A $20 k g$ bullet pierces through a plate of mass $M_{1} = 1 k g$ and then comes to rest inside a second plate of mass $M_{2} = 2.98 k g$ as shown in the figure. It is found that the two plates initially at rest and 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).

6

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Solution

The correct option is D $6$ %
Let initial velocity of bullet $= V_{1} ㎧$
Velocity with which each plate moves $= V_{2} ㎧$
According to the law of conservation of momentum the initial momentum of the bullet is equal to the sum of the final momentum of the second plate including the bullet
$∴ m V_{1} = M_{1} V_{2} + (M_{2} + m) V_{2}$
$20 V_{1} = 1 \times V_{2} + (2.98 + 20) V_{2}$
$20 V_{1} = V_{2} + 22.98 V_{2}$
$V_{1} = \frac{23.98}{20} V_{2}$
$V_{1} = 1.199 V_{2} \to 1$
Let the velocity of bullet when it comes out of first plate $= V_{3}$
The momentum of the bullet the first and second, a plate is equal to the sum of the momentum of the second plate and the bullet.
$20 V_{3} = (20 + 2.980 V_{2})$
$20 V_{3} = 22.98 V_{2}$
$∴ V_{3} = 1.149 V_{2} \to 2$
Loss Percentage in the initial velocity of the bullet when it is moving between $m_{1} & m_{2}$ is expressed as the following
Loss % $= \frac{V_{1} - V_{3}}{V_{1}} \times 100$
Loss % $= \frac{1.199 V_{2} - 1.149 V_{2}}{1.199 V_{2}} \times 100$
Loss % $= \frac{0.05}{1.199} \times 100$
$= 5.995$ %

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