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.

Answer

25

Explanation

Let u = intial velocity of bullet
`v_1` = velocity of bullet after crossing the first plate
v = velocity of each plate
Applying law of conservation of linear momentum to system of bullet and first plate.
`mu + M_1 xx 0 = M_1 v + mv_1`
or `0.02 u = v + 0.02 v_1 " ".....(I)`
Applying the law of conservation of linear momentum to system of bullet and second plate,
`mv_1 + M_2 xx 0 = (m + M_2)v`
or `0.02 v_1 = (0.02 + 2.98) v = 3v " "....(ii)`
Putting equation (ii) in equation (i), we have ,
`0.02 u = (0.02 v_1)/3 + 0.02 v_1`
`0.02 u = 0.02 v_1(1/3 + 1) or u/(v_1) = 4/3 or (v_1)/u = 3/4 or (u - v_1)/(u) xx 100 = 25%` .