A bullet of mass 9 g moving with initial speed 300 m/s passes through a block of mass 4 kg, initially at rest on a frictionless, horizontal surface. The block is connected to a spring of force constant 904 N/m. If the block moves 2 cm to the right after the bullet passed through it, find the speed at which the bullet emerges from the block and the magnitude of the energy lost in the collision.
Conservation of momentum.
The Attempt at a Solution
Well initial mometum is just (.009)(300), and it must equal (.009)(Vf)=(4)(Vblk)
to solve for the block velocity immediately after the bullet passes, I used Vf = sqrt(2ad). I said acceleration was (904)(.02)/(4)...(may be where I went wrong). I took that velocity and plugged it back into the original momentum equation and got velocity of the bullet to be 111 m/s, which was wrong.