Chapter 9, Problem 58A

(a)

To determine

The impulse needed to stop the child.

Answer to Problem 58A

  $-2.0\times {10}^2\text{ kg}\text{.m/s}$

Explanation of Solution

Given:

Initial velocity of the car and the child is $v_i=10.0\text{ m/s}$

The car crashes in to a barrier and stops in time, $\Delta t=0.050\text{ s}$

Since the child is in the car, time period for the child is also $\Delta t=0.050\text{ s}$ .

Mass of the child is $m\text{ = 20}\text{.0 kg}$

Final velocity of the car and the child is $v_f=0.0\text{ m/s}$

Formula used:

Impulse experienced by a body must be the change in momentum of that body. This is given by the expression,

  $F\Delta t=p_f-p_i$  $\left(1\right)$

Where,

$F$ is force on a body,

$\Delta t$ is the time period,

$p_f$ is the final momentum of a body, $p_f=m{v}_f$

$p_i$ is the initial momentum of a body, $p_i=m{v}_i$

$m$ is mass of a body,

$v_i$ and $v_f$ are the initial and final velocity of a body

Substituting for $p_f$ and $p_i$ in equation $\left(1\right)$ ,

  $F\Delta t=m{v}_f-m{v}_i=m\left(v_f-v_i\right)$  $\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $\text{F}\Delta \text{t = }\left(\text{20}\text{.0 kg}\right)\left[\left(0.0\text{ m/s}\right)-\left(10.0\text{ m/s}\right)\right]$

  $=-2.0\times {10}^2\text{ kg}\text{.m/s}$

Conclusion:

The impulse needed to stop the child is $-2.0\times {10}^2\text{kg}\cdot \text{m/s}$ .

(b)

To determine

The average force on the child.

Answer to Problem 58A

  $-4.0\times {10}^3\text{ N}$

Explanation of Solution

Given:

Initial velocity of the car and the child is $v_i=10.0\text{ m/s}$

The car crashes in to a barrier and stops in time $\Delta t=0.050\text{ s}$

Since the child is in the car, time period for the child is also $\Delta t=0.050\text{ s}$ .

Mass of the child is $m\text{ = 20}\text{.0 kg}$

Final velocity of the car and the child is $v_f=0.0\text{ m/s}$

Formula used:

From the equation $\left(2\right)$ , average force on the child can be calculated as,

  $F=\frac{m\left(v_f-v_i\right)}{\Delta t}$  $\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$ ,

  $F=\frac{\left(\text{20}\text{.0 kg}\right)\left[\left(0.0\text{ m/s}\right)-\left(10.0\text{ m/s}\right)\right]}{\left(\text{0}\text{.050 s}\right)}$

  $=-4.0\times {10}^3\text{ kg}\cdot {\text{m/s}}^2$

  $=-4.0\times {10}^3\text{ N}$

Conclusion:

The average force on the child is $-4.0\times {10}^3\text{ N}$ .

(c)

To determine

The approximate mass of an object whose weight equals the force in part b.

Answer to Problem 58A

  $4.1\times {10}^2\text{ kg}$

Explanation of Solution

Given:

Force, $F=4.0\times {10}^3\text{ kg}\cdot {\text{m/s}}^2$

Formula used:

Force $\left(F\right)$ and mass $\left(m\right)$ is connected by Newton’s second law of motion,

  $F=ma$  $\left(4\right)$

For weight calculation, acceleration $\left(a\right)$ is can be replaced by acceleration due to gravity $\left(g\right)$ , $g=9.8{\text{ m/s}}^2$

Hence the equation can be written as,

  $F=mg$

  $\Rightarrow m=\frac{F}{g}$  $\left(5\right)$

Calculation:

Substituting the numerical values in equation $\left(5\right)$ ,

  $m=\frac{4.0\times {10}^3\text{ kg}\cdot {\text{m/s}}^2}{9.8{\text{ m/s}}^2}$

  $=4.1\times {10}^2\text{ kg}$

Conclusion:

The approximate mass of an object whose weight equals the force in part b is $4.1\times {10}^2\text{ kg}$ .

(d)

To determine

Whether a person can lift a weight obtained in part c with his arm.

Answer to Problem 58A

No

Explanation of Solution

The approximate mass of an object obtained in the part c is $4.1\times {10}^2\text{ kg}$ . This much mass is too heavy to lift. Hence, a person cannot lift this much weight with his arm.

(e)

To determine

To Explain: It is advisable to use a proper restraining seat rather than hold a child on person’s lap.

Answer to Problem 58A

Because it is difficult to protect a child during an accident.

Explanation of Solution

Introduction:

People hold their children in their lap while travelling by vehicle. But it is not the safe way of travelling with children. Nowadays, a specially designed seat called child restraint seat or baby seat are available to protect children from getting injury or death during an accident. Child restraint seat offers passive restraint and it is very essential to use it properly to be effective.

A person will not be able to protect a child on his lap during an accident. Hence it is better to use a proper restraining seat rather than hold a child on person’s lap.