91Ó°ÊÓ

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

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

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

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

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,$m_{bullet}\stackrel{→}{v_i}=m_{bullet}\stackrel{→}{v_1}+m_{block}\stackrel{→}{v_2}$

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

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

Choosealong the initial direction of motion and apply momentum conservation,

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

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

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,

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

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

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.