. A bullet of mass 30 g moving with velocity x, hits a wooden target with a

force of 187.5 N. If the bullet penetrates 80 cm, find the value of x.

Ans. m = 30 g 0.03 kg;

u = x;

v = 0;

F = –187.5 N;

S = 80 cm = 0.80 m;

a = ?

∴ a F – 187.50.03m

= =

= – 6250 ms–2.

Applying, v2 – u2 = 2aS

(0)2 – (x)2 = 2 × – 6250 × 0.8

– x2 = –10000

∴ x = 100 ms–1.

19. A car of 1000 kg develops a force of 500 N, over a distance of 49 m. If

initially the car is at rest, find :

(i) final velocity, (ii) time for which car accelerates.

Ans. m = 1000 kg; F = 500 N; S = 49 m; u = 0; v = ?; t = ?;


Applying, a = F 500 Nm 1000 kg

= = 0.5 ms–2

(i) Applying, v2 – u2 = 2aS ∴ v2 – (0)2 = 2 × 0.5 × 49 ⇒ v2 = 49 ∴ v = 7 ms–1 (ii) Applying, v = u + at 7 = 0 + 0.5 × t ∴ t = 14 s.

20. A bullet of mass 25 g is fired from the barrel of a gun of length 0.80 m, with

a velocity of 270 kmh–1. Find

(i) acceleration acting on bullet, (ii) force acting on the bullet.

Ans. m = 0.025 kg;

S = 0.80 m;

v = 270 kmh–1 = 75 ms–1;

F = ?;

a = ?;

u = 0

(i) Applying, v2 – u2 = 2aS

(75)2 – (0)2 = 2 × a × 0.8

∴ a = 75 751.6×

= 3515.625 ms–2.


(ii) Force acting on bullet, F = ma

= 0.025 × 3515.625

= 87.89 N.

21. What extra force should an engine of a car develop, such that its velocity

changes from 18 kmh–1 to 72 kmh–1, over a distance of 20 m, when mass of

car is 900 kg?

Ans. u = 18 kmh–1 = 5 ms–1;

v = 72 kmh–1 = 20 ms–1;

S = 20 m;

F = ?;

m = 900 kg.

Applying, v2 – u2 = 2aS.

(20)2 – (5)2 = 2 × a × 20

∴ 375 = 40 a

or a = 375 ÷ 40

= 9.375 ms–2.

∴ Extra force developed, F = ma = 900 × 9.375

= 8437.5 N.


22. A bullet of mass 80 g, moving with a velocity of 360 kmh–1, is brought to rest

by wooden target in 0.01 s. Find (a) impulse, (b) average retarding force.

Ans. m = 0.08 kg;

u = 360 kmh–1 = 100 ms–1;

v = 0;

t = 0.01 s.

(i) Impulse = m(v – u)

= 0.08 (0 – 100)

= – 8 Ns.

(ii) Retarding force = ( ) 8 N0.01s

m v – ut

s = 800 N.

23. A cricket ball of mass 100 g strikes the hand of a player with a velocity of

20 ms–1 and is brought to rest in 0.01 s. Calculate the force applied by the

hand of the player.

Ans. Mass of the cricket ball (m) = 100 g = 0.1 kg

Initial velocity of ball (u) = 20 ms–1

Final velocity of ball (v) = 0

Time (t) = 0.01 s


0 = 20 + a × .01


a = –2–20 – 2000 ms.01

Retardation = – (a) = – (–2000 ms–2)

= 2000 ms–2

Retarding force acting on hand

= mass × retardation

= 0.1 kg × 2000 ms–2 = 200 N

24. A lead bullet of mass 20 g, travelling with a velocity of 350 ms–1, comes to

rest after penetrating 40 cm in a still target. Find the resistive force offered by

the target.

