Chapter 9, Problem 68A

To determine

Velocity of concrete ball.

Answer to Problem 68A

  $v_{2f}=0.367\text{ m/s}$

Explanation of Solution

Given:

Mass of the rubber bullet, $m_1=12\text{ g = 0}\text{.012 kg}$

Velocity of rubber bullet before collision, $v_{1i}=150\text{ m/s}$

Mass of the concrete block, $m_2=8.5\text{ kg}$

Velocity of rubber bullet after collision, $v_{1f}=-100\text{ m/s}$

Formula used:

Momentum of any object is given by,

  $P=mv$

Here, P is momentum, m is mass and v is velocity

If no external force is applied on the object,then momentum is conserved.

  $P_i=P_f$

Calculation:

As the concrete block is stationary before the collision, therefore, velocity of concrete block before the collision, $v_{2i}=0\text{ m/s}$

According to conservation of momentum, initial momentum will be equal to final momentum.

Therefore,

  $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$

Now, substitute the values of $m_1,v_{1i},m_2,v_{2i}$ and $v_{1f}$ and solve.

  $\begin{array}{c}0.012\times 150+8.5\times 0=0.012\times -110+8.5\times v_{2f}\\ 8.5v_{2f}-1.32=1.8\\ 8.5v_{2f}=3.12\\ v_{2f}=\frac{3.12}{8.5}\\ =0.367\text{ m/s}\end{array}$

Conclusion:

Therefore, velocity of concrete ball will be 0.367 m/s after the collision.