Chapter 9, Problem 47A

To determine

The average force that stops the bullet if the officer fires a $6.0\text{g}$ bullet at $350\text{m/s}$ into a container that stops it in $1.8\text{ms}$ .

Answer to Problem 47A

  $1166.67\text{N}$

Explanation of Solution

Given:

The mass of the bullet, $m=6\text{g = 0}\text{.006}\text{kg}$

Initial velocity of the bullet, $v_i=350\text{m/s}$

Since, the bullet stops, thus final velocity of the bullet, $v_f=0\text{m/s}$

Time taken, $\Delta t=1.8\text{ms = 0}\text{.0018 s}$ .

Formula used:

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

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Using all the given values in the formula mentioned, the average force on the bullet will be,

  $\begin{array}{c}F\Delta t=m\left(v_f-v_i\right)\\ F\left(0.0018\right)=0.006\left(0-350\right)\\ F=-1166.67\text{N}\end{array}$

Negative sign denotes that this force acts in opposite direction of the motion.

Conclusion:

Thus, the magnitude of the average force applied on the bullet is $1166.67\text{N}$ .