Question

A bullet of mass10 gstrikes a ballistic pendulum of mass 2.0 kg. The centre of mass of the pendulum rises a vertical distance of12 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet’s initial speed.

Short answer

The initial speed of the bullet is $v=3.0\times {10}^2\mathrm{m}/\mathrm{s}$

Step-by-step solution

Step 1: Given Data

The mass of the bullet is,$m=10\mathrm{g}=0.010\mathrm{kg}$.

The mass of the ballistic pendulum is,$M=2.0\mathrm{kg}$.

The center of mass of the pendulum rises a vertical distance, $h=12\mathrm{cm}=0.12\mathrm{m}$

Determining the concept

By using mechanical conservation of energy and linear momentum conservation, find the initial speed of the bullet. According tothe conservation of mechanical energy, if an isolated system is subject only to conservative forces, then the mechanical energy is constant. According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formulae are as follow:

The mechanical energy conservation,
$\frac{1}{2}\left(m+M\right)v^2=\left(m+M\right)gh$

The linear energy conservation,
$v=\frac{m}{m+M}v$

Where, m, Mare masses, v, Vare velocities, g is an acceleration due to gravity and h is height.

Determining the initial speed of the bullet

Applying mechanical energy conservation,

$\frac{1}{2}\left(m+M\right)V^2=\left(m+M\right)gh$

But, according to linear momentum,

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

Substituting,

$$\begin{gathered} v=\frac{\left(m+M\right)}{m}\sqrt{2gh} \\ v=\frac{\left(0.010+2.00\right)}{0.010}\times \sqrt{2\times 9.8\times 0.12} \\ v=3.1\times {10}^2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the initial speed of the bullet is$v=3.1\times {10}^2\mathrm{m}/\mathrm{s}$

Therefore, by using mechanical conservation of energy and linear momentum conservation, the velocity of the bullet can be found.