Kinematicsof Particles - Rectilinear Motion

Types of Plane motionAs we know a particle has translation motion only, not rotation about a point or an axis. Further

the translation motion of a particle can be classified into two types.1. Rectilinear motion2. Curvilinear motion

Kinematics:Kinematics is the study of motion of a moving body without considering the forces which causes the

motion.Kinetics:

It is the study of motion of a moving body with also considering the external forces which cause the motion.

Rectilinear motion:The motion of a particle along a straight line is known as rectilinear motion (or straight line motion).

Examples:1. A car moving on a straight road,2. A stone falling vertically downward3. A ball thrown vertically upward etc.,

Curvilinear motion:The motion of a particle along a curved path is known as curvilinear motion.

Characteristics of kinematics:

1.Displacement:The displacement of a moving particle is the change in its position, during which the particle

remains in motion. It is a vector quantity, has both magnitude and sense of direction.It is denoted by the letter 'S'.

2.Speed:It is the distance travelled by the particle or body along its path per unit time. It is a scalar quantity has magnitude only. It is measured in m/s, km/hr etc.

Speed =Distance travelled/time taken

3.Velocity:It is the rate of change of displacement. It is a vector quantity.Velocity = Distance travelled in particular direction/time takenIn SI system, it is measured in m/s.

4. Acceleration:It is the rate of change of velocity. It is denoted by 'a'. In SI system, it is measured in m/s2.It is a vector quantity.acceleration, a = change of velocity/time taken

=(Final velocity – Initial velocity)/time takenNegative acceleration is called as Retardation (or) Deceleration.

5. Average velocity:It is the ratio of displacement and time interval.ie., Average velocity = Change in position /Change in timeIt can be either positive or negative.

Equations of Linear motion:

Sn=u+(a/2(2n-1))Where,u- Initial velocity,V- Final velocity,a- Acceleration , t-time taken ,S- Distance travelled.

Note:1)When a particle moves in one direction & it covers equal distance in equal intervals of time,then the velocity of particle is known as Uniform velocity.

2) When there is a change in direction (or) change in magnitude (or) change in magnitude &direction,then the velocity is known as variable velocity.

3) If a body start from the reset , its initial velocity is zero(i.e, u=0)

4) If a body comes to rest , its final velocity is zero (i.e, v=0)

1.A car is moving with a velocity of 20m/s. The car is brought to rest by applying brakes in 6 seconds. Find (i) Retardation (ii) Distance travelled by the car after applying the brakes.Solution :when, Initial velocity, u =20m/sFinal velocity, v = 0 (car is brought to rest); time, t = 6 sec

Retardation:Retardation is the negative acceleration.using the equation of motion, v = u + atie, 0 = 20 + (a x 6)

a = -3.33 m/s2

Retardation = 3.33 m/s2

Distance travelled:Let s = Distance travelled by the car after applying the brakes using the equation, s = ut + 1/2 at²s = (20 x 6) +1/2(-3.33) x 6²=60 m

2)A Train starts from rest and attains a velocity of 45 km per hour in 2 minutes, with uniform acceleration. Calculate,(i) acceleration (ii) Distance travelled in this time, 2 miniii) time required to reach a velocity of 36 km/hr.

Given, Initial velocity, u = 0 ( '.' starts from rest)Final velocity, v = 45 km/hr

=((45 x 1000)/3600) m/s = 12.5 m/stime taken, (t)= 2 minutes = 2 x 60 = 120 secs.

1km=1000 m,1hr=60 mins,1min=60 sec

(i) Acceleration (a):Using the equation, v = u + at

12.5 = 0 + (a * 120) a = 12.5/120= 0.104 m/s2

(ii) Distance travelled in 2 minutesUsing the eqn, s = ut + 1/2at²

=(0 x 120) + (1/2 * 0.104 *120²)= 748.8 m

(iii) Time required to attain the velocity of 36 km/hrinitial velocity, u = 0, Final velocity V=36 km/hr=((36*1000)/3600)=10m/s.

Now, using the equation v = u + at10 = 0 + (0.104 x t) t = 96.15 sec.

3)A motor is travelling at 80 km/hr when he observes a traffic light 200m ahead of him turns red.The traffic light is timed to stay red for 10Secs.If the motor is wishes to pass the light without stop.Just of its turn green, determine,

i)The Required uniform deceleration of the motor.

ii)The Speed of the motor as it passes the light.

Solution:

Initial Velocity u=80 km/hr=((80*1000)/(60*60))=22.22 m/s.

Distance S=200m , t=10 Secs.

To find:

i)Deceleration of the motor, 2)Speed of the motor

Solution:

i)Distance travelled

s = ut + 1/2at²

200=(22.22*10)+(0.5*a*102);200=222.2+50a

50a=200-222.2;a=-0.445 m/s2.

2)Speed v = u + at=22.22-(0.445*10) =17.75 m/s.

4)A Car initially moving at 45 km/hr has to cover a distance of 7.6 km ,in 8 minutes with what acceleration,should it be speeded up so that it covers a distance exactly in time, ii)if the same car moves with the uniform speed of 45 km/hr how much the time it would have taken to cover the distance of 7.6 km.Also calculate the time saved due to the accelerated car?

