A 50g bullet traveling horizontally at an unknown speed hits

A 5.0g bullet traveling horizontally at an unknown speed hits and embeds itself in a 0.195kg block resting on a frictionless table. The block slides into and compresses a 180 N/m spring a distance of 7.0*10^-2 m before stopping the block and bullet.

Determine the initial speed of the bullet. Express answer to two significant figures.

Solution

after collision, Applying energy conservation,

(M + m) v^2 / 2 = k x^2 / 2

(0.195 + 0.005) (v^2) = (180)(0.07^2)

v = 2.1 m/s

Now applying momentum conservation for the collision,

m v0 + (M x 0) = (m + M) v

0.005 v0 = (0.195 + 0.005)(2.1)

v0 = 84 m/s .....Ans