Chapter 8, Problem 59A

Assume an 8-kg bowling ball moving at 2 m/s bounces off a spring at the same speed that it had before bouncing.

a. What is its momentum of recoil?

b. What is its change in momentum? (Hint: What is the change in temperature when something goes from $\text{1°}$ to $-\text{1°}$?)

c. If the interaction with the spring occurs in 0.5 s, calculate the average force the spring exerts on it.

To determine

To Calculate: The momentum of recoil of bouncing ball.

Answer to Problem 59A

  $-16\text{kg}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Explanation of Solution

Given information:

The mass of bowling ball is 8kg.

Its velocity is $2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ and its recoil velocity is $\left(\text{-2}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)$ .

Formula used:

The recoil momentum is:

  $p=mv$

Where, m is the mass and v is the velocity.

Calculation:

By substituting the given values in formula, then the recoil momentum is,

  $\begin{array}{l}p=mv\\ =8\text{kg}\times \left(-2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\\ =-16\text{kg}\cdot \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Conclusion:

Thus, the recoil momentum is $\text{-16}\text{kg}\cdot \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ .

To determine

To Calculate: The change in momentum of bouncing ball.

Answer to Problem 59A

  $-32\text{N·s}$

Explanation of Solution

Given information:

The weight of bowling ball is 8kg.

Its velocity is $2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ and its recoil velocity is $\left(-2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)$ .

The calculated recoil momentum is $-16\text{kg}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ .

Formula used:

The change in momentum is given as,

  $\begin{array}{c}\Delta p=p_0-p\\ =mv-m{v}_0\\ =m\left(v-v_0\right)\end{array}$

Where, $p_0\text{and}p$ are the momentum initial and final momentum respectively, m is the mass and $v\text{and}{v}_0$ are the final and initial velocities.

Calculation:

By substituting the given values in the formula then the change in momentum is given as,

  $\begin{array}{c}\Delta p=p_0-p\\ =mv-m{v}_0\\ =m\left(v-v_0\right)\\ =8\text{kg}\left(-2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.-2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\\ =-32\text{N}\cdot \text{s}\end{array}$

Conclusion:

Thus, the change in momentum is $-32\text{N·s}$ .

To determine

To Calculate: The average force that exerts on the spring.

Answer to Problem 59A

The average force is $-64\text{N}$ .

Explanation of Solution

Given information:

The mass of bowling ball is 8kg.

Its velocity is $2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ and its recoil velocity is $\left(-2\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)$ .

The calculated recoil momentum is $-16\text{kg}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ .

The calculates change in momentum is $-32\text{N·s}$ .

Formula used:

The average force is:

  $F=\frac{\Delta p}{t}$

Where, t is the time period and $\Delta p$ is the change in momentum.

Calculation:

By substituting the values in the above formula, so the average force is calculated as,

  $\begin{array}{l}F=\frac{\Delta p}{t}\\ =\frac{\text{-32N·s}}{\text{0}\text{.5s}}\\ =-64\text{N}\end{array}$

Conclusion:

Thus, the average force is $-64\text{N}$ .