Question

A bullet of mass 7 gram is fired into a block of metal weighing 7 kg the block is free to move after the impact the velocity of the bullet and the block is 0.7ms what is initial velocity of the bullet

Answer 1

Apply the law of conservation of momentum
It says that,
Momentum of system before collision = momentum of the system after collision
(so the momentum get conserved along the collision)
momentum of particle = mass × velocity
say, initial velocity of the bullet is V
therefore,
momentum of bullet before collision = 7gm×V
and momentum of block before collision = 7000gm×0=0
Now
momentum of block as well as bullet = total mass × velocity
total mass= mass of bullet + mass of block
=( 7+7000)gm
therefore momentum after collision will be 7007×0.7 = 4904.9
Now according to low of momentum we will equate both initial as well as final momentum
7×V=4904.9

$v = \frac{4904.9}{7}$
V=700.7m/s
Therefore initial velocity of bullet =700.7m/s