Chapter 9.2, Problem 26PP

To determine

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

Answer to Problem 26PP

  $v_f=11.15\text{ m/s}$ and are directed north of west.

Explanation of Solution

Given:

Mass of the car moving towards north (+ y-direction) is $m_y=925\text{ kg}$

Initial speed of the car moving towards north (+ y-direction) is $v_{i,y}=20.1\text{ m/s}$

Mass of the car moving towards west (- x-direction) is $m_x=1865\text{ kg}$

Initial speed of the car moving towards west (- x-direction) is $v_{i,x}=13.4\text{ m/s}$ or $v_{i,x}=-13.4\text{ m/s}$ (towards +x-direction)

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 north can be given by the equation

  $p_{i,y}=m_y{v}_{i,y}$$\left(2\right)$

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

  $p_{i,x}=m_x{v}_{i,x}$$\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,y}\right)}^2+{\left(p_{i,x}\right)}^2}$$\left(4\right)$

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

  $p_f=\left(m_y+m_x\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_y+m_x\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,y}}{p_{i,x}}\right)$$\left(7\right)$

Calculation:

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

  $p_{i,y}=\left(925\text{ kg}\right)\left(20.1\text{ m/s}\right)$

  $=1.86\times {10}^4\text{ kg}\text{.m/s}$

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

  $p_{i,x}=\left(1865\text{ kg}\right)\left(-13.4\text{ m/s}\right)$

  $=-2.5\times {10}^4\text{ kg}\text{.m/s}$

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

  $p_f=\sqrt{{\left(1.86\times {10}^4\text{ kg}\text{.m/s}\right)}^2+{\left(-2.5\times {10}^4\text{ kg}\text{.m/s}\right)}^2}$

  $=\sqrt{3.45\times {10}^8{\text{kg}}^2{\text{.m}}^2{\text{/s}}^2+6.25\times {10}^8{\text{kg}}^2{\text{.m}}^2{\text{/s}}^2}$

  $=\sqrt{9.7\times {10}^8{\text{kg}}^2{\text{.m}}^2{\text{/s}}^2}$

  $=3.11\times {10}^4\text{ kg}\text{.m/s}$

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

  $v_f=\frac{3.11\times {10}^4\text{ kg}\text{.m/s}}{\left(925\text{ kg}+1865\text{ kg}\right)}$

  $=\frac{3.11\times {10}^4\text{ kg}\text{.m/s}}{\left(2790\text{ kg}\right)}$

  $=11.15\text{ m/s}$

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

  $\theta ={\tan }^{-1}\left(\frac{1.86\times {10}^4\text{ kg}\text{.m/s}}{-2.5\times {10}^4\text{ kg}\text{.m/s}}\right)$

  $={\tan }^{-1}\left(-0.744\right)$

  $={36.6}^0$ , north of west

Conclusion:

Speed of the cars after collision is $11.15\text{ m/s}$ and are directed north of west.