Given data:

Initial velocity u= 45 km/hr=((45*1000)/3600)=12.5 m/s.

Distance S=7.6 km=7.6*103 m, time t=8mins =8*60=480 secs.

To find:

1)Acceleration,2) Time,3)Saved time.

Solution:

Distance s = ut + 1/2at²

7600=(12.5*480)+(1/2*a*4802 )

a=0.0139 m/s2

Distance travelled S= uniform velocity*time

S=u*t; u=S/t =Distance/time ;t=S/u=7.6*103/12.5=608 Sec.

Time saved t= 608-480=128 Secs.

5)A motor moving with an uniform acceleration covers a distance of 20m in 4 seconds and 40m in 6 seconds. Find the uniform acceleration of the motor?Solution:Given,

S1=20m , T1=4secS2=40m , T2=6sec

Case (i) From equation, s = ut + 1/2at²

1

Case (ii) From equation, s = ut + 1/2at²

(ii)

On solving the equations (i) and (ii), we get a = 1.667 m/s2

and u =1.667 m/s.uniform acceleration of the motor is l.667 mls2

6)A motor starts from rest and uniformly accelerated to speed of 20 km/hr over a distance of 200m. Calculate the acceleration and time taken. If further acceleration raises the speed to 50 km/hr in 8 seconds, find the acceleration and the further distance moved?Solution:case (i)

Given: Initial velocity, u = 0 ; Distance, s = 200mFinal velocity, v = 20 km/hr =((20*103)/3600)= 5.555 mlsTo find acceleration using the equation,

v²=u2 + 2as(5.555)² =0+ (2 * a*200)

a = 0.077 m/s2

To find the time taken use the eqn, v = u + at5.555= 0 + (0.077 x t):. t= 72.14 sec

Case (ii) Given,Initial velocity, u= Final velocity of case (i)Final velocity, v = 50 km/hr = (50*1000)/3600= 13.89 mlstime taken, t = 8 secs,

To find accelerationusing the equation, v = u + atl3.89 =5.555 + (a*8)a = 1.0418 m/s2

To find the further distance moveduse the eqn, s = ut + 1/2at²=(5.555 x 8) + (1/2*1.0418* 8²)

= 77.78 m

7)A train is travelling from A to D along the track shown in fig.Its initial velocity at A is zero. The train takes 5 min to cover the distance AB, 2250 m length and 2.5minutes to cover the distance BC, 3000 m in length, on reaching the station C, the brakes are applied and the train stops 2250 m beyond at D (i) Find the retardation on CD, (ii) the time it takes the train to get from A to D (iii) its average speed for the whole distance.

Solution:Given, Distance AB' = 2250 m; t = 5 min=(5*60)=300 SecsBC = 3000 m; t = 2.5 min=(2.5*60)=150SecsCD = 2250 m; t = ?

Brakes are applied at C.Train starts from A and stops at D.(i) Retardation of the train on CD: let us consider the motion of the train on AB.

Initial velocity u (at A) = 0Distance, s = 2250 mLet final velocity (at B) = vtime, t= 5 min = 300 secusing the equation, s=ut + 1/2at²

2250=0 + (1/2* a*300²) a=0.05 mls2

using the eqn, v=u + at =o + (0.05 x 300) =15 m/s

:. velocity at B = 15 mls

consider the motion of the train on BC.Initial velocity (at B), u = 15 mls (as in previous case)Final velocity at C = vDistance, S = 3000 mtime, t = 2.5 min = 150 secThe train is moving on constant acceleration from A to C. So on BC region also, a = 0.05 m/s2. Now, using the relation, v = u + at

= 15 + (0.05 x 150)=22.5 mls consider the motion of the train on CD.Initial velocity. (at 9, . .u = '27.5 m/s (as in previous case)Final velocity (at D) = v = 0 ( '.' stops at D)Distance, s 2250 mtime, t = ?

Using the relation, v² =u²+ 2as0 = (22.5²) + (2*a*2250):. a =0.1125 m/s2 (retardation)

from the equation, v = u + at0 = 22.5 - (0.1125 *t) t = 200 sec or 3.333 minRetardation on CD =0.1125 m/s2

(ii)Total time taken from A to DTotal time taken by the train from A to D, T= t1 + t2 + t3

=5 + 2.5 + 3.333= 10.833 min

Average speed of the train for the whole distance=Total distance travelled/Total time taken=(2250 + 3000 + 2250)/(10.833 x 60)=11.538 m/s=41.53km/hr.

8)The velocity of a particle, which is moving in a straight line with constant retardation, decreases 10m/s while the particle travels 10m and 15m/s while it travels 12.5 m from the starting point. Find the distance the particle travels from the starting point till it comes to rest.Solution:

Consider the motion from o to A.using v² = u²+ 2as , u=u,v=(u-10), s=10m.(u -10) ² = u² +2a(10) 100-20u = 20a

u+a= 5Consider the motion from 0 to Busing v² = u²+ 2as(u -15) ² = u² +2a(12.5) 225-30u = 25a

6u+5a = 45Solving the equations (i) and (ii) u = 20m/s and a = -15m/s2

i.e., retardation = 15m/s2

1

2

Consider the motion from o to c:Velocity at 0, u= 20m/sVelocity at c, v =0 ( .. comes to rest)retardation, a = -15m/s2

Distance travelled (OC), s = ? using v² = u²+ 2as

0 = 20²+ [2*(- 15s)] s =13.33m.

