Chapter 9.1, Problem 2PP

(a)

To determine

The change in momentum, or equivalently, the magnitude and the direction of the impulse on the car.

Answer to Problem 2PP

The magnitude of the impulse is $1.0\times {10}^4\text{ kg}\cdot \text{m/s}$ and is directed westward.

Explanation of Solution

Given:

The average force exerted on the car to slow it downis $F=-5.0\times {10}^3\text{ N}$ (negative sign is taken because the force is applied to decrease the speed of the car)

The driver applies the brakes hard for, $\Delta \text{t = 2}\text{.0 s}$

  

Figure 1

Formula used:

Impulse is equal to the force $\left(F\right)$ times the time $\left(\Delta \text{t}\right)$ which is given by the equation,

  $I=F\Delta t$$\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$

  $I\text{= }\left(-5.0\times {10}^3\text{ N}\right)\left(\text{2}\text{.0 s}\right)$

  $=-1.0\times {10}^4\text{ N}\cdot \text{s = }-1.0\times {10}^4\text{ kg}\cdot \text{m/s}$

  $\Rightarrow$ The magnitude of the impulse is $1.0\times {10}^4\text{ kg}\text{.m/s}$ and is directed westward.

Conclusion:

The magnitude of the impulse is $1.0\times {10}^4\text{ kg}\text{.m/s}$ and is directed westward.

(b)

To determine

To complete the “before” and “after” sketches, and to determine the momentum and the velocity of the car now.

Answer to Problem 2PP

  $p_i=2.3\times {10}^4\text{ kg}\cdot \text{m/s}$ toward the east

  $p_f=1.3\times {10}^4\text{ kg}\cdot \text{m/s}$ toward the east

  $v_f=17.93\text{ m/s}$ toward the east

Explanation of Solution

Given:

The mass of the compact car is $\text{m = 725 kg}$

The initial velocity of the car is $v_i=115\text{ km/h = }\frac{115\times 1000}{3600}=32\text{ m/s}$ toward east.

Formula used:

Momentum of a car can be calculated by using the equation,

  $p=mv$$\left(2\right)$

Where,

  $m$ is mass of an object and

  $v$ is velocity of an object

From equation $\left(2\right)$ ,initial momentum of the car is obtained by substituting $v=v_i$ ,

Where, $v_i$ is the initial velocity of the car. That is,

  $p_i=m{v}_i$$\left(3\right)$

Final momentum of the car can be calculated using the equation,

  $p_f=F\Delta t+p_i$$\left(4\right)$

Where, $F\Delta t$ is the impulse on the car From equation $\left(2\right)$ , final momentum of the car can also be obtained by substituting $v=v_f$ ,

Where, $v_f$ is the final velocity of the car. That is,

  $p_f=m{v}_f$

  $\Rightarrow v_f=\frac{p_f}{m}$$\left(5\right)$

Calculation:

Sketches of “before” and ‘after” the brake is applied on the car is as shown in figure $\text{2}\left(a\right)$ and $\text{2}\left(b\right)$ respectively.

  

Figure $\text{2}\left(a\right)$

  

Figure $\text{2}\left(b\right)$

Initial momentum of the car is calculated by substituting the numerical values in equation $\left(3\right)$ ,

  $p_i=\left(\text{725 kg}\right)\left(32\text{ m/s}\right)=2.3\times {10}^4\text{ kg}\text{.m/s}$ toward the east.

Substituting the values of impulse from part (a) and initial momentum in equation $\left(4\right)$ , final momentum of the car is,

  $p_f=-1.0\times {10}^4\text{ kg}\text{.m/s}+2.3\times {10}^4\text{ kg}\text{.m/s}$

  $=1.3\times {10}^4\text{ kg}\text{.m/s}$

Final velocity of the car is calculated by substituting the value of final momentum and mass of the car in equation $\left(5\right)$ ,

  $v_f=\frac{1.3\times {10}^4\text{ kg}\text{.m/s}}{\text{725 kg}}=17.93\text{ m/s}$ toward the east.

Conclusion:

Initial momentum of the car is $2.3\times {10}^4\text{ kg}\cdot \text{m/s}$ toward the east.

Final momentum of the car is $1.3\times {10}^4\text{ kg}\cdot \text{m/s}$ toward the east.

Final velocity of the car is $17.93\text{ m/s}$ toward the east.