Chapter 11, Problem 9STP

To determine

To Calculate: The decrease in the kinetic energy of the system.

Answer to Problem 9STP

  $1520\text{J}$

Explanation of Solution

Given:

Mass of the hockey player, $m_1=90.0\text{kg}$

Mass of another player, $m_2=110\text{kg}$

Initial velocity of the first player, $u_1=5.0\text{m/s}$

Initial velocity of the second player, $u_2=-3.0\text{m/s}$

Final velocity of both the players, $v_f=1.0{\text{m/s}}^{\text{2}}$

Formula Used:

Kinetic energy can be obtained by:

  $K.E=\frac{1}{2}m{v}^2$

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

Calculations:

Initial kinetic energy of the system:

  $\begin{array}{c}\frac{1}{2}{m}_1{u}_1{}^2+\frac{1}{2}{m}_2{u}_2{}^2=\frac{1}{2}\left(90.0\right){\left(5\right)}^2+\frac{1}{2}\left(110\right){\left(-3\right)}^2\\ =1125\text{J}+495\text{J}\\ \text{=1620}\text{}\text{J}\end{array}$

Final kinetic energy of the system:

  $\begin{array}{c}\frac{1}{2}\left(m_1+m_2\right)v_f{}^2=\frac{1}{2}\left(90.0+110\right){\left(1.0\right)}^2\\ =100\text{J}\end{array}$

Decrease in the kinetic energy of the system:

  $\begin{array}{c}\Delta K.E.=10\text{0}\text{J}-1620\text{J}\\ =-1520\text{J}\end{array}$

Conclusion:

Thus, the decrease in the kinetic energy of the system is $1520\text{J}$ .