Chapter 3, Problem 56E

A 40-kg skater moving at 4 m/s overtakes a 60-kg skater moving at 2 m/s in the same direction and collides with her. The two skaters stick together. (a) What is their final speed? (b) How much kinetic energy is lost?

To determine

The final speed of the system.

Answer to Problem 56E

The final speed of the system is $2.8\text{m/s}$.

Explanation of Solution

Given info: Mass of the first skater is $40\text{kg}$, Mass of the another skater is $60\text{kg}$, and speed of the first skater is $4\text{m/s}$, speed of the second skater is $2\text{m/s}$.

Write the expression from conservation of momentum.

$m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v$

Here,

$m_1$ is the mass of the first skater

$m_2$ is the mass of the second skater

$v_1$ is the speed of the first skater

$v_2$ is the speed of the second skater

$v$ is the final speed of the system.

Substitute $40\text{kg}$ for $m_1$, $60\text{kg}$ for $m_2$, $4\text{m/s}$ for $v_1$ and $2\text{m/s}$ for $v_2$ in the above expression to get final speed.

$\begin{array}{l}\left[\left(40\text{kg}\right)\left(4\text{m/s}\right)\right]+\left[\left(60\text{kg}\right)\left(2\text{m/s}\right)\right]=\left[\left(40\text{kg}\right)+\left(60\text{kg}\right)\right]v\\ v=2.8\text{m/s}\end{array}$

Conclusion:

Therefore, the final speed of the system is $2.8\text{m/s}$.

To determine

The amount of kinetic energy lost.

Answer to Problem 56E

The amount of kinetic energy lost is $48\text{kJ}$.

Explanation of Solution

Given info: Mass of the first skater is $40\text{kg}$, Mass of the another skater is $60\text{kg}$, and speed of the first skater is $4\text{m/s}$, speed of the second skater is $2\text{m/s}$, final speed of the system is $2.8\text{m/s}$.

Write the expression for kinetic energy lost.

$\left(\Delta KE\right)=\left[\frac{1}{2}\left(m_1{v}_1^2+m_1{v}_2^2\right)\right]-\left[\frac{1}{2}\left(m_1+m_2\right)v^2\right]$

Here,

$\left(\Delta KE\right)$ is the lost in kinetic energy

$m_1$ is the mass of the first skater

$m_2$ is the mass of the second skater

$v_1$ is the speed of the first skater

$v_2$ is the speed of the second skater

$v$ is the final speed of the system.

Substitute $40\text{kg}$ for $m_1$, $60\text{kg}$ for $m_2$, $4\text{m/s}$ for $v_1$ and $2\text{m/s}$ for $v_2$, $2.8\text{m/s}$ for $v$ in the above expression to get lost in kinetic energy.

$\begin{array}{c}\left(\Delta KE\right)=\left[\frac{1}{2}\left(\left\{\left(40\text{kg}\right){\left(4\text{m/s}\right)}^2\right\}+\left\{\left(60\text{kg}\right){\left(2\text{m/s}\right)}^2\right\}\right)\right]-\left[\frac{1}{2}\left\{\left(40\text{kg}\right)+\left(60\text{kg}\right)\right\}{\left(2.8\text{m/s}\right)}^2\right]\\ =48\text{kJ}\end{array}$

Conclusion:

The amount of kinetic energy lost is $48\text{kJ}$.