Question

A bullet of mass m moving horizontally with speed u hits a wooden block of mass M that is suspended from a massless string.The bullet gets lodged into the block and comes into halt.If the block-bullet combine swings to a maximum height h then how much of the initial kinetic energy of the bullet is lost in the collision?
Ans: mM u²/2(m+M)

Answer 1

The law of conservation of momentum gives

$$mu=(m+M)v$$

Hence, the initial velocity of block and bullet is

$$v=\frac{mu}{m+M}$$

The change of energy

$$(m+M)v^2/2-mu^2/2\\ =m^2u^2/(2(m+M))-mu^2/2=mMu^2/2(m+M)$$