Chapter 7, Problem 27P

(a)

To determine

The velocity of the cars after collision.

Answer to Problem 27P

The velocity of the car A after collision is $\text{4}\text{.8 }\text{m}/\text{s}$ .

The velocity of the car B after collision is $\text{4}\text{.8 }\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the car A is $m_1=450\text{ kg}$ .

The mass of the car B is $m_2=550\text{ kg}$ .

The initial velocity of the car A is $u_1=4.50\text{m}/\text{s}$ .

The initial velocity of the car B is $u_2=3.70\text{m}/\text{s}$ .

Formula used:

The expression for relative velocity is,

  $u_1-u_2=-\left(v_1-v_2\right)$

Here, $u_1$ is the initial velocity of car B, $u_2$ is the initial velocity of the car B, $v_1$ is the final velocity of car B, and $v_2$ is the final velocity of the car B.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the car B, $m_2$ is the mass of car B, $u_1$ is the initial velocity of car B, $u_2$ is the initial velocity of the car B, $v_1$ is the final velocity of car B, and $v_2$ is the final velocity of the car B.

Calculation:

The velocity of car A after collision is,

  $\begin{array}{l}{u}_1-u_2=-\left(v_1-v_2\right)\\ \left(\text{4}\text{.5 }\text{m}/\text{s}\right)-\left(\text{3}\text{.7 }\text{m}/\text{s}\right)=v_2-v_1\\ \text{0}\text{.8 }\text{m}/\text{s}=v_2-v_1\\ v_1=v_2-\text{0}\text{.8}\end{array}$     ......(1)

The velocity of car B after collision is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left(\begin{array}{l}\left(450\text{ kg}\right)\left(4.50\text{m}/\text{s}\right)\\ +\left(550\text{ kg}\right)\left(3.70\text{m}/\text{s}\right)\end{array}\right)=\left(450\text{ kg}\right)v_1+\left(550\text{ kg}\right)v_2\\ 4,060=450\left(v_2-0.8\right)+550v_2\\ 4,420=1,000v_2\\ v_2=4.42\text{m}/\text{s}\end{array}$

So, the velocity of car A using equation (1) is,

  $\begin{array}{l}{v}_1=v_2-0.8\\ v_1=\left(4.42\text{m}/\text{s}\right)-0.8\\ v_1=3.62\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of car A after collision is $v_1=3.62\text{m}/\text{s}$ and the velocity of car B after collision is $v_2=\text{4}\text{.42 }\text{m}/\text{s}$ .

(b)

To determine

The change in momentum of each car.

Answer to Problem 27P

The change in momentum of car A is $396\text{kg}\cdot \text{m}/\text{s}$ .

The change in momentum of car B is $-396\text{kg}\cdot \text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the car A is $m_1=450\text{ kg}$ .

The mass of the car B is $m_2=550\text{ kg}$ .

The initial velocity of the car A is $u_1=4.50\text{m}/\text{s}$ .

The initial velocity of the car B is $u_2=3.70\text{m}/\text{s}$ .

Formula used:

The expression for change in momentum is,

  $\Delta p=mu-mv$

Here, m is the mass of the object, u is the initial velocity of the object, and v is the final velocity of the object.

Calculation:

From part (a), the velocity of car A after collision is $v_1=3.62\text{m}/\text{s}$ and the velocity of car B after collision is $v_2=\text{4}\text{.42 }\text{m}/\text{s}$ .

The change in momentum of car A is,

  $\begin{array}{l}\Delta p_{\text{A}}=m_1{u}_1-m_1{v}_1\\ \Delta p_{\text{A}}=\left(450\text{ kg}\right)\left(4.50\text{m}/\text{s}\right)-\left(450\text{ kg}\right)\left(3.62\text{m}/\text{s}\right)\\ \Delta p_A=396\text{kg}\cdot \text{m}/\text{s}\end{array}$

The change in momentum of car B is,

  $\begin{array}{l}\Delta p_{\text{B}}=m_2{u}_2-m_2{v}_2\\ \Delta p_{\text{B}}=\left(550\text{ kg}\right)\left(3.70\text{m}/\text{s}\right)-\left(550\text{ kg}\right)\left(4.42\text{m}/\text{s}\right)\\ \Delta p_B=-396\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Thus, the change in momentum of car A is $\Delta p_A=396\text{kg}\cdot \text{m}/\text{s}$ and the change in momentum of car B is $\Delta p_B=-396\text{kg}\cdot \text{m}/\text{s}$ .