Question

A bullet with mass 120g hits a ballistic pendulum with mass 30kg and lodges in it. When the bullet hits the pendulum it swings up reaching the height of 0.25m. Neglecting air resistance, calculate the speed of the bullet. 

Answer 1

1. $mv=(m+M)u\to u=mv/(M+m)$

2. $(M+m)u^2/2=(M+m)gh\to (M+m)(\frac{mv}{M+m})^2/2=(M+m)gh\to$

$m^2v^2=2(M+m)^2gh\to v=\sqrt{\frac{2(M+m)^2gh}{m^2}}=\frac{M+m}{m}\cdot\sqrt{2gh}=$

$=\frac{30+0.12}{0.12}\cdot\sqrt{2\cdot9.8\cdot 0.25}=555.6\ (m/s)$ . Answer