Question

A bullet of mass $20g$ moving with velocity $v$ is embedded into a block of mass $10kg$. As a result of this collision, the block and the bullet rise to a height of $25cm$ from the equilibrium position. Find the original velocity $v$ of the bullet.

Answer 1

Using momentum conservation

$mv=(M+m)v^1{v}^1=\frac{mv}{M+m}\Delta K.EatA=\Delta P.EatB\frac{1}{2}(M+m)v^{12}=(M+m)gh{v}^{12}=2gh(\frac{mv}{M+m}{)}^2=2\times 10\times 0.25(\frac{20\times {10}^{-3}v}{10+20\times {10}^{-3}}{)}^2=5\frac{0.020v}{10.02}=\sqrt{5}v=1120m/s$