Question

a. A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance $d$across a horizontal surface. The coefficient of kinetic friction is ${\mu }_k$. Find an expression for the bullet’s speed $v_{bullet}$

b. What is the speed of a $10$g bullet that, when fired into a $10$kg stationary woodblock, causes the block to slide 5.0 cm across a wood table?

Short answer

a. The expression of bullet's speed,

$v_{bullet}=\frac{m_{bullet}+m_{block}}{m_{bullet}}\sqrt{2{\mu }_{k}gd}$

b. The speed of the bullet, $v_{bullet}=443m/s$

Step-by-step solution

Part (a). Step 1. Given information

Mass of bullet is $10g$and block is $10kg$is given along with the initial velocities.

Step 2. Analyze and apply

Using the law of conservation of momentum we have,

$$\begin{gathered} m_{bullet}{v}_{bullet}+m_{block}{v}_{block}=\left(m_{bullet+block}\right)v_f \\ {v}_{bullet}=\frac{\left(m_{bullet+block}\right)v}{m_{bullet}} \end{gathered}$$

For the velocity after collision, we use the equation of motion,

$$\begin{gathered} {v_{s,y}}^2={v_f}^2+2ad \\ {v}_f=\sqrt{-2ad}=\sqrt{2{\mu }_{k}gd} \end{gathered}$$

Step 3. Substitute and solve

On substitution of the values we have,$v_{bullet}=\frac{m_{bullet}+m_{block}}{m_{bullet}}\sqrt{2{\mu }_{k}gd}$

Part (b). Step 1. Substituting the values of the known parameter

On substituting the values of the known parameter we get the value of the velocity,

$v_{bullet}=\frac{0.01kg+10kg}{0.01kg}\sqrt{2·0.2·9.81m/s^2·0.05m}=443m/s$