Question

A 5.20 gbullet moving at 672 m/sstrikes a 700 gwooden block at rest on a frictionless surface. The bullet emerges, travelling in the same direction with its speed reduced to 428 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet–block centre of mass?

Short answer

1. The resulting speed of the block is,$v_2=1.81m/s$
2. The speed of the bullet-block center of mass is,$v_{com}=4.96m/s$

Step-by-step solution

Step 1: Given Data

The mass of the bullet is,$m_{bullet}=5.20g=0.0052kg$.

The initial velocity of the bullet is,$\overrightarrow{v_i}=672\mathrm{m}/\mathrm{s}$.

The mass of the wooden block is,$m_{block}=700g=0.700kg$.

The reduced speed of the bullet is, ${\overrightarrow{v}}_1=428m/s$.

Determining the concept

By using conservation of momentum, we can find the resulting speed of the block and the speed of the bullet-block center of mass.

Formulas are as follow:

The conservation of momentum,$m_{bullet}\overrightarrow{v_i}=m_{bullet}\overrightarrow{v_1}+m_{block}\overrightarrow{v_2}$

The velocity of the center of mass,
$\overrightarrow{v_{com}}=\frac{m_{bullet}\overrightarrow{v_i}}{m_{bullet}+m_{block}}$

Where, are masses of bullet and block, role="math" localid="1661487619547" $\overrightarrow{v_{com}},\overrightarrow{v_i},\overrightarrow{v_1},\overrightarrow{v_2}$are velocity vectors.

(a) Determining the resulting speed of the block

Choosealong the initial direction of motion and apply momentum conservation,

$$\begin{gathered} m_{bullet}\overrightarrow{v_i}=m_{bullet}\overrightarrow{v_1}+m_{block}\overrightarrow{v_2} \\ 0.0052\times 672=0.0052\times 428+0.700\times \overrightarrow{v_2} \\ \overrightarrow{v_2}=\frac{\left(0.0052\times 672\right)-\left(0.0052\times 428\right)}{0.700} \\ \overrightarrow{v_2}=1.81m/s \end{gathered}$$

Hence, the resulting speed of the block is 1.81 m/s.

(b) Determining the speed of the bullet-block center of mass

It is a consequence of momentum conservation that the velocity of the center of mass is unchanged by the collision. Choosing to evaluate it before the collision,

$$\begin{gathered} {\overrightarrow{v}}_{com}=\frac{m_{bullet}\overrightarrow{v_i}}{m_{bullet}+m_{block}} \\ {\overrightarrow{v}}_{com}=\frac{0.0052\times 672}{0.0052+0.700} \\ {\overrightarrow{v}}_{com}=4.96m/s \end{gathered}$$

$$\begin{gathered} \overrightarrow{v_{com}}=\frac{m_{bullet}\overrightarrow{v_i}}{m_{bullet}+m_{block}} \\ \overrightarrow{v_{com}}=\frac{0.0052\times 672}{0.0052+0.700} \\ \overrightarrow{v_{com}}=4.96m/s \end{gathered}$$

Hence,the speed of the bullet-block center of mass is 4.96 m/s.

Therefore, by using conservation of momentum, the resulting speed and speed of a center of mass of the bullet block system can be found.