Chapter 8, Problem 52TP

To determine

The center of mass velocity before and after collision.

Answer to Problem 52TP

The centre-of-mass velocity of the system will not change after collision as the there is no net external force acting on the system of the two cars. The velocity of the system is $42.1\text{m}/\text{s}$ at an angle of $45.9°$ in the north of east.

Explanation of Solution

Given:

The mass of car A is, $m_1=2000\text{kg}$.

The initial velocity of car A is, $v_1=38\text{m}/\text{s}$ in the east direction.

The mass of car B is, $m_2=3500\text{kg}$.

The initial velocity of car B is, $v_2=53\text{m}/\text{s}$ in $63°$ north of east direction.

Formula used:

Write the expression for the conservation of momentum for the system.

  $m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v_{\text{cm}}$

Here, $v_{cm}$ is the center of mass velocity of the system after collision.

The collision of both the cars and their directions is illustrated in the following figure.

  

  Figure (1)

The sine and cosine component of the velocity of car B can be understood from the following diagram.

  

  Figure (2)

Calculation:

Obtain the center of mass for the system after the collision. Calculation has to be done for both the horizontal and vertical components of the direction of motion.

The vertical component of the center of mass, $v_{cm\text{v}}$ is calculated as

  $\begin{array}{c}{m}_1{v}_{1\text{v}}+m_2{v}_{2\text{v}}=\left(m_1+m_2\right)v_{\text{cm v }}\\ \left(2000\text{kg}\right)\left(\left(38\text{m}/\text{s}\right)\sin 0°\right)+\left(3500\text{kg}\right)\left(\left(53\text{m}/\text{s}\right)\sin 63°\right)=\left[\left(2000\text{kg}\right)+\left(3500\text{kg}\right)\right]v_{\text{cm v }}\\ \left(3500\text{kg}\right)\left(47.22\text{m}/\text{s}\right)=\left(5500\text{kg}\right)v_{\text{cm v}}\\ v_{\text{cm v}}=30.05\text{m}/\text{s}\end{array}$

The horizontal component of the center of mass, $v_{cm\text{v}}$ is calculated as

  $\begin{array}{c}{m}_1{v}_{1\text{h}}+m_2{v}_{2\text{h}}=\left(m_1+m_2\right)v_{\text{cm h}}\\ \left(2000\text{kg}\right)\left(\left(38\text{m}/\text{s}\right)\cos 0°\right)+\left(3500\text{kg}\right)\left(\left(53\text{m}/\text{s}\right)\cos 63°\right)=\left[\left(2000\text{kg}\right)+\left(3500\text{kg}\right)\right]v_{\text{cm h}}\\ \left(2000\text{kg}\right)\left(38\text{m}/\text{s}\right)+\left(3500\text{kg}\right)\left(24.06\text{m}/\text{s}\right)=\left(5500\text{kg}\right)v_{\text{cm h}}\\ v_{\text{cm h}}=29.1\text{m}/\text{s}\end{array}$

The resultant of horizontal and vertical component of center of mass is calculated as

  $\begin{array}{c}{v}_{cm\text{h}}=\sqrt{{\left(v_{\text{cm h }}\right)}^2+{\left(v_{\text{cm v }}\right)}^2}\\ =\sqrt{{\left(29.1\text{m}/\text{s}\right)}^2+{\left(30.05\text{m}/\text{s}\right)}^2}\\ =41.83\text{m}/\text{s}\end{array}$

The angle $\theta$ at which the system is moving after collision is calculated as

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{\text{cm v}}}{v_{\text{cm h }}}\right)\\ ={\tan }^{-1}\left(\frac{30.05\text{m}/\text{s}}{29.1\text{m}/\text{s}}\right)\\ =45.9°\end{array}$

Conclusion:

Therefore, the centre-of-mass velocity of the system will not change after collision as the there is no net external force acting on the system of the two cars. The velocity of the system is $42.1\text{m}/\text{s}$ at an angle of $45.9°$ in the north of east.