Question

Q A bullet of mass m=20g pierces through a plate ofmass

M1=1kg then comes to rest inside a second plate ofmass
M2=2.98kg. It is found that the two platesinitially at
rest , now move with equal velocities.Find theprecentage loss in
the initial velocity of the bullet whern it isbetween M1
and M2 .Neglect any loss ofmaterial of the plates due to
action of bullet.

Answer 1

Percentage loss is 99.5%

Solution:

As per the question:

Mass of the bullet, m = 20 g = 0.020 kg

Mass of plate,
`M_(1) = 1\ kg`

Mass of the second plate,
`M_(2) = 2.98\ kg`

Now,

To calculate the percentage loss:

By using the principle of conservation of energy:

Since, the velocity with which the bullet is fired is
`v_(b)` and then after piercing through the plates the bullet gets embed in it and hence moves with the same velocity, v as that of the plates:

`mv_(b) = (M_(1) + M_(2) + m)v`

`v_(b)` = initial velocity of the bullet

v = final velocity of the combination

Thus by substituting the appropriate values in the eqn to find 'v':

`v = (0.020)/(1 + 2.98 + 0.020)v_(b) = 0.005v_(b)`

Now, the loss in the velocity of the bullet:

`v_(b) - 0.005v_(b) = 0.995v_(b)`

Percentage loss =
`(0.995v_(b))/(v_(b))* 100` = 99.5%