Chapter 7, Problem 5SP

(a)

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 $30,000\text{kg}\cdot \text{m}/\text{s},\text{ south}$.

Explanation of Solution

Given info: The mass of the car is $1,500\text{kg}$ and its speed is $\text{25 }\text{m}/\text{s},\text{ north}$. The mass of the truck is $4,500\text{ kg}$ with a speed of $15\text{m}/\text{s},\text{ south}$.

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,

$p_{\text{total}}$ is the total momentum prior to the collision

$p_t$ 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 $1500\text{kg}$ for $m_c$,$20\text{m}/\text{s},\text{ north}$ for $v_c$,$4500\text{ kg}$ for $m_t$ and $15\text{m}/\text{s},\text{ south}$ for $v_t$ in the above equation to find $p_{\text{total}}$.

$\begin{array}{c}{p}_{\text{total}}=\left(4,500\text{kg}\right)\left(15\text{m}/\text{s},\text{south}\right)-\left(1,500\text{kg}\right)\left(25\text{m}/\text{s},\text{north}\right)\\ =30,000\text{kg}\cdot \text{m}/\text{s},\text{south}\end{array}$

Conclusion:

Therefore, the total momentum of the system prior to the collision is $30000\text{kg}\cdot \text{m}/\text{s},\text{ south}$.

(b)

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 $5\text{m}/\text{s}\text{ south}$.

Explanation of Solution

Total momentum before collision is equal to the total momentum after collision according to the principle of conservation of momentum.

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

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

Substitute $30000\text{kg}\cdot \text{m}/\text{s},\text{ south}$ for $p_{\text{total}}$,$1500\text{kg}$ for $m_c$ and $4500\text{ kg}$ for $m_t$ in the above equation to find $v$.

$\begin{array}{c}v=\frac{30,000\text{kg}\cdot \text{m}/\text{s},\text{south}}{1,500\text{kg}+4,500\text{kg}}\\ =5\text{m}/\text{s}\text{south}\end{array}$

Conclusion:

Therefore the velocity of the two vehicles just after the collision is $5\text{m}/\text{s}\text{ south}$.

(c)

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 $9.75\times {10}^5\text{ J}$.

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,

$K{E}_{\text{before}}$ is the total kinetic energy of the system before collision

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

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

Substitute $1500\text{kg}$ for $m_c$,$\text{25 }\text{m}/\text{s},\text{ north}$ for $v_c$, $4500\text{ kg}$ for $m_t$ and $15\text{m}/\text{s},\text{ south}$ for $v_t$ in the above equation to find $K{E}_{\text{before}}$.

$\begin{array}{c}K{E}_{\text{before}}=\frac{1}{2}\left(4,500\text{kg}\right){\left(15\text{m}/\text{s},\text{south}\right)}^2+\frac{1}{2}\left(1,500\text{kg}\right){\left(\text{25}\text{m}/\text{s},\text{north}\right)}^2\\ =9.75\times {10}^5\text{J}\end{array}$

Conclusion:

Therefore the total kinetic energy of the system before collision is $9.75\times {10}^5\text{ J}$.

(d)

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 $7.5\times {10}^4\text{ J}$.

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,

$K{E}_{\text{after}}$ is the total kinetic energy after collision

Substitute $30,000\text{kg}\cdot \text{m}/\text{s},\text{south}$ for $p_{\text{total}}$,$1500\text{kg}$ for $m_c$ and $4500\text{ kg}$ for $m_t$ in the above equation to find $K{E}_{\text{after}}$.

$\begin{array}{c}K{E}_{\text{after}}=\frac{{\left(30,000\text{kg}\cdot \text{m}/\text{s},\text{south}\right)}^2}{2\left(4,500\text{ kg}+1,500\text{kg}\right)}\\ =7.5\times {10}^4\text{J}\end{array}$

Conclusion:

Therefore, the total kinetic energy of the system just after collision is $7.5\times {10}^4\text{ J}$.

(e)

To determine

Whether the collision is elastic.

Answer to Problem 5SP

The collision is not elastic.

Explanation of Solution

Elastic collision is associated with no change in the kinetic energy. This implies for an elastic collision to occur, there should be no change in the kinetic energy before and after collision or the kinetic energy should be conserved.

In the given situation the vehicles stick together after collision. The total kinetic energy before collision is $9.75\times {10}^5\text{ J}$ and that after collision is $7.5\times {10}^4\text{ J}$. The difference in the value of the total kinetic energy of the system shows that kinetic energy is not conserved during the head-on-collision. This means that the collision is inelastic.

Conclusion:

Thus, the collision is not elastic.