Chapter 9.2, Problem 21PP

To determine

The speed of the ball when the bullet bounces backward.

Answer to Problem 21PP

  $6.72\text{m/s}$

Explanation of Solution

Given:

From the previous problem, the mass of the bullet, $m_b=35.0\text{g}=0.035\text{kg}$

From the previous problem, the initial speed of the bullet $v_{bi}=475\text{m/s}$ .

The mass of the stationary steel ball, $m_B=2.5\text{kg}$

The bullet bounces backward after the collision at the speed, $v_{bf}=5.0\text{}\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

And, law of conservation of linear momentum: Momentum is conserved in a closed and isolated system and is given by,

  $p_i=p_f$

Calculation:

Consider the bullet and ball as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Now, since the ball was stationary initially, thus initial velocity of ball is,

  $v_{Bi}=0\text{m/s}$

Assume the backward direction as negative, thus the final speed of the bullet will be,

  $v_{bf}=-5.0\text{}\text{m/s}$

Let, the final speed of the ball be $v_{Bf}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_B{v}_{Bi}+m_b{v}_{bi}=m_B{v}_{Bf}+m_b{v}_{bf}\end{array}$

Using given values in above,

  $\begin{array}{c}\left(2.5\right)\left(0\right)+\left(0.035\right)\left(475\right)=\left(2.5\right)\left(v_{Bf}\right)+\left(0.035\right)\left(-5\right)\\ v_{Bf}=6.72\text{m/s}\end{array}$

Positive sign of the above velocity denotes that after the collision, the ball moves in the initial direction of the bullet’s motion.

Conclusion:

Thus, the speed of the ball when the bullet bounces backward is $6.72\text{m/s}$ .