Chapter 7, Problem 19P

(a)

To determine

The original momentum of the fullback.

Answer to Problem 19P

The original momentum of the fullback is $380\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

The mass of the fullback is $m=95\text{ kg}$ .

The speed of the fullback is $v=4\text{ m/s}$ .

The time is $t=0.75\text{ s}$ .

Formula used:

The expression for momentum is,

  $p=mv$

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

Calculation:

The original momentum is,

  $\begin{array}{l}p=mv\\ p=\left(95\text{ kg}\right)\left(4\text{ m/s}\right)\\ p=380\text{ kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the original momentum of the fullback is $380\text{ kg}\cdot \text{m/s}$ .

(b)

To determine

The impulse exerted on the fullback.

Answer to Problem 19P

The impulse exerted on the fullback is $-380\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

The mass of the fullback is $m=95\text{ kg}$ .

The speed of the fullback is $v=4\text{ m/s}$ .

The time is $t=0.75\text{ s}$ .

Formula used:

The expression for impulse is,

  $I=\Delta p$

Here, $\Delta p$ is the change in momentum.

Calculation:

Consider the final momentum of the fullback $p_2=0$ .

Refer to part (a).

The original momentum of the fullback is $p_1=380\text{ kg}\cdot \text{m/s}$ .

Find the impulse exerted on the fullback as follows:

  $\begin{array}{l}I=\Delta p\\ I=p_2-p_1\\ I=0-380\text{ kg}\cdot \text{m/s}\\ I=-380\text{ kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse exerted on the fullback is $-380\text{ kg}\cdot \text{m/s}$ .

(c)

To determine

The impulse exerted on the tackler.

Answer to Problem 19P

The impulse exerted on the tackler is $380\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

The mass of the fullback is $m=95\text{ kg}$ .

The speed of the fullback is $v=4\text{ m/s}$ .

The time is $t=0.75\text{ s}$ .

Formula used:

The expression for impulse is,

  $I_{\text{tackler}}=-I_{\text{fullback}}$

Here, $I_{\text{fullback}}$ is the impulse exerted on fullback.

Calculation:

Refer to part (b).

The impulse exerted on the fullback is $I_{\text{fullback}}=-380\text{ kg}\cdot \text{m/s}$ .

The impulse exerted on the tackler is,

  $\begin{array}{l}{I}_{\text{tackler}}=-I_{\text{fullback}}\\ I_{\text{tackler}}=-\left(-380\text{ kg}\cdot \text{m/s}\right)\\ I_{\text{tackler}}=380\text{ kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse exerted on the tackler is $380\text{ kg}\cdot \text{m/s}$ .

(d)

To determine

The average force exerted on the tackler.

Answer to Problem 19P

The average force exerted on the tackler is 507 N.

Explanation of Solution

Given:

The mass of the fullback is $m=95\text{ kg}$ .

The speed of the fullback is $v=4\text{ m/s}$ .

The time is $t=0.75\text{ s}$ .

Formula used:

The expression for force is,

  $F=\frac{I_{\text{tackler}}}{t}$

Here, $I_{\text{tackler}}$ is the impulse exerted on tackler and t is the time.

Calculation:

Refer to part (b).

The impulse exerted on the tackler is $I_{\text{tackler}}=380\text{ kg}\cdot \text{m/s}$ .

The average force exerted on the tackler is,

  $\begin{array}{l}F=\frac{I_{\text{tackler}}}{t}\\ F=\frac{380\text{ kg}\cdot \text{m/s}}{0.75\text{ s}}\\ F=506.7\text{ kg}\cdot {\text{m/s}}^2\times \frac{1\text{ N}}{1\text{ kg}\cdot {\text{m/s}}^2}\\ F=507\text{ N}\end{array}$

Conclusion:

Thus, the average force exerted on the tackler is 507 N.