Chapter 6, Problem 24P

To determine

Original speed of the car if one car has twice the mass of a second car, but only half the as much kinetic energy. When both cars increase their speed by 9.3 m/s, they then have the same kinetic energy.

Answer to Problem 24P

Solution:Speed of the first and second cards $5.65\text{ m/s}$ and $11.3\text{ m/s}$ respectively.

Explanation of Solution

Given:

Mass of the first car, $m_1=m$

Mass of the second car, $m_2=2m$

Formula Used:

The kinetic energy of a moving particle is given by

  $K.E=\frac{1}{2}m{v}^2$

Where K.E is the kinetic energy, m is the mass of the particle, v is the velocity of the particle

Calculations: The kinetic energy of the first car is

  $K.E_1=\frac{1}{2}{m}_1{v}_1{}^2$

Here $m_1$ is mass of the firstcar; $v_1$ is the speed of the firstcar.

The kinetic energy of the second car is

  $K.E_2=\frac{1}{2}{m}_2{v}_2{}^2$

Where $m_2$ is mass of the firstcar, $v_2$ is the speed of the first car.

The kinetic energy of the first car is half of the second car

  $\begin{array}{l}K.E_1=\frac{1}{2}K.E_2\\ \frac{1}{2}{m}_1{v}_1{}^2=\frac{1}{2}\times \frac{1}{2}{m}_2{v}_2{}^2\\ 2m_1{v}_1{}^2=m_2{v}_2{}^2\end{array}$

  $m_1=2m_2$ , substitute this in above equation

  $\begin{array}{l}2m_1{v}_1{}^2=m_2{v}_2{}^2\\ 2\times 2m{v}_1{}^2=m{v}_2{}^2\\ 4v_1{}^2=v_2{}^2\\ v_2=2v_1.\mathrm{...}(1)\end{array}$

Now when the cars increase their speeds by $8\text{ m/s}$ , then their K. E’s are equal which means 

  $\begin{array}{l}K.E_1=K.E_2\\ $ \frac{1}{2}2m{(v_1+8)}^2=\frac{1}{2}m{(v_2+8)}^2$$ 2{(v_1+8)}^2={(v_2+8)}^2$\end{array}$

Substitute $v_2$ value from equation 1

  $\begin{array}{l}$ 2{(v_1+8)}^2={(2v_1+8)}^2$2(v_1{}^2+64+16v_1)=(4v_1{}^2+64+32v_1)\\ 2v_1{}^2+128+32v_1=4v_1{}^2+64+32v_1\\ 2v_1{}^2=128-64\\ 2v_1{}^2=64\\ v_1{}^2=32\\ v_1=\sqrt{32}\\ v_1=5.65\text{ m/s}\end{array}$

The initial speed of the first car is $5.65\text{ m/s}$

From equation 1

  $\begin{array}{l}{v}_2=2v_1\\ v_2=2\times 5.65\\ v_2=11.3\text{ m/s}\end{array}$

The initial speed of the second car is $11.3\text{m/s}.$

Conclusion:Initial speed of the first and second car is $5.65\text{ m/s}$ and $11.3\text{ m/s}$ respectively if one car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by $9.3\text{ m/s}$ , they then have the same kinetic energy.