9)A car starts from rest with a constant acceleration of 4m/s . Determine the distance travelled in the 7th second.

Solution:Given: Initial velocity, u =0 ( starts from rest)acceleration, a = 4m/s2; n = 7

using the relation, Distance travelled in nth second, S(nth) = a+u/2(2n- 1)S7=0+ 4 /2(2 x 7 -1)= 26m

Motion of the particle under gravity:

Important result: Upward motion.(i) Maximum height attained by upward particle

(ii) Time taken by the particle to reach maximum height

Total time taken to return the surface(T) = 2 x Time up

Important result :- Downward motionStriking-velocity of the particle, moving downwards,

from the position of rest

1)A stone is dropped from the top of a tower, reaches the ground in 8 seconds, find(i) the height of the tower(ii) velocity of the particle, when it reaches the ground.

Solution:Given: Initial velocity, u = 0; time taken, t = 8 seconds

1.Height of the towerusing the relation, h = ut + 1/2gt²

h =0+(1/2*99.81*8²)=313.92 m.2.Velocity of the particle, while it reaches the groundLet v be the velocity (ie, Final velocity) of the particle, while striking the ground.

using the relation, v = u + gtv= 0 + (9.81 x 8)V=78.48 m/s

3)A Stone is thrown vertically upwards.It reaches the max. height 12m.Determine (i) The velocity with which the stone was thrown ii)The time taken to reach max.height.(iii)Total time taken by the stone, to return to the ground surface,after projected upwards.

Given data:

Max.height, hmax=12 m.

Solution:

i)Velocity with which the stone was thrown

w.k.t, At max height,velocity V=0,

V2=u2-2gh; 0=u2-(2gh)

u= 2ghmax

u=(2*9.81*12) = 15.34 m/s.

ii) Time taken to reach max. height

At max height,velocity V=0,

V=u-gt;0=15.34-(9.81*t); t=1.56 Secs.

iii) Total time taken , T=2t=(2*1.56)=3.12 Sec.

4)A particle moves along a straight line with variable acceleration. If the displacement is measured in m & given by the relation in terms of time taken, as below,S = 3t³ + 2t² + 7t + 3Determine (i) the velocity of the particle at start and after 3 seconds.(ii) the acceleration of the particle at start and after 3 seconds.Solution:

velocity v= ds/dt

=9t² + 4t + 7 (i)

(i) Velocity at startTo find the velocity of the particle at start, substitute t = 0 in equation (i)velocity at start, v (9 * 0) + (4 *0) + 7 = 7 m/s.Velocity after 3 secondsSubstitute t = 3 in equation (i)Velocity after 3 seconds, V = (9*32) + (4 *3) + 7

=81 +12 + 7 = 100 m/s.(ii) AccelerationDifferentiate the equation (i) w.r.t 't' to get acceleration.

ii

Acceleration at startt=0

acceleration at start, a=(18*0) + 4 = 4 m/s²Acceleration after 3 seconds

t=3acceleration at start, a3=(18*3) + 4 = 58 m/s²

5)The equation of motion of a particle is given, acceleration (a) in terms of time (t) as below.a =3t³ + 2t² + 4, in which acceleration is in m/s2 and time 't' is in seconds. It is observed that the velocity of the particle is 12m/s after 4 seconds,' and the displacement of the particle is 8 m after 4 seconds. Determine(i) Velocity after 8 seconds (ii) Displacement after 2 secondsSolutionGiven, a == 3t2 + 2t + 4 Boundary conditions:

at t = 4 sec; v =12m/sand at t = 4 sec; S =8m

Velocity:Velocity, V = ∫a dt

=∫ (3t³ + 2t² + 4 )dt V= t³ +t² +4t + C1

To find the constant of integration C1, apply the boundary condition of velocityie, at t = 4 see, V = 12 m/s.12 = 4³ +2² +(4*4) +C1

12=96+C1

C1=-84V=t³ +t² +4t - 84

i

DisplacementDisplacement, S = ∫v dt

=∫ (t³ +t² +4t - 84)dt

To find the constant of integration C2, apply the boundary condition ofDisplacement ie, at t = 4 sec, S = 8m

C2=226.66

Velocity after 8 secSubstitute t = 8 in velocity equation

V = t³ +t² + 4t - 84 = 83+82+(4x8)-84 =524m/s

Displacement after 2 secSubstitute t = 2 in displacement equation

S = 73.32m

6)A particle moves along a straight line with an acceleration described by the equation, a = (3t² - 6), in which 'a' is in m/s2; and t is in seconds. It is observed that distance travelled by the particle is 6 m at the end of 2 seconds and 10 m at the end of 3 seconds. Calculate,(i) Velocity after 10 secs (ii) Displacement after 8secs(iii)Determine the time

at which the velocity is max & Calculate the max velocity.Solution:a = (3t² - 6)

