Question

A bullet with mass $m$ hits a ballistic pendulum, having length $L$ and mass $M$, and gets embedded in it. After the bullet hits the pendulum, it swings up, with the bullet embedded into it, and reaches to a maximum height $h$ with respect to its initial position. Determine the initial velocity of the bullet.

• A. $\left(\frac{M+m}{2m}\right)\sqrt{2gh}$
• B. $\left(\frac{M-m}{2m}\right)\sqrt{2gh}$
• C. $\left(\frac{M+m}{m}\right)\sqrt{2gh}$
• D. $\left(\frac{M-m}{m}\right)\sqrt{2gh}$

Answer 1

The correct option is C $\left(\frac{M+m}{m}\right)\sqrt{2gh}$

According to the conservation of energy principle for $(M+m)$ system, in the interval after collision and reaching to the maximum height by pendulum,

$\frac{1}{2}(M+m)v^2=(M+m)gh$

$\therefore v=\sqrt{2gh}...\left(1\right)$

According to the conservation of momentum principle, in the interval before and after the collision of bullet with pendulum,

$m\left(u\right)+M\left(0\right)=(M+m)v$

$\therefore u=\frac{(M+m)v}{m}=\left(\frac{M+m}{m}\right)\sqrt{2gh}\left[\text{From}1\right]$

Hence, option $\left(C\right)$ is the correct answer.