Chapter 7, Problem 5SP

A 1600-kg car traveling due east with a speed of 30 m/s collides head-on with a 4800-kg truck traveling due west with a speed of 15 m/s. The two vehicles stick together after the collision.

1. a. What is the total momentum of the system prior to the collision?
2. b. What is the velocity of the two vehicles just after the collision?
3. c. What is the total kinetic energy of the system before the collision?
4. d. What is the total kinetic energy just after the collision?
5. e. Is the collision elastic? Explain.

To determine

The total momentum of the system prior to the collision.

Answer to Problem 5SP

The total momentum of the system prior to the collision is $24000\text{kg}\cdot \text{m}/\text{s},\text{ west}$.

Explanation of Solution

Given info: The mass of the car is $1600\text{kg}$ and its velocity is $30\text{m}/\text{s},\text{ east}$. The mass of the truck is $4800\text{ kg}$ with a velocity of $15\text{m}/\text{s},\text{ west}$.

Write the expression for total momentum of the system prior to the collision.

$\begin{array}{c}{p}_{\text{total}}=p_t-p_c\\ =m_t{v}_t-m_c{v}_c\end{array}$

Here,

$p_t$ is the momentum of the truck

$m_t$ is the mass of the truck

$v_t$ is the velocity of the truck

$p_c$ is the momentum of the car

$m_c$ is the mass of the car

$v_c$ is the velocity of the car

Substitute $1600\text{kg}$ for $m_c$, $30\text{m}/\text{s},\text{ east}$ for $v_c$, $4800\text{ kg}$ for $m_t$ and $15\text{m}/\text{s},\text{ west}$ for $v_t$ in the above equation.

$\begin{array}{c}{p}_{\text{total}}=\left(4800\text{ kg}\right)\left(15\text{m}/\text{s},\text{ west}\right)-\left(1600\text{kg}\right)\left(30\text{m}/\text{s},\text{ east}\right)\\ =24000\text{kg}\cdot \text{m}/\text{s},\text{ west}\end{array}$

Conclusion:

Therefore, the total momentum of the system prior to the collision is $24000\text{kg}\cdot \text{m}/\text{s},\text{ west}$.

To determine

The velocity of the two vehicles just after the collision.

Answer to Problem 5SP

The velocity of the two vehicles just after the collision is $3.75\text{m}/\text{s}\text{ west}$.

Explanation of Solution

Write the expression to find the velocity of the two vehicles just after collision.

$v=\frac{p_{\text{total}}}{m_t+m_c}$

Substitute $24000\text{kg}\cdot \text{m}/\text{s},\text{ west}$ for $p_{\text{total}}$, $1600\text{kg}$ for $m_c$ and $4800\text{ kg}$ for $m_t$ in the above equation.

$\begin{array}{c}v=\frac{24000\text{kg}\cdot \text{m}/\text{s},\text{ west}}{1600\text{kg}+4800\text{ kg}}\\ =3.75\text{m}/\text{s}\text{ west}\end{array}$

Conclusion:

Therefore, the velocity of the two vehicles just after the collision is $3.75\text{m}/\text{s}\text{ west}$.

To determine

The total kinetic energy of the system before collision.

Answer to Problem 5SP

The total kinetic energy of the system before collision is $12.6\times {10}^5\text{ J}$.

Explanation of Solution

Write the expression to find the kinetic energy of the system before collision.

$\begin{array}{c}K{E}_{\text{before}}=K{E}_t+K{E}_c\\ =\frac{1}{2}{m}_t{v}_t^2+\frac{1}{2}{m}_c{v}_c^2\end{array}$

Here,

$K{E}_t$ is the kinetic energy of the truck

$K{E}_c$ is the kinetic energy of the car

Substitute $1600\text{kg}$ for $m_c$, $30\text{m}/\text{s},\text{ east}$ for $v_c$, $4800\text{ kg}$ for $m_t$ and $15\text{m}/\text{s},\text{ west}$ for $v_t$ in the above equation.

$\begin{array}{c}K{E}_{\text{before}}=\frac{1}{2}\left(4800\text{ kg}\right){\left(15\text{m}/\text{s},\text{ west}\right)}^2+\frac{1}{2}\left(1600\text{kg}\right){\left(30\text{m}/\text{s},\text{ east}\right)}^2\\ =12.6\times {10}^5\text{ J}\end{array}$

Conclusion:

Therefore, the total kinetic energy of the system before collision is $12.6\times {10}^5\text{ J}$.

To determine

The total kinetic energy of the system just after collision.

Answer to Problem 5SP

The total kinetic energy of the system just after collision is $4.5\times {10}^4\text{ J}$.

Explanation of Solution

Write the expression to find the total kinetic energy of the system just after collision.

$K{E}_{\text{after}}=\frac{p_{\text{total}}^2}{2\left(m_t+m_c\right)}$

Here,

$p$ is the momentum of the system after collision

Substitute $24000\text{kg}\cdot \text{m}/\text{s},\text{ west}$ for $p_{\text{total}}$, $1600\text{kg}$ for $m_c$ and $4800\text{ kg}$ for $m_t$ in the above equation.

$\begin{array}{c}K{E}_{\text{after}}=\frac{{\left(24000\text{kg}\cdot \text{m}/\text{s},\text{ west}\right)}^2}{2\left(4800\text{ kg}+1600\text{kg}\right)}\\ =4.5\times {10}^4\text{ J}\end{array}$

Conclusion:

Therefore, the total kinetic energy of the system just after collision is $4.5\times {10}^4\text{ J}$.

To determine

Whether the collision is elastic.

Answer to Problem 5SP

No, the collision is not elastic.

Explanation of Solution

Elastic collision associates with it no change in the kinetic energy, that means for an elastic collision to occur, there should be no change in the kinetic energy before and after collision should be same.

When the vehicles stick together after collision, the kinetic energy before and after collision are the same. This results in inelastic collision case. Thus the head on collision of a car and truck is in elastic in nature since after the collision, they stick together.

Conclusion:

Therefore, the collision is not elastic.