Question

A 2.25g bullet imbeds itself in a 1.50-kg block, which is attatched to a spring of force constant 785N/m. If the maximum compression of the spring is 5.88cm, find (a) the initial speed of the bullet and (b) the time for the bullet block system to come to rest.

Answer 1

 first find the block's velocity after the collision occurs. 
1/2 (1.5 + 0.00225) V^2 = 1/2 (785)(0.0588)^2 
so V = 1.344 m/s 
using conservation of momentum we get: 
(mv)bullet + 0 = (1.50225) (1.344) 
0.00225 v = 2.02 
v = 2.02 / 0.00225 = 897.3 m/s

Using distance = average velocity* time t = 0.0588 / (1.344/2) = 0.0875 s