A bullet of mass 10g plerces through a plane A of mass 500g and then gets embedded into a second plate B of mass 1.49kg as shown in the figure. Initially, the two plates A and B are at rest and move with same velocity after collision. The percentage loss in the initial kinetic energy of the bullet, when it is between the plates A and B is ______(Neglect any loss of material of the plates during the collision)

Options

A. 25
B. 56.25
C. 43.75
D. 75

Answer

C

Explanation

Let `v_(1)` be the initial velocity of bullet and `v_(2)` be the velocity with which each plate moves. So, applying law of conservation of momentum,
`mv_(1) = M_(1) v_(2) (M_(2) + m) v_(2)`
`0.01v_(1) = 0.5v_(2) + (1.49 + 0.01) v_(2)`
`v_(1) = (2)/(0.01) v_(2) = 200v_(2)`
Let `v_(3)` be the velocity of bullet, when it comes out of plate M.
So, momentum of bullet between plate `M_(1) and M_(2)` = Sum of momentum of plate `M_(2)` and bullet.
`mv_(3) = (M_(2) + m) v_(2)`
`0.01v_(3) = (1.49 + 0.01) v_(2) = 1.5v_(2)`
or `v_(3) = 150v_(2)` ..(ii)
% loss in KE `=((1//2)mv_(1)^(2) - (1//2)mv_(3)^(3))/((1//2) mv_(1)^(2)) xx 100`
`= (v_(1)^(2) - v_(3)^(2))/(v_(1)^(2)) xx 100`
`={1-((v_(3))/(v_(1)))^(2)} xx 100`
`={1-((150)/(200))^(2)} xx 100`
`=(1-(9)/(16)) xx 100 = (7)/(16) xx 100 = (7)/(4) xx 25`
= 43.75%