## If a ball is dropped from rest, it bounces from the floor repeatedly. The coefficient of restitution….

Q: If a ball is dropped from rest, it bounces from the floor repeatedly. The coefficient of restitution is 0.5 and the
speed just before the first bounce is 5 m/s . The total time taken by the ball to come to rest finally is :

(A) 1.5 s

(B) 1 s

(C) 0.5 s

(D) 0.25 s

Ans: (A)

Solution: v = 0 + gt
t = 0.5 sec
After first collision :
Speed becomes 5 (0.5) = 2.5 m/s
t1 = 2 (0.25) = 0.5

t2 = 2 (0.125) = 0.25

t3 = 0.125 and so on
[where ti is the time taken to complete the ith to and fro motion after collision]

Total time = 0.5 + [0.5 + 0.25 + 0.125 + …]

$= 0.5 + \frac{0.5}{1-0.5}$

(Since above is a G.P. with a = 0.5 and r = 0.5)

= 0.5 + 1 = 1.5

## Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side smooth surface…..

Q: Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side smooth surface. A bullet of mass m = 20 gm strikes the block M1 and pierces through it, then strikes the second block and sticks to it. Consequently both the blocks move with equal velocities. Find the percentage change in speed of the bullet when it escapes from the first block.

Sol: Conservation of linear momentum of the system yields

$\displaystyle m v_0 = M_1 v + m v’ = (M_2 + m ) v$

$\displaystyle v = \frac{m v_0}{M_2 + m}$

and mv’ = m v0 – M1 v

$\displaystyle m v’ = m v_0 – \frac{M_1 m v_0}{M_2 + m}$

$\displaystyle m v’ = m v_0(1 – \frac{M_1}{M_2 + m})$

Now find , $\displaystyle \frac{|v’-v_0|}{v_0} \times 100 = \frac{M_2 – M_1 + m}{M_2 + m} \times 100$

## A cannon and a supply of cannon balls are inside a sealed railroad car. The cannon fires to the right, …

Q: A cannon and a supply of cannon balls are inside a sealed railroad car. The cannon fires to the right, the car recoils to the left. The cannon balls remain in the car after hitting the far wall. Show that no matter how the cannon balls are fired, the railroad car cannot travel more than l , assuming it starts from rest .

Sol: The speed of the car = vc The speed of the balls = vb

m & M are the mass of each ball and the rest of the given system respectively.

$\displaystyle \vec{m v_b} + \vec{M v_c} = 0$ ; Since $F_{ext} = 0$

$\displaystyle m (\vec{v_{bc}} + \vec{v_c}) + \vec{M v_c} = 0$

$\displaystyle \vec{v_c} = – \frac{\vec{m v_{bc}}}{M+m}$ ; ; vbc =velocity of ball relative to car.

$\displaystyle x_c = \int_{0}^{t}v_c dt$

$\displaystyle x_c = \frac{m}{M+m} \int_{0}^{t}v_{bc} dt$

$\displaystyle x_c = \frac{m l}{M+m}$

⇒ xc < l

Where xc distance travelled by the car and

$\displaystyle \int_{0}^{t}v_{bc} dt$ = total distance travelled by each ball relative to the car = l

## Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane…

Q: Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane. There exists friction between A and B. If a bullet of mass m hits the lower block with a horizontal velocity v and gets embedded into it. Find the work done by friction between A and B.

Sol: Let the velocity of lower block just after collision be v1,

Applying COLM between bullet and lower block as impulse due to friction is negligible.

$\displaystyle m v = (M + m)v_1$

$\displaystyle v_1 = \frac{m v}{M+m}$

After collision frictional force acts on both and will act till velocities of both blocks become equal.

$\displaystyle (M+m)v_1 = (2M+m)v_2$

$\displaystyle v_2 = \frac{(M+m)v_1}{2M+m}$

Applying work energy theorem ,

$\displaystyle W_{friction }= K_f -K_i$

$\displaystyle W_{friction }= \frac{1}{2}(2M+m)v_2^2 – \frac{1}{2}(M+m)v_1^2$

$\displaystyle W_{friction }= -\frac{M m^2 v^2}{2(m+2M)(m+M)}$

## A block of mass M is hanging from a rigid support by an in-extensible light string…..

Q: A block of mass M is hanging from a rigid support by an in-extensible light string. A ball of mass m hits it with a vertical velocity v at its bottom. Find the change in momentum of the ball assuming inelastic collision.

