A bullet with mass 4.91 g is fired horizontally into a 1.981-kg block attached to a horizontal spring. The spring has a constant 6.25 102 N/m and reaches a maximum compression of 6.38 cm.


a) find the initial speed of the bullet-block system.
b) find the speed of the bullet

1 answer

a) By "initial speed", they really mean "right after the bullet hits the block and gets embedded". Call that speed v. The bullet-block speed is zero before that.

Assume energy is conserved during spring compression.

(1/2)(M+m) v^2 = (1/2)k*X^2
v = sqrt [k/(M+m)]*X
= sqrt[625/1.985] s^-1 *0.0638 m
= 1.132 m/s

b) For the initial bullet speed V, apply conservation of momentum to the bullet-stopping process. Mechanical energy is NOT conserved then.
m*V = (m+M)*v

(0.00491 kg)* V = (1.985 kg)* v

V = 458 m/s