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

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