Chapter 5, Problem 8TE

In answering the preceding exercise, perhaps you stated that the automobile requires more stopping force. Make an argument that the skateboard could require more stopping force, depending on how quickly you want to stop it.

To determine

The argument that the skateboard requires more stopping force.

Answer to Problem 8TE

By reducing the time taken to stop, such that the ratio $\frac{P_{skate}}{t_{skate}}$ becomes larger than the ratio $\frac{P_{auto}}{t_{auto}}$, then force needed to stop the skateboard would be larger than the force needed to stop the automobile.

Explanation of Solution

Write the expression for the change in momentum.

$\begin{array}{l}\Delta P=0-P_f\\ \Delta P=-P_f\end{array}$

Here,

$P_f$ is the momentum

$\Delta P$ is the change in momentum

The change in momentum, which is also known as impulse is equal to the product of force and time over which the force acts on the body.

Write the expression for the relation of impulse with momentum.

$\begin{array}{c}\Delta P=-F\times t\\ F=\frac{\Delta P}{t}\end{array}$

Here,

$F$ is the force

$t$ is the time over which the force acts on the body

Write the expression for force needed stop the skateboard

$F_{skate}=\frac{P_{skate}}{t_{skate}}$.......(I)

Here,

$P_{skate}$ is the momentum of the skateboard

$F_{skate}$ is the force on skateboard

$t_{skate}$ is the time over which the force acts on the skateboard

Write the expression for force needed stop the skateboard

$F_{auto}=\frac{P_{auto}}{t_{auto}}$.......(II)

Here,

$P_{auto}$ is the momentum of the automobile

$F_{auto}$ is the force on the automobile

$t_{auto}$ is the time over which the force acts on the automobile

From equations (I) and (II) it is clear that both the automobile and the skateboard are moving with same velocity and both are to be stopped in same time. The force required to stop the automobile will be larger, since momentum of the automobile is larger than that of skateboard. If the time taken to stop is reduced, such that $\frac{P_{skate}}{t_{skate}}$ becomes larger than $\frac{P_{auto}}{t_{auto}}$, then force needed to stop the skateboard would be larger than the force needed to stop the automobile despite the fact that automobile has larger momentum.

Conclusion:

Therefore, by reducing the time taken to stop, such that the ratio $\frac{P_{skate}}{t_{skate}}$ becomes larger than the ratio $\frac{P_{auto}}{t_{auto}}$, then force needed to stop the skateboard would be larger than the force needed to stop the automobile.