Boundary Condns s=6m ,t=2sec: s=10m,t=3 secVelocity, V = ∫a dt

=∫ (3t² - 6)dt V = t³ - 6t + C1

Displacement,S = ∫v dt = ∫ (t³ - 6t + C1) dt

To find the constants of integration C1 and C2 apply the given boundary conditions.at t = 2 sec, S = 6m

6 = 4 - 12 + 2C1 + C2

6 = 2C1 + C2 – 8or 2C1 +C2 = 14

at t = 3 sec, S = 10m

10 =20.25 - 27 + 3C1 + C2

10 =6.75+ 3C1 + C2

or 3C1 + C2 = 16.75On Solving equations (i) and (ii), we getC1 = 2.75 and C2 = 8.5Substitute C1 and C2 in velocity and

(i)

(ii)

Velocity, V = t³ - 6t + 2.75

Velocity after 10 secondsSubstitute t = 10 in velocity equationV = t³ -6t+2.75 = 10³ -(6*l0)+2.75 = 942.75 m/sDisplacement after 8 secondsSubstitute t = 8 in displacement equation

=862.5 miii) dv/dt=0=a, 3t2-6=0;3t2=6; t=1.414 SecSub t=1.414 Sec in velocity eqn, we get V = t³ - 6t + 2.75

=(1.414)3-(6*1.414)+2.75=-2.90 m/s;Vmin=2.90 m/s.

Kinematics of particles -Newton's Law of Motions.Important Definitions:1. Mass

The quantity of matter, contained in a body is known as mass. It is denoted by the letter ‘m’. It is a scalar quantity. It’s unit is kg.

2. Weight Weight of a body is defined as the force, by which the body is attracted

towards the centre of the earth. It is denoted by the letter 'W'. It is a vector quantity.Mathematically,

Weight = mass*acceleration due to gravityW = mg.

In SI system of units, weight is expressed in Newton (N)

Note:

Momentum:The quantity of motion possessed by a moving body is called 'Momentum'.

For an exact measure of motion, momentum of any moving body is defined as the product of its mass and velocity. It is denoted by M (Capital letter).

If m mass of a moving body,v = Velocity of the moving body

then, momentum = mass x velocity (or) M = m x vNote that, in SI system of units, mass 'm' is in kg and velocity 'v' is in mlsHence, the unit of momentum is 'kg. m/s'.

Newton's second law of motion:The change of momentum= Final momentum- Initial momentum

=mv-mu=m(v-u):. Rate of change of momentum= Change of momentum/time taken

=m (v-u)/t P=m.a (v-u)/t=a

P=m.a called as Fundamental eqn of dynamics.

Frictional force:The maximum frictional force developed at the contact area is called as

‘limiting friction’Co-efficient of friction:

The ratio of limiting friction to the normal reaction is known as "Co-efficient of friction“.

Co-efficient of friction = Limiting friction/Normal reaction

Case 1 : Body moving on rough horizontal surface

If µ = Co-efficient of frictionFӎ = µNʀ

Nʀ = Weight 'w‘ Fӎ = µW

Case 2: Body pulled up on an inclined surfaceTo find Fm:Resolving the forces normal to the plane, andequated to zero.

Fm = µNʀ ;NR - W cosƟ= 0Fm= µ W cosƟ

1)A body of. mass 16 kg is pulled along a rough horizontal table by a constant force. It describes 3m from rest in 3 seconds. Find the magnitude of this force. Take co-efficient of friction as 0.25.Solution:

Given:u = 0 (starts from rest); s = 3m; t = 3 sec; µ= 0.25

By applying Static equilibrium CondnFrom ƩV = 0, Nʀ = W = mg = 16*9.81 = 156.96 NLimiting friction, Fm = µNʀ = 0.25*156.96 = 39.24 Nusing the equation; s = ut + 1/2 at²

3 =0 + (1/2*a*3*3)a =0.667 m/s2

Now, applying the equation of dynamics,ƩF =ma (System of forces on a single body) but ƩF = P-Fm

P-Fm=maP = ma + Fm =(16*0.667) + 39.24P=49.912 N.

D'Alembert's Principle:D'Alembert's principle states that, "the system of forces acting on a body

in motion is in dynamic equilibrium, with the inertia force of the body".

1)A Block of mass 10 kg, rests on a horizontal plane, as shown in fig.Find the magnitude of the force P, required to move the block at an acceleration of 2m/s2 towards right .Take μ= 0.25.

Solution:Given, m = 10 kg :. W = mg= 10*9.81 =98.1Na = 2 m/s2 ; µ = 0.25With the Inertia force ,the body is in the Condn of Static equilibrium ƩV = 0, ƩH= 0 Applying ƩV = 0

Nʀ-W-PsinƟ = 0Nʀ-98.1-Psin30 = 0Nʀ= 98.1+0.5P (1)

Applying ƩH= 0P cos Ɵ -F –ma=0

but, the frictional force, F= µNʀ= µ (98.1 + O.5P)

=0.25 ( 98.1 + 0.5 P)F=24.52 + 0.125 P

:.P cos 30 - (24.52 + 0.125 P) - (10 x 2) = 00.741 P = 44.52

P = 60 N:. The force required to move the block is 60 N.

