A bullet mass(b) .008 kg is fired (and embedded) into a block mass(B) 0.992 kg on a horizontal frictionless surface. The spring is compressed 0.15 m on impact. The spring's constant is 300 N/m.
Q: What is the block's velocity after impact
Q: What is the initial speed of the bullet?
I want to say that this is a completely inelastic collision since the two masses have the same final velocity (2)? so in that case.... (k2/k1) = m(b)/[(m(b)+m(B)]
m(b)v(b) + m(B)v(B) = [(m(b)+m(B)]v2
and U(el) ...potential energy of an elastic... is 1/2kx^2
W(el) ....work done by spring.... is -(change in potential energy).
The Attempt at a Solution
Im not sure what to do with so many unknowns. Also not sure if I need to use the work done by a spring or the potential energy of a spring. Any help to get me in the right direction is much appreciated.