Chapter 9.1, Problem 1PP

a.

To determine

To Find: Magnitude and direction of momentum.

To Show: Arrow direction for momentum.

Answer to Problem 1PP

Magnitude of momentum is $p=2.32\times {10}^4\text{kg}\cdot \text{m/s}$ and direction is eastward.

Explanation of Solution

Given:

Mass of car, $m=725\text{kg}$

Velocity of car, $v=115\text{km/h}$

Formula used:

Momentum can be obtained by:

  $p=mv$

Where, $p$ is momentum od object, $m$ is mass and $v$ is velocity.

Calculation:

The sketch of arrow showing momentum is shown below;

  

As velocity is in east, so the momentum will also be in east.

Now, on putting the values:

  $\begin{array}{l}p=725\times \left(115\times \frac{1000}{3600}\right)\\ =2.32\times {10}^4\text{kg}\cdot \text{m/s towards east}\end{array}$

Conclusion:

Therefore, magnitude of momentum is $p=2.32\times {10}^4\text{kg}\cdot \text{m/s}$ and direction is eastward.

b.

To determine

Velocity of car second car.

Answer to Problem 1PP

Velocity of second car is $v=38.4\text{km/h}$

Explanation of Solution

Given:

Mass of second car, $m=2175\text{kg}$

Momentum of car, $p=2.32\times {10}^4\text{kg}\cdot \text{m/s}$

Formula used:

Momentum can be obtained by:

  $p=mv$

Where, $p$ is momentum od object, $m$ is mass and $v$ is velocity.

Calculation:

On putting the values,

  $\begin{array}{l}2.32\times {10}^4=2175\times v\\ \Rightarrow v=\frac{2.32\times {10}^4}{2175}\\ =10.65\text{m/s}\\ \text{=10}\text{.65}\times \frac{3600}{1000}\text{km/h}\\ v=38.4\text{km/h}\end{array}$

Conclusion:

Therefore, velocity of second car is $v=38.4\text{km/h}$ .