Chapter 9, Problem 44A

To determine

The change in ball’s momentum and its change in velocity if a force of $186\text{N}$ acts on a $7.3\text{kg}$ bowling ball for $0.40\text{s}$ .

Answer to Problem 44A

  $74.4\text{kg}\cdot \text{m/s}$and $10.19\text{m/s}$ , respectively.

Explanation of Solution

Given:

The mass of the ball, $m=7.3\text{kg}$

The applied force on the ball, $F=186\text{N}$

The time for which the force is applied on the ball, $\Delta t=0.40\text{s}$

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 the above values in above formula, the change in the momentum of the ball will be,

  $\begin{array}{c}{p}_f-p_i=F\Delta t\\ \Delta p=\left(186\right)\left(0.40\right)\\ =74.4\text{kg}\cdot \text{m/s}\end{array}$

Now, using the formula of the momentum and the given value of the mass of the ball, the change in velocity of the ball is,

  $\begin{array}{c}\Delta p=m\Delta v\\ 74.4=7.3\Delta v\\ \Delta v=10.19\text{m/s}\end{array}$

Conclusion:

Thus, the change in momentum and the velocity of the ball is $74.4\text{kg}\cdot \text{m/s}$ and $10.19\text{m/s}$ , respectively.