A `20 ` gm 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 , it is found that the two plates initily atrest , now move with equal velocity . Find the percentage loss in the initial velocity of the bullet when it is between `M` and `M_(2)` neglet any loss of material of the pletes due to the action of the bullet

Answer

B

Explanation

Let `v` be the velocity of bollet befor striking `A` , appling consveration of linear momention for the system of bullet and plate `A` , we get
`(0.02v = 0.02v_(1) + 1 xx v_(2)`

Again applying conservation for collision at `B`
`0.02v_(1) = (2.98 - 0.02)v_(2) = 3v_(2)`
`rArr v_(2) = (0.02v_(2))/(3)` ...(i)
from (i) and (ii)
`0.02v = 0.02v_(1) + (0.02v_(1))/(3) , v = (4)/(3) v_(1) rArr (v)/(v_(1)) = (4)/(5) `
`(v_(1))/(v) = (3)/(4) rArr (v_(1))/(v) = 1 - (3)/(4) = (1)/(4) = 0.25 rArr (v - v_(1))/(v) = 0.25`
` :. % `loss is velocity `= (1)/(2) xx 100 = 25 %`