Sol: Let the velocity of combined mass of bullet and block just after collision be v1
Applying COLM,
$\displaystyle m v = (M + m) v_1$

$\displaystyle v_1 = \frac{m v}{M+m}$

Change in the momentum of bullet = Pf – Pi = mv1 – mv

$\displaystyle = \frac{m^2 v}{M+m} – m v$

$\displaystyle = – \frac{M m }{M+m}v$

## A small empty bucket of mass M is attached to a long inextensible cord of length l . The bucket is released from rest ….

Q: A small empty bucket of mass M is attached to a long inextensible cord of length l . The bucket is released from rest when the cord is in a horizontal position. In its lowest position the bucket scoops up a mass m of water, what is the height of the swing above the lowest position ?

Sol: Let the bucket of the mass reach the lowest position with a speed v. As it scoops up water of mass m, at the lowest position, its speed decreases to v’ The speed of the bucket just after scooping of water is given by conserving momentum of the system just before and after scooping of water we obtain
Mv = (M + m) v’

$\displaystyle v’ = \frac{M v}{M+m}$

Applying Energy conservation of bucket ,

$\displaystyle \frac{1}{2}M v^2 = M g l$

$\displaystyle v = \sqrt{2 g l}$

Height $\displaystyle H = \frac{v’^2}{2 g}$

$\displaystyle H = (\frac{M}{M+m})^2 l$

## A block of mass m moving with a velocity v hits a light spring of stiffness K attached rigidly to a stationary

Q: A block of mass m moving with a velocity v hits a light spring of stiffness K attached rigidly to a stationary sledge of mass M. Neglecting friction between all contacting surface, find the maximum compression of the spring.

Sol: The velocity of the combination at the time of maximum compression of the spring , obtained by conserving the momentum of the system.

(M+ m)v’ = m v

$v’ = \frac{m v}{M+m}$

Let x0 = Max compression of the spring

$\displaystyle \Delta K.E + \Delta P.E = 0$

$\displaystyle \frac{1}{2}(M+m) v’^2 – \frac{1}{2}m v^2 + \frac{1}{2}k x_0^2 = 0$

$\displaystyle \frac{1}{2}(M+m) (\frac{m v}{M+m})^2 – \frac{1}{2}m v^2 + \frac{1}{2}k x_0^2 = 0$

$\displaystyle x_0 = \sqrt{\frac{M m}{(M+m)k}} v_0$

## A ball collides with an inclined plane of inclination θ after falling through a distance h….

Q: A ball collides with an inclined plane of inclination θ after falling through a distance h. If it moves horizontally just after the impact, find the coefficient of restitution.

Sol: Let the ball make an angle α with perpendicular to the incline plane.

From geometry

θ + α = π/2 ….(i)

Momentum along plane will remain conserve

mv1 sinθ = mv2sinα

v1 sinθ = v2sinα …(ii)

v2 = v1 tanθ …(iii)

Vertical to the plane

$\displaystyle e = \frac{v_2 cos\alpha}{-v_1 cos\theta}$

$\displaystyle e = \frac{v_2 cos\alpha}{v_1 cos\theta}$

$\displaystyle e = \frac{v_1 tan\theta cos(\pi/2 – \theta)}{v_1 cos\theta}$

$\displaystyle e = tan^2 \theta$

## A steel ball with a mass of m = 20 g falls from a height of h1 = 1 m onto a steel plate and rebounds to a height of h2 ….

Q: A steel ball with a mass of m = 20 g falls from a height of h1 = 1 m onto a steel plate and rebounds to a height of h2 = 81 cm. Find: the impulse of the force received by the plate during the impact.

Sol: Impulse = Change in momentum

$\displaystyle \vec{J} = \vec{P_f} – \vec{P_i}$

$\displaystyle \vec{J} = \vec{m v_2} – \vec{m v_1}$

$\displaystyle J = m v_2 + m v_1$ ; Since v1 & v2 are in opposite direction

$\displaystyle J = m ( v_2 + v_1)$

$\displaystyle J = m ( \sqrt{2 g h_2} + \sqrt{2 g h_1})$

## A skater of mass m standing on ice throws a stone of mass M with a velocity of v m/s in a horizontal direction….

Q: A skater of mass m standing on ice throws a stone of mass M with a velocity of v m/s in a horizontal direction. Find the distance over which the skater will move back if the coefficient of friction between the skaters and the ice is μ.

Sol: Conservation of momentum yields
$\displaystyle 0 = M \vec{v} + m \vec{u}$ ; where u = recoil velocity of the skater

$\displaystyle 0 = M v – m u$

$\displaystyle u = \frac{M v}{m}$

since the skater stops after covering a distance S

$\displaystyle u^2 = 2 a S$

$\displaystyle S = \frac{u^2}{2 a}$

$\displaystyle S = \frac{u^2}{2 \mu g}$ ; Where a = μ g