Chapter 3, Problem 58E

A 1000-kg car moving east at 80 km/h collides head-on with a 1500-kg car moving west at 40 km/h, and the two cars stick together. (a) Which way does the wreckage move and with what initial speed? (b) How much KE is lost in the collision?

To determine

The final speed of the system.

Answer to Problem 58E

The final speed of the system is $2.2\text{m/s}$.

Explanation of Solution

Given info: Mass of the first car is $1000\text{kg}$, Mass of the another car is $1500\text{kg}$, and speed of the first car is $\text{80km/h}$, speed of the second car is $\text{40km/h}$.

Write the expression from conservation of momentum.

$m_1{v}_1+m_2\left(-v_2\right)=\left(m_1+m_2\right)v$

Here,

$m_1$ is the mass of the first car

$m_2$ is the mass of the second car

$v_1$ is the speed of the first car

$v_2$ is the speed of the second car

$v$ is the final speed of the combined system.

Substitute $1000\text{kg}$ for $m_1$, $1500\text{kg}$ for $m_2$, $\text{80km/h}$ for $v_1$ and $\text{40km/h}$ for $v_2$ in the above expression to get final speed.

$\begin{array}{l}\left[\left(1000\text{kg}\right)\left[\left(80\text{km/h}\right)\frac{\left(0.2778\text{m/s}\right)}{\left(1\text{km/h}\right)}\right]\right]+\left[\left(1500\text{kg}\right)\left[\left(-40\text{m/s}\right)\frac{\left(0.2778\text{m/s}\right)}{\left(1\text{km/h}\right)}\right]\right]=\left[\left(1000\text{kg}\right)+\left(1500\text{kg}\right)\right]v\\ v=2.2\text{m/s}\end{array}$

Conclusion:

Therefore, the final speed of the system is $2.2\text{m/s}$.

To determine

The amount of kinetic energy lost.

Answer to Problem 58E

The amount of kinetic energy lost is $302\text{kJ}$.

Explanation of Solution

Given info: Mass of the first car is $1000\text{kg}$, Mass of the another car is $1500\text{kg}$, and speed of the first car is $\text{80km/h}$, speed of the second car is $\text{40km/h}$, final speed of the system is $2.2\text{m/s}$.

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 car

$m_2$ is the mass of the second car

$v_1$ is the speed of the first car

$v_2$ is the speed of the second car

$v$ is the final speed of the combined system.

Substitute $1000\text{kg}$ for $m_1$, $1500\text{kg}$ for $m_2$, $\text{80km/h}$ for $v_1$ and $\text{40km/h}$ for $v_2$, $2.2\text{m/s}$ for $v$ in the above expression to get lost in kinetic energy.

$\begin{array}{c}\left(\Delta KE\right)=\left[\frac{1}{2}\left(\left\{\left(1000\text{kg}\right){\left[\left(80\text{km/h}\right)\frac{\left(0.2778\text{m/s}\right)}{\left(1\text{km/h}\right)}\right]}^2\right\}+\left\{\left(1500\text{kg}\right){\left[\left(40\text{km/h}\right)\frac{\left(0.2778\text{m/s}\right)}{\left(1\text{km/h}\right)}\right]}^2\right\}\right)\right]-\\ \left[\frac{1}{2}\left\{\left(1000\text{kg}\right)+\left(1500\text{kg}\right)\right\}{\left(2.2\text{m/s}\right)}^2\right]\\ =302\text{kJ}\end{array}$

Conclusion:

The amount of kinetic energy lost is $302\text{kJ}$.