Motion of a lift:Let us consider the motion of a lift with some acceleration

(i) the lift is moving upwards (ii) the lift is moving downwardsCase(i) : Lift is moving upwardsW = Weight carried by the lift.a = uniform acceleration of the lift.N = Reaction of the lift (which is also equal to the tension T in the cable).

When the lift is moving upwards the net force is acting upwards and hence the inertia force is acting downwards.

Applying ƩV = 0N1-W-ma = 0N1 - W - (W/g *a) =0N1=W(1+a/g)

Applying ƩV = 0N2+W-ma = 0

Or N2 + W - (W/g *a) =0N2=W(1-a/g)

Case (ii) : Lift is moving downwards

2)A man weighing 600 N gets into a lift. calculate the force exerted by him on the floor of the lift, when it is(i) moving upwards with an acceleration of 3m/s2, and(ii) moving downwards with the same acceleration.Solution:(i) When the lift is moving upwardsForce exerted by the man on the floor of lift N1=W(1+a/g)

=600(1+ 3/9.81) =783.48N

(ii) When the lift is moving downwardsForce exerted by the man on the floor of lift N2=W(1-a/g)

=600(1 - 3/9.81) =416.51N

3)An elevator of weight (including the weight of man) 4.5 KN starts moving upwards with a constant acceleration and acquires a velocity of 1.8 m/s, after travelling a distance of 2m. Find the pull in the cable during accelerated motion.Solution:Given: W = 4.5 KN = 4500 Nu = 0 ('.' starts moving from rest)v = 1.8 m/s ,s = 2mUsing the equation, v² = u² + 2as1.8² =0 + (2*a*2) a=0.81 m/s2 .Applying ƩV = 0 ( +)T-4500-ma = 0T - 4500 - ((4500/9.81) x 0.81)=0T = 4871.5 N

Motion on an inclined surface:

using p= mahere P = WsinƟW sinƟ = ma; mg sinƟ = ma a= gsinƟ.

1)A body of mass 15 kg is initially at rest on a 10◦ inclined plane. Then it slides down. Calculate the distance moved by the body, on the inclined plane, when the velocity reaches to 6m/s. The coefficient of friction between the body and the plane is 0.1Solution:

Given, m = 15 kg; :. weight, W = (15*9.81) N = 147.l5 Nu = 0 (. starts from rest) v = 6 m/s ,Ɵ=10˚Resolving the forces normal to the planeNʀ-147.15 cos 10 =0Nʀ = 147.15 cos 10 = 144.91 N.

Resolving the forces along the plane,F + ma - WsinƟ=0µN + ma - W sinƟ =0(0.1 x 144.91) + ((147.15/9.81)a) - (l47.15sin10)=0a=0.737 m/s2

Now, using the equation, v² = u² + 2as6² =0 + (2 x 0.737 x s)S=24.42 mDistance moved by the body is 24.42 m

2)Two Blocks A and B of weight 100 N and 200 N respectively are initially at rest on a 30 ْ inclined plane as shown in fig. The distance between the blocks is 6 m. The co efficient of friction between the block A and the plane is 0.25 and that between the block B and the plane is 0.15.If they are released at the same time, in what time the upper block (B) reaches the lower Block (A). Given:WA = 100 N; µA= 0.25WB = 200 N; µB= 0.15

Consider Block B:Resolving the forces normal to the planeNB - 200 cos30 = 0NB = 200 cos 30 = 173.2 NResolving the forces along the plane,FB - 200sin30 + maB = 0

Consider Block A:Resolving the forces normal to the plane,NA - 100 cos30 = 0 :. NA= 86.6 NResolving the forces along the plane,FA + maA - 100sin30 = 0

Let t =time at which the blocks A and B touches each other, after released at same time from rest.SA = Distance travelled by Block A in time ‘t'.SB = Distance travelled by Block B in time ‘t'.To find SA ;Using the equation s= ut + 1/2at²uA = 0; aA = 2.78 m/s2

SA = 0+(1*2.78t²)SA = 1.39t²

To find SB :Using the equation s= ut + 1/2at²uB = 0; aB = 3.63 m/s2

SB = 0+(1/2X3.63xt²)SB = 1.851t²

When the two blocks touch each other, then SB = SA + 6ie, l.815t² = l.392t² + 6 t=3.75sec

Motion of Connected Bodies:Motion of two bodies connected by a string and passing over a smooth pulley.

Fig. shows two bodies of masses ml and m2connected by an inextensible light string (ie, self weight is neglected), passing over a smooth pulley. Let us assume the mass m1is less than m2. Due to this condition, the greater mass m2, will move downwards and the smaller mass ml will move upwards. But, both the bodies will have the same acceleration.

