A 5.0-g bullet traveling horizontally at an unknown speed hits and embeds itself in a 0.195-kg block resting on a frictionless table. The block slides into and compresses a 180-N/m spring a distance of 5.0×10^−2 m before stopping the block and bullet.

Answer 1

energy in the block and bullet=spring energy
1/2 (.005+.195)V=1/2 180*(5E-2)^2

solve for V, the velocity of the block with bullet.

Now consider the bullet and block impact, conservation of momentum applies.

.05Vb=(.195+.005)Vabove
solve for the velocity of the bullet, Vb