Question

A 20g bullet pierces through a plate of mass M1=1 kg and then comes to rest inside a second plate of mass M2= 2.98kg. It is found that the 2 plates are initially at rest now move with equal velocities. Find the percentage loss in initial velocity of bulletwhen it is betwen M1 and M2. Neglect any loss of material in plate due to action of plate. a)50% b)100% c)25% d)70%

Answer 1

Let initial velocity of bullet $V_1$ m/s
velocity with which each plate moves $V_2$ m/s

according to law of conservation of linear momentum,
mV1 = M1V1 + (M2 + m)V2
0.02V1 = 1 × V1 + (2.98 + 0.02)V2
0.02V1 = V1 + 3V2 = 4V2
V1 =200V2 .......(1)

Let $V_3$ is the velocity of bullet when it comes out of the first plate.
now, according to law of conservation of linear momentum,
mV3 = mV2 + M2V2
0.02V3 = (0.02 + 2.98)V2
V3 = 150V2.........(2)

now, loss percentage in initial velocity of bullet when it is moving between M1 and M2 is expressed as follow :
% loss = (V1 - V3)/V1 × 100
= (200V2 - 150V2)/200V2 × 100
= 50/200 × 100
= 25 %

Answer 2

Let initial velocity of bullet  Am/s

velocity with which each plate moves  m/s

according to law of conservation of linear momentum,

mA = MA + (M2 + m)B

0.02A = 1 × A + (2.98 + 0.02)A

0.02A = B + 3B = 4B

A =200B .......(1)

Let C be the velocity of bullet when it comes out of the first plate.

now, according to law of conservation of linear momentum,

mC = mB + M2B

0.02C = (0.02 + 2.98)B

C = 150B.........(2)

now, loss percentage in initial velocity of bullet when it is moving between M1 and M2 is expressed as follow :

% loss = (A - C)/A × 100

= (200B - 150B)/200B × 100

= 50/200 × 100

= 25 %

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