Let, a = acceleration of the bodies in m/s2

T = Tension in the string in N.Consider the mass m1:m1is less than m2, hence it moves upwardsapplying ƩV = 0.T-m1g-m1a=O or T-m1g=m1a (1)Consider the mass m2:[m2 is greater than m1,hence it moves downwards.]applying ƩV = 0.T+m2a-m2g=O m2a=m2g-T (2)solving the two equations (i) and (ii), we get the twounknowns, acceleration 'a' and the tension in the string 'T'

1)Two blocks A and B of weight 80 N & 60 N are connected by a string, passing through a smooth pulley, as shown in fig . Calculate the acceleration of the body and the tension in the string.Solution:Given; Weight of block, A, WA=80 NWeight of block, B, WB=60NConsidering the Block A (moving downwards):Applying ƩV = 0 T-80 + ma = 0T - 80 + ((80/9.81) a) = 0T+ 8.155 a = 80 1

Considering the Block B (moving upwards)Applying ƩV = 0T-60-ma = 0T - 60 + ((60/9.81) a) = 0T+ 6.116 a = 60 (2)solving the equations (1) and (2), we get 'a' and T.T + 8.155 a = 80T - 6.116 a = 60(i)- (ii) :- 14.271 a = 20

a =1.401 m/s2

To find T, substitute 'a' in equation (i)T + 8.155 a=0T +.8.155 (1.401) = 80:. T = 68.57 N

2)Two weights 80 N and 20 N are connected by a thread and move along a rough horizontalPlane under the action of a force 40 N, applied to the first weight of 80 N as shown in fig .The coefficient of friction between the sliding surfaces of the weights and the plane is 0.3. Determine the acceleration of the weights and the tension in the thread using D'Alembert's principle.Solution:Let a = acceleration of the weights, T = Tension in the threadConsider 80 N block:Applying ƩV = 0N1 - 80N =0or N1 = 80NApplying ƩH = 040 -T- F1 +ma=040-T-(µN1)-ma = 040 -T - (0.3 x 80) – ((80/9.81)x a) = 0T+8.155a =16 1

Consider 20N block:Applying ƩV = 0N2 - 20 = 0 :. N2=20NApplying ƩH = 0T - F2 - ma =0T-(µN2)-ma = 0T - (0.3 x 20) - ((20/9.81)*a) = 0T - 2.038a = 6 (2)solving the two equations (i) and (ii), the two unknowns 'a' and T' can be determined.a = 0.981 m/s2 and T = 8N

3)Fig.shows two blocks of weight 60 N and 140 N, placed on two inclined surfaces and connected by an inextensible string. Calculate the acceleration of the system and the tension in the string. Take, µ= 0.2Solution;Let a = acceleration of the system, andT = Tension in the string.Consider 60 N Block (moving upwards)Resolving the forces, normal to the plane. ƩV = 0NR - 60 cos 60 = 0or NR = 60 cos 60 = 30 N.Resolving the forces along the plane ƩH = 0 T - 60 sin 60 - F - ma = 0 T - 60sin60 - (0.2 x 30)-((60/9.81)a) = 0 T-51.96-6- 6.116 a= 0 T-6.116a = 57.96 .....(i)

Consider 140 N Block ( moving downwards)Resolving the forces normal to the plane, ƩV = 0NR - 140 cos 30 = 0NR = 140 cos 30 = 121.24 N.Resolving the forces along the plane ƩH = 0 140 sin 30 - T - F – ma= 0; 140sin30-T-(O.2 x 121.24)-((140/9.81) a) = 0

70 - T - 24.24 –((140/9.81)a)= 0 T + 14.27 a = 45.76 .....(ii)Solving the equations (i) and (ii), we get a = -0.598 m/s2

(-) sign shows that, the body will move in the' opposite direction of the assumed direction.ie, 60 N block moves downwards and the 140 N block moves upwards.

Substitute 'a' in equation (i)T-6.116 a = 57.96;T - (6.116 x -0.598) = 57.96or T + 3.657= 57.96T =54.3 N..'. Acceleration of the system is 0.598 mls2 and the tension in the string is 54.3 N

Kinetics Of Particles – Work Energy Method:Work:Work is defined as the product of force and displacement of the body.Work done by the force = Force x Distance moved

=p x s

Work done Component of force in the direction of motion X Distance moved= P cosƟ x s

Unit of work:In SI system of units, force is in Newton and the distance is in m.:. unit of work is (Newton x meter) = 1 Nm = 1 Joule:. In SI system of units, unit of work is joule.

1)Determine the total workdone on a 5 kg body, which is pulled 6m up on a rough inclined plane, as shown in fig. Take the coefficient of kinetic friction between the body and the plane is 0.2.Solution:To find ƩFx

ƩFx=70cos 10 - (5 x 9.81 sin 30) - FƩFx=70cos 10 - (5 x 9.81 sin 30) - µN

To find the normal reaction (N), resolve the forces normal to the plane and equate to zero.NR + 70sin 10 - (5 x 9.81cos 30) = 0NR + 12.15 - (42.47) = 0. . NR = 30.32 NewtonSubstitute the value of 'N' in ƩFx equation,ƩFx=70cos 10 - (5 x 9.81 sin 30) - FƩFx=70cos 10 - (5 x 9.81 sin 30) – (0.2*30.32)

=68.93 - 24.52 - 6.06= 38.35 N.:.Workdone = ƩFx *s

= 38.35 x 6 = 230 N = 230 Joule.

