First slide

The steps in collision

Question

A 20 g bullet pierces through a plate of mass $M_{1}$ = 1 kg and then comes to rest inside a second plate of mass $M_{2}$ = 2.98 kg as shown in Fig. It is found that the two plates, initially at rest, now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between $M_{1}$ and $M_{2}$ . Neglect any loss of material of the plates due to the action of bullet.

Moderate

A

50%
B

25%
C

100%
D

75%

Solution

Apply conservation of momentum after piercing M1 :
$m u = M_{1} v + m u_{1}$
After it enters into M2, finally bullet also moves with velocity v, so
$m u_{1} = (M_{2} + m) v$
hence, $m u = (M_{1} + M_{2} + m) v$
Required percentage loss, $= \frac{u - u_{1}}{u} \times 100 = \frac{m (u - u_{1})}{m u} \times 100$ $= \frac{M_{1} v}{(M_{1} + M_{2} + m) v} \times 100$ $= \frac{1}{1 + 2.98 + 0.02} \times 100 = 25 %$

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