Chapter 9.1, Problem 16SSC

To determine

The arrow that will produce bigger impulse on target.

Answer to Problem 16SSC

The arrow that bounces off will produces a bigger impulse.

Explanation of Solution

Given:

The mass of the arrows and their velocities are same.

Formula Used:

The impulse and momentum are related by the following formula,

  $I=\Delta p$

The equation for momentum is

  $p=mv$

Calculation:

Consider two cases.

Case1:

In the first case a few arrows are stuck on the target .Therefore the final velocity of those arrows are zero.

The initial momentum of the arrow is $p_i=mv$

The final momentum of the arrow is $p_f=0$

The change in momentum of the arrow which is stuck on the target

  $\begin{array}{l}\Delta p=p_f-p_i\\ \Delta p=-mv\end{array}$

(Negative sign shows the direction of the change in momentum)

Impulse experienced is equal to the change in momentum.

Therefore the magnitude of impulse I = mv

Case 2:

In the second case, few arrows are bouncing back from the target. Since the velocity reverses, the final velocity is − v

The initial momentum of the arrow $p_i=mv$$$

The final momentum of the arrow $p_f=-mv$

The change in momentum of the arrow which is bounced off

  $\begin{array}{l}\Delta p=p_f-p_i=-mv-mv\\ \Delta p=-2mv\end{array}$

Therefore the magnitude of impulse I = 2mv

Conclusion Therefore, the arrow which bounces back from the target produces a bigger impulse.