Question

A bullet of mass 12.4 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.15 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 4.00 m

Answer 1

The question does not mention , what is to be found. from the question, the quantity that can be calculated is the initial speed of the bullet by applying conservation of momemtum.

bullent momentum =mv

                       =(12.4/1000)v kg.m/s

final momentum of block and bullet= (1.15+(12.4/1000))X4 kg.m/s

initial momentum of block is zro.

(12.4/1000)v kg.m/s=(1.15+(12.4/1000))X4 kg.m/s

   v= 374.96 m/s