Ans. Mass of bullet (m) = 20 g = 0.02 kg Initial velocity (u) = 350 ms–1 Final velocity (v) = 0 Distance covered (S) = 40 cm = 0.4 m Applying, v2 – u2 = 2aS (0)2 – (350)2 = 2 × a × 0.4

a = –350 3502 0.4

××

= –153125 ms–2 Retardation = – (a) = – (–153125) ms–2

= 153125 ms–2 Resistive force F = mass × retardation = 0.02 kg × 153125 ms–2 = 3062.5 N


25. A body of mass 50 g is moving with a velocity of 10 ms–1. It is brought to

rest by a resistive force of 10 N. Find the distance that the body will travel

after the resistive force is applied.

Ans. Mass of body (m) = 50 g = 0.05 kg

Initial velocity (u) = 10 ms–1

Final velocity (v) = 0

Resistive force (F) = 10 N

Applying, resistive force = mass × retardation

10 N = 0.05 kg × retardation

Retardation = 10 N0.05 Kg

= 200 ms–2

Acceleration (a) = – 200 ms–2

Applying, v2 – u2 = 2aS

(0)2 – (10)2 = 2 × – 200 × S

S = 100400

= 0.25 m


26. A toy car of mass 500 g travels with a uniform velocity of 25 ms–1 for 5 s.

The brakes are then applied and the car is then uniformly retarded which

comes to rest in further 10 s. Calculate :

(a) retardation.

(b) the distance which the car travels after the brakes are applied.

(c) the force exerted by the brakes.

Ans. Mass of the toy car (m) = 500 g = 0.5 kg

Initial velocity (u) = 25 ms–1

Final velocity (v) = 0

Time (t) = 10 s

Applying,

v = u + at 0 = 25 + a × 10 s

a = ∴ 25–10

= –2.5 ms–2 Applying, v2 – u2 = 2aS (0)2 – (25)2 = 2 × – 2.5 × S

∴ S = 25 252 2.5××

= 125 m Applying, F = ma = 0.5 kg × 2.5 ms–2 = 1.25 N


27. Write approximate weight of a body of mass 10 kg. What assumption have

you made above?

Ans. The approximate weight of body is 100 N. It is assumed that ‘g’ is 10 ms–2

and hence weight is 10 kg × 10 ms–2 = 100 N.

28. Calculate the weight of a body of mass 20 kg in

(a) kgf, (b) newton. Take g = 9.8 ms–2. State direction of force acting on the

body.

Ans. (a) Weight in kgf = 20 kg × g = 20 kgf, acting downward

(b) Weight in newton = 20 kg × 9.8 ms–2 = 196 N, acting downward

29. The weights of two bodies are 2.0 N and 2 kgf respectively. What is the mass

of each body? [g = 10 ms–2]

Ans. (i) F = m.a

2.0 N = m × 10 ms–2

m = –2

2.0 N10 ms

= 0.2 kg

(ii) F = ma

2 kgf = m × 10 ms–2

2 kg × 10 ms–2 = m × 10 ms–2

m = 2 kg


30. The weight of a body on the earth is 98 N, where acceleration due to gravity is

9.8 ms–2. What will be its (a) mass, (b) weight on the moon, where

acceleration due to gravity is 1.6 ms–2 ?

Ans. Force on earth= Mass on earth × gEarth

98 N= Mass on earth × 9.8 ms–2

Mass on earth = –298 N

9.8 ms

= 10 kg Now, as mass of a body is a constant quantity Mass of body on moon = 10 kg Weight of body on moon= mass × gmoon = 10 kg × 1.6 ms–2 = 16 N 31. A man weigh 600 N on the earth. What would be his approximate weight on

the moon? What is the reason for your answer?

Ans. The acceleration due to gravity on moon is 1/6th compared to acceleration due

to gravity on the earth.

Weight of man on moon = 16 × 600 N = 100 N

32. What force and what weight act on a body of mass 10.5 kg, being pulled by

the earth? [Take g = 10 ms–2]

Ans. Weight of body

= Force due to gravity acting on the body

= 10.5 kg × 10 ms–2 = 105 N.

Question Bank Force - The Assam Valley School Today/WORKSH… · · 2013-07-24Class-X Question 1...

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