Chapter 9, Problem 49A

a.

To determine

The impulse acted on the object if, after the collision, it moved at the velocity $+8.0\text{m/s}$ .

Answer to Problem 49A

  $-2600\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the object, $m=25\text{kg}$

Before collision, the object was moving at $112\text{m/s}$ . Thus, the initial velocity of the object, $v_i=112\text{}\text{m/s}$ .

Final velocity of the object, $v_f=+8\text{}.0\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }{p}_f-p_i$

where $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Using the given values, the impulse on the object will be,

  $\begin{array}{c}\text{Impulse}=p_f-p_i\\ =m\left(v_f-v_f\right)\\ =25\left(8-112\right)\\ =-2600\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse on the object is $-2600\text{kg}\cdot \text{m/s}$ .

b.

To determine

The impulse acted on the object if, after the collision, it moved at the velocity $-8.0\text{m/s}$ .

Answer to Problem 49A

  $-3000\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the object, $m=25\text{kg}$

Before collision, the object was moving at $112\text{m/s}$ . Thus, the initial velocity of the object, $v_i=112\text{}\text{m/s}$

Final velocity of the object, $v_f=-8\text{}.0\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }{p}_f-p_i$

Where $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Using the given values, the impulse on the object will be,

  $\begin{array}{c}\text{Impulse}=p_f-p_i\\ =m\left(v_f-v_f\right)\\ =25\left(-8-112\right)\\ =-3000\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse on the object is $-3000\text{kg}\cdot \text{m/s}$ .