PowerThe rate of doing work is known as power:. Power =workdone/ time= (Force x Distance)/ time

= Force x VelocityUnit of powerIn SI system of units, unit of work is Newton-metre, and the unit of time is sec .so, um.t 0f power = -Nm/sec = 1 watt :. In SI system, unit of power is watt.Energy:The capacity of doing work is known as Energy.Unit of energy is same as that of work.Potential Energy:It is the capacity to do work by virtue of position of thebody. It is. denoted by P.E.

Work done by the body = Force*Distance= mg*h= mgh

Potential Energy, P.E = mghKinetic Energy:It is the capacity to do work by virtue of motion of the body. It is denoted by K.EKinetic Energy, K.E. = 1/2mv²1)An aero plane of mass 8 tonne is flying at a rate of 250 kmph, at a height of 2km above the ground level; Calculate the "total energy possessed by the aero plane.Solution:Mass= 8t = 8 x 1000 = 8000 kgvelocity, v = 250 km/hr = 250*1000 /3600= 69.44m/sheight, h =2 km = 2000 m.

Potential Energy, P.E. = mgh= 8000 x 9.81 x 2000=15696 x 104 Nm= 15696 x 104 Joules.

Kinetic Energy,K.E. = 1/2mv²= t1/2x 8000 x (69.44) ²= 19287654 Nm= 19287654 Joules.

Total Energy = P.E+ K.E.= .(15696 x 104) + (19287654)= 176247654 Joules

=176.247 x 106 Joules or l76.24 MJ

Work-Energy Equation:Work done = Final Kinetic Energy - Initial Kinetic EnergyNote:If a system of forces are acting on a moving body, then the work-energy equation as modified as below. ƩF*s = W/2g [V² - u²]If a system of forces are acting on a system of bodies, then the work-energy equation is,ƩF*s = ƩW/2g [V² - u²]

1)A body of weight 600N, is placed on a 20° inclined plane and pulled by a 500N force, applied parallel to the plane as shown in fig.If the initial velocity of the body is 2m/s, calculate the final velocity of the body, when it has travelled a distance of 2m. Take the coefficient of kinetic friction between the body and the plane is 0.2.

Solution:The free body diagram of the body is shown in fig.Work energy equation is, Work done = Change in Kinetic EnergyƩF*s = W/2g [V² - u²]ƩFx = Net force in the direction of motionIn this problem ƩFx = Net force along the plane, ie, xx axis.ƩFx = 500 - F - 600sin 20 = 500 - (µNR)- 600sin 20To find NR, resolve the forces, normal to the planeNR - 60Ocos 20 = 0:. NR = 600cos 20 =563.81 N'.. Ʃ Fx = 500 - (0.2 x 563.81) - 600sin 20 = 182N

Substituting in the work energy equation

solving, v = 3.987 m/s.

Motion of connected bodies - Work-Energy Equation:

To apply the work-energy equation, first find the net force along the direction of motion of the system.

Applying work-energy principle,Total work done of the system =Total change in kinetic energy of the system

1)Two weights 80N and 20N are connected by a thread and move along a rought horizontal V plane under the action of a force 40N, applied-to the first weight of 80N as shown in fig. The coefficient of friction between the sliding surfaces of the weights and the plane is 0.3. Determine the acceleration of the weights and the tension in the thread using work-energy equation.

Solution:The forces acting on the bodies are shown in fig

ƩFx = 40 - (0.3 x 20) - (0.3 x 80) =10Nnow, applying the work-energy equation,

or v²=1.962s

Substitute v2 = 1.962s in the equation, v2 =u2 + 2as1.962s = 0+ 2as ;a= 0.981m/s2

To find T:Apply work-energy equation on anyone body.Consider 20N block,Free body diagram of 20N weight is shown in fig.Workdone = Net force x Distance

={T - (0.3 x 20)}s =(T - 6)s ........................(i)Change in kinetic energy=W/2g(v² - u²)

Equating (i) and (ii), we get

T=8N

Kinetics of Particles – Impulse and Momentum

The Impulse-Momentum method is based on integration of kinetic equations of motion with respect to time.This method is particularly suitable for analysing the kinetic problems, when (i) The force varies with time(ii) Large force, acts for a short period of time in the case of sudden blow or impact.Impulse of a force:-When a large force acts for a short period of time, that force is called an Impulsive force.It is denoted by the symbol I. It is a vector quantity.

Linear Impulse = Force x Time; Unit is N.

Momentum:-Momentum = Mass x Velocity

M = mvImpulse - Momentum equation:Impulse = Final momentum - Initial momentumor Final momentum = Initial momentum + Impulse.

1)A 500 N block is in contact with a level plane, the coefficient of friction between two contact surfaces being 0.25. If the block is acted upon by a horizontal force of 1300 N, what time it' will elapse before the block reaches a velocity of 24 m/s.

Given, W = 500 N µ= 0.25; v = 24 m/s,u=0.Normal reaction, NR = 500 NFrictional force, F = µNR

= 0.25 x 500 = 125 NThe Net force along the direction of motion,ƩF =1300 – F = 1300-125 = 1175NThe Net force along the direction of motion,ƩF x t = m(v-u)

t =1.04 sec.

