Chapter 7, Problem 68P

(a)

To determine

The velocity of the cars after the collision.

Answer to Problem 68P

The velocity of the cars after the collision is $11\text{m}/\text{s}$ and the direction is $47°$ south of east.

Explanation of Solution

The mass of the car moving east ward is $1500\text{kg}$, velocity of the car moving east ward is $17\text{m}/\text{s}$, the mass of the car moving south ward is $1800\text{kg}$ and velocity of the car moving south ward is $15\text{m}/\text{s}$.

Write the expression for the conservation of momentum in east direction.

$mv=\left(m+M\right)v_f$

Here, $v_f$ is the final velocity of the cars in east direction, m is the mass of the car moving in east ward direction, M is the mass of the car moving south ward and v is the velocity of the car moving towards east.

Rewrite the above equation in terms of $v_f$.

$v_f=\frac{mv}{\left(m+M\right)}$. (I)

Write the expression for the conservation of momentum in south direction.

$MV=\left(m+M\right)v_f{}^{\prime }$

Here, $v_f{}^{\prime }$ is the final velocity of the cars after collision in south direction and V is the velocity of the car moving south ward direction.

Rewrite the above equation in terms of $v_f{}^{\prime }$.

$v_f{}^{\prime }=\frac{MV}{\left(m+M\right)}$ (II)

Write the expression to calculate the final velocity of the cars.

$V_f=\sqrt{v_f{}^2+{\left(v_f{}^{\prime }\right)}^2}$

Here, $V_f$ is the final velocity of the cars.

Substitute (I) and (II) in the above equation to calculate $V_f$.

$\begin{array}{c}{V}_f=\sqrt{{\left(\frac{mv}{\left(m+M\right)}\right)}^2+{\left(\frac{MV}{\left(m+M\right)}\right)}^2}\\ =\frac{1}{m+M}\left(\sqrt{{\left(mv\right)}^2+{\left(MV\right)}^2}\right)\end{array}$

Substitute $1500\text{kg}$ for m, $17\text{m}/\text{s}$ for v, $1800\text{kg}$ for M and $15\text{m}/\text{s}$ for V in the above equation to calculate $V_f$.

$\begin{array}{c}{V}_f=\frac{1}{1500\text{kg}+1800\text{kg}}\left(\sqrt{{\left(1500\text{kg}\right)}^2{\left(17\text{m}/\text{s}\right)}^2+{\left(1800\text{kg}\right)}^2{\left(15\text{m}/\text{s}\right)}^2}\right)\\ =11.3\text{m}/\text{s}\\ \sim 11\text{m}/\text{s}\end{array}$

Write the expression to calculate the direction of the final velocity of the cars.

$\varphi ={\tan }^{-1}\left(\frac{v_f{}^{\prime }}{v_f}\right)$

Here, $\varphi$ is the direction.

Rewrite the above equation using (I) and (II).

$\varphi ={\tan }^{-1}\left(\frac{MV}{mv}\right)$

Substitute $1500\text{kg}$ for m, $17\text{m}/\text{s}$ for v, $1800\text{kg}$ for M and $15\text{m}/\text{s}$ for V in the above equation to calculate $\varphi$.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{1800\text{kg}\left(15\text{m}/\text{s}\right)}{1500\text{kg}\left(17\text{m}/\text{s}\right)}\right)\\ =46.6°\\ \sim 47°\end{array}$

Conclusion:

Therefore, The velocity of the cars after the collision is $11\text{m}/\text{s}$ and the direction is $47°$ south of east.

(b)

To determine

The amount of kinetic energy converted during the collision.

Answer to Problem 68P

The amount of kinetic energy converted during the collision is $220\text{kJ}$.

Explanation of Solution

Write the expression to calculate the amount of kinetic energy converted during collision.

$\Delta K=\frac{1}{2}\left(m+M\right)V_f{}^2-\frac{1}{2}m{v}^2-\frac{1}{2}M{V}^2$

Here, $\Delta K$ is the change in kinetic energy.

Substitute $1500\text{kg}$ for m, $17\text{m}/\text{s}$ for v, $1800\text{kg}$ for M, $11\text{m}/\text{s}$ for $V_f$ and $15\text{m}/\text{s}$ for V in the above equation to calculate $\Delta K$.

$\begin{array}{c}\Delta K=\frac{1}{2}\left(1500\text{kg}+1800\text{kg}\right){\left(11\text{m}/\text{s}\right)}^2-\frac{1}{2}\left(1500\text{kg}\right){\left(17\text{m}/\text{s}\right)}^2-\frac{1}{2}\left(1800\text{kg}\right){\left(15\text{m}/\text{s}\right)}^2\\ =199650\text{J}-216750\text{J}-202500\text{J}\\ =-\text{219600}\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ \sim -220\text{kJ}\end{array}$

Thus, $220\text{kJ}$ of initial kinetic converted to another form of energy.

Conclusion:

Therefore, the amount of kinetic energy converted during the collision is $220\text{kJ}$.