Chapter 9.2, Problem 27PP

To determine

The direction of motion and speed of the cars immediately after the collision.

Answer to Problem 27PP

  $v_f=18.07\text{ m/s}$ and are directed south of east.

Explanation of Solution

Given:

Mass of the car moving towards the south (- y-direction) is $m_1=1383\text{ kg}$

Initial speed of the car moving towards the south (- y-direction) is $v_{1i}=11.2\text{ m/s}$ or itcan be taken as $v_{1i}=-11.2\text{ m/s}$ (+ y-direction)

Mass of the car moving towards east (x-direction) is $m_2=1732\text{ kg}$

Initial speed of the car moving towards east (x-direction) is $v_{2i}=31.3\text{ m/s}$

After the collision two cars are stuck together.

Formula used:

Momentum $\left(p\right)$ is the product of mass $\left(m\right)$ and velocity $\left(v\right)$ of a moving body. This can be given by the equation

  $p=mv$$\left(1\right)$

Initial momentum of the car moving towards the south can be given by the equation

  $p_{i,1}=m_1{v}_{1i}$$\left(2\right)$

Initial momentum of the car moving towards west can be given by the equation

  $p_{i,2}=m_2{v}_{2i}$$\left(3\right)$

Since momentum is conserved, final momentum of the system is equal to the initial momentum of the system. That is

  $p_f=p_i$

  $p_f=\sqrt{{\left(p_{i,1}\right)}^2+{\left(p_{i,2}\right)}^2}$$\left(4\right)$

Final momentum of the system can also be calculated using the equation,

  $p_f=\left(m_1+m_2\right)v_f$$\left(5\right)$

Where the sum of the masses of the two cars is taken since after collision two cars are stuck together.

  $v_f$ is the final speed of the cars

From equation $\left(5\right)$ final velocity of the cars can be calculated as,

  $v_f=\frac{p_f}{\left(m_1+m_2\right)}$$\left(6\right)$

Direction of motion of the cars after collision can be calculated using the equation,

  $\theta ={\tan }^{-1}\left(\frac{p_{i,1}}{p_{i,2}}\right)$$\left(7\right)$

Calculation:

Initial momentum of the car moving towards the south can be calculated by substituting the numerical values in equation $\left(2\right)$ ,

  $p_{i,1}=\left(1383\text{ kg}\right)\left(-11.2\text{ m/s}\right)$

  $=-1.55\times {10}^4\text{ kg}\text{.m/s}$

Initial momentum of the car moving towards the east can be calculated by substituting the numerical values in equation $\left(3\right)$ ,

  $p_{i,2}=\left(1732\text{ kg}\right)\left(31.3\text{ m/s}\right)$

  $=5.42\times {10}^4\text{ kg}\text{.m/s}$

Substituting the values of $p_{i,1}$ and $p_{i,2}$ in equation $\left(4\right)$ , final momentum of the system is

  $p_f=\sqrt{{\left(-1.55\times {10}^4\text{kg}\cdot \text{m/s}\right)}^2+{\left(5.42\times {10}^4\text{kg}\cdot \text{m/s}\right)}^2}$

  $=\sqrt{2.37\times {10}^8{\text{kg}}^2\cdot {\text{m}}^2{\text{/s}}^2+29.37\times {10}^8{\text{kg}}^2\cdot {\text{m}}^2{\text{/s}}^2}$

  $=\sqrt{31.7\times {10}^8{\text{kg}}^2\cdot {\text{m}}^2{\text{/s}}^2}$

  $=5.63\times {10}^4\text{ kg}\cdot \text{m/s}$

The final velocity of the cars can be calculated by substituting the numerical values in equation $\left(6\right)$ ,

  $v_f=\frac{5.63\times {10}^4\text{ kg}\cdot \text{m/s}}{\left(1383\text{ kg}+1732\text{ kg}\right)}$

  $=\frac{5.63\times {10}^4\text{ kg}\cdot \text{m/s}}{\left(3115\text{ kg}\right)}$

  $=18.07\text{ m/s}$

Direction of motion of the cars after collision can be calculated by substituting the numerical values in equation $\left(7\right)$ ,

  $\theta ={\tan }^{-1}\left(\frac{-1.55\times {10}^4\text{ kg}\text{.m/s}}{5.63\times {10}^4\text{ kg}\text{.m/s}}\right)$

  $={\tan }^{-1}\left(-0.273\right)$

  $=15.4°$ , south of east

Conclusion:

Speed of the cars after collision is $18.07\text{ m/s}$ and are directed south of east.