Chapter 11.2, Problem 22PP

a.

To determine

Total mechanical energy and momentum of the system before collision.

Answer to Problem 22PP

Total mechanical energy before collision is 4361.6 J and momentum before collision is $1237.6\text{ kg}\cdot \text{m/s}.$

Explanation of Solution

Given:

Mass of both hockey player, $m_1=m_2=91\text{ kg}$

Velocity of first hockey player, $v_1=5.50\text{ m/s}$

Velocity of second hockey player, $v_2=8.1\text{ m/s}$

Formula used:

Kinetic energy can be obtained by:

  $K.E=\frac{1}{2}m{v}^2$

Momentum can be obtained by:

  $p=mv$

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

Calculation:

Total kinetic energy before collision is:

  $\begin{array}{l}KE=K{E}_1+K{E}_2\\ =\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ =\frac{1}{2}\times 91\times {5.50}^2+\frac{1}{2}\times 91\times {8.1}^2\\ =\frac{1}{2}\times 91\times \left({5.50}^2+{8.1}^2\right)\\ =\frac{1}{2}\times 91\times 95.86\\ =4361.6\text{ J}\end{array}$

Momentum before collision is

  $\begin{array}{l}{P}_i=m_1{v}_1+m_2{v}_2\\ =91\times 5.50+91\times 8.1\\ =91\left(5.50+8.1\right)\\ =91\times 13.6\\ =1237.6\text{ kg}\cdot \text{m/s}\end{array}$

b.

To determine

Velocity of players after collision.

Answer to Problem 22PP

6.8 m/s

Explanation of Solution

Given:

Mass of both hockey player, $m_1=m_2=91\text{ kg}$

Velocity of first hockey player, $v_1=5.50\text{ m/s}$

Velocity of second hockey player, $v_2=8.1\text{ m/s}$

Formula Used:

Let velocity of players after collision is v m/s

As there is no external force is applied therefore momentum will be conserved.

  $P_i=P_f$

Calculations:

  $\begin{array}{l}{P}_i=\left(m_1+m_2\right)v\\ 1237.6=\left(91+91\right)v\\ v=\frac{1237.6}{182}\\ =6.8\text{ m/s}\end{array}$

Conclusion:

So, velocity of hockey players after collision is $6.8\text{ m/s}$ .

c.

To determine

Kinetic energy decreased in collision.

Answer to Problem 22PP

Kinetic energy decreased by 153.76 J in collision.

Explanation of Solution

Given:

Mass of both hockey player, $m_1=m_2=91\text{ kg}$

Velocity of first hockey player, $v_1=5.50\text{ m/s}$

Velocity of second hockey player, $v_2=8.1\text{ m/s}$

Formula Used:

Kinetic energy can be obtained by:

  $K.E=\frac{1}{2}m{v}^2$

Calculations:

Velocity of hockey players after collision is 6.8 m/s.

Therefore, kinetic energy after collision is

  $\begin{array}{l}K{E}_f=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2\\ =\frac{1}{2}\times 91\times {6.8}^2+\frac{1}{2}\times 91\times {6.8}^2\\ =2103.92+2103.92\\ =4207.84\text{ J}\end{array}$

From part (a) kinetic energy before collision is 4361.6 J

Therefore, loss of kinetic energy in collision is:

  $\begin{array}{l}\Delta KE=K{E}_i-K{E}_f\\ =4361.6-4207.84\\ =153.76\text{ J}\end{array}$