A2.25g 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.

 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

Ibelieve the answer would be 7, 13, 23

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