Question

A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it. If the centre of the pendulum rises by a distance of 10 cm, find the speed of the bullet.

Answer 1

Given:
Mass of bullet, m = 25 g = 0.025 kg
Mass of ballistic pendulum, M = 5 kg
Vertical displacement, h = 10 cm = 0.1 m

Let the bullet strikes the pendulum with a velocity u.
Let the final velocity be v.

Using the law of conservation of linear momentum, we can write:
$$\begin{gathered} mu=(M+m)v \\ \Rightarrow v=\frac{m}{(M+m)}u \\ \Rightarrow v=\frac{0.25}{5.025}\times u=\frac{u}{201} \end{gathered}$$

Applying the law of conservation of energy, we get:
$$\begin{gathered} \left(\frac{1}{2}\right)(M+m)v^2=(M+m)gh \\ \Rightarrow \frac{u^2}{(201{)}^2}=2\times 10\times 0.1 \\ \Rightarrow u=201\times \sqrt{2}=280\mathrm{m}/\mathrm{s} \end{gathered}$$

The bullet strikes the pendulum with a velocity of 280 m/s.