Kinetics of Particles - Impact of Elastic bodiesThe time taken by the bodies in compression, after the instant of

collision is known as time of compression.The time taken by the bodies to regain the original shape and size after

compression is known as time of restitution.The sum of time of compression and time of restitution is known as

time of collision (or) time of impact.

Direct impact:In direct impact, the velocities of the two colliding bodies, before

collision are collinear with the line of impact.

Oblique Impact:In oblique impact, the velocities of the two colliding .bodies, before

collision are not collinear with the line the impact.

Period of restitution:-The time elapsed from the instant of initial contact to the max.

deformation is known as period of deformation and the time elapsed from the instant of max. deformation to the instant of just separation of particle is known as Period of restitution.

Co-efficient of restitution:The ratio of the magnitude of restitution impulse to the magnitude of

deformation impulse is known as co-efficient of restitution, denoted by the letter 'e'.Co-efficient of restitution=

impulse during restitution/ impulse during deformation

e=(v2-v1)/(u1-u2)Note:Law of conservation of momentum:m1u1 + m2u2 = m1v1 + m2v2

Newton's law of collision of elastic bodies:(v2 – v1) == e(ul - u2)

ie, velocity of separation =e *(velocity of approach)

1)A sphere of mass 1kg moving with a velocity 2 m/s impinges directly on a sphere of mass 2 kg at rest. If the first sphere comes to rest after the impact, find the velocity of the second sphere and the co-efficient of restitution.Solution:Given:m1 = lkg; u = 2 m/s v1 = 0 m2 = 2 kg ; u= 0 v2 =? e=?From law of conservation of momentum,m1u1 + m2u2 = m1v1 + m2v2

(1 *2) + (2*0) = (1 x 0) + (2 x v2)v2=1 m/s.

From Newton's law of collision of bodies,e=(v2-v1)/(u1-u2)(1 - 0) = e (2 - 0)e = 1/2 = 0.5

The velocity of the second sphere after impact is 1 m/s; and the co-efficient of restitution is 0.5.

2)A ball strikes centrally on another ball of mass twice the mass of first ball but moving with a velocity 1/7 th of the velocity of first ball and in the same direction. Show that, the first ball comes to rest after impact. The co-efficient of restitution between them is 3/4.

Solution:Given, Mass of second body Twice the mass of first body

Velocity of second body =1/7 x velocity of first bodye= ¾ ; both are moving in the same direction

Let, mass of first body, ml =mmass of second body, m2 =2mInitial velocity of first body, u1 = uInitial velocity of second body, u2 =1/7 x u=u/7v1 and v2 be the final velocities of the bodies, after impact.

Applying law of conservation of momentum,m1u1 + m2u2 = m1v1 + m2v2

Dividing by m on either side,u+2u/7=v1+v2

2v2 +v1 = 9u/7 …….(i)Applying newton's law of collision,(v2 – v1) =e(u1 - u2)

v2 – v1 == 9u/14 ..... (i)Solving the equations (i) and (ii) we get v1 and v2

Substitute v2=9u/14 in equation (ii)

Final velocity of first body after impact is zero.

Oblique Central Impact:

law of conservation of momentum

Newton’s Law of Collision:

1)A Ball of mass 500 grams, moving with a velocity of 1m/s impinges on a ball of mass 1 kg, moving with a velocity of O.75 m/s. At the time of impact, the velocities of the balls are parallel and inclined at 600 to the line joining their centres. Determine the velocities and directions of the balls after impact. Take, e = 0.6.Solution:Given: m1 =0.5 kg; ul =1 m/s ; α1 =60˚m2 =1kg; u2 == 0.75 m/s; α2 =60˚Calculate, v1, v2 , θ1 and θ2

The vertical components of velocities before and after impact are the same.u1 sinα1 =v1 sin Ɵ1

u2 sinα2 =v2 sin Ɵ2

u1 sinα1 =1x sin60=0.866 v1 sin Ɵ1=0.8661 …..(i)u2 sinα2 =0.75x sin 60= 0.649 v2 sin Ɵ2=0.649 …..(ii)Applying law of conservation of momentum along the line of impact,

(0.5 x 1 x cos 60) + (1 x 0.75 x cos 60) = (O.5v1 cosƟ1) + (1 x v2 cosƟ2)O.5v1 cosƟ1 + v2 cosƟ2 =0.625 ......(iii)Applying Newton's law of collision along the line of impact

=0.6 ( 1 x cos 60 - 0.75 cos 60)v2cosƟ1-v1cosƟ1 =0.75 ……(iv)

Solving the equations (iii) & (iv)eqn (iii) - eqn (iv) =1.5v1 cosƟ1 =0.55v1 cosƟ1 = 0.376 ……..(v)

tanƟ1= 2.359;Ɵ1=1/tan (2.359)=67˚Substituting the value of Ɵ1 in eqn (v)

From equation (iv),v2cosƟ1-v1cosƟ1=0.75 v2cosƟ1 =0.075 + 0.367=0.442

tanƟ2=1.468;Ɵ2=1/tan (1.468)Ɵ2=55.74˚.

Substituting Ɵ2 =55.74 ° in eqn (vi)v2 cosƟ2 = 0.442v2 COS 55.74° = 0.442v2 = 0.442/cos55.7v2=0.785m/s