## Chapter 9

### Check Your Understanding

To reach a final speed of ${v}_{\text{f}}=\frac{1}{4}\left(3.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)$ at an acceleration of 10*g*, the time

required is

$\begin{array}{ccc}\hfill 10g& =\hfill & \frac{{v}_{\text{f}}}{\text{\Delta}t}\hfill \\ \hfill \text{\Delta}t& =\hfill & \frac{{v}_{\text{f}}}{10g}\phantom{\rule{0.2em}{0ex}}\frac{\frac{1}{4}\left(3.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}{10g}=7.7\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{s}=8.9\phantom{\rule{0.2em}{0ex}}\text{d}\hfill \end{array}$

If the phone bounces up with approximately the same initial speed as its impact speed, the change in momentum of the phone will be $\text{\Delta}\overrightarrow{p}=m\text{\Delta}\overrightarrow{v}-\left(\text{\u2212}m\text{\Delta}\overrightarrow{v}\right)=2m\text{\Delta}\overrightarrow{v}$. This is twice the momentum change than when the phone does not bounce, so the impulse-momentum theorem tells us that more force must be applied to the phone.

If the smaller cart were rolling at 1.33 m/s to the left, then conservation of momentum gives

$\begin{array}{ccc}\hfill \left({m}_{1}+{m}_{2}\right){\overrightarrow{v}}_{f}& =\hfill & {m}_{1}{v}_{1}\widehat{i}-{m}_{2}{v}_{2}\widehat{i}\hfill \\ \hfill {\overrightarrow{v}}_{f}& =\hfill & \left(\frac{{m}_{1}{v}_{1}-{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}\right)\widehat{i}\hfill \\ & =\hfill & \left[\frac{\left(0.675\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(0.75\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)-\left(0.500\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(1.33\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}{1.175\phantom{\rule{0.2em}{0ex}}\text{kg}}\right]\widehat{i}\hfill \\ & =\hfill & \text{\u2212}\left(0.135\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}\hfill \end{array}$

Thus, the final velocity is 0.135 m/s to the left.

If the ball does not bounce, its final momentum ${\overrightarrow{p}}_{2}$ is zero, so

$\begin{array}{cc}\hfill \text{\Delta}\overrightarrow{p}& ={\overrightarrow{p}}_{2}-{\overrightarrow{p}}_{1}\hfill \\ & =\left(0\right)\widehat{j}-\left(\mathrm{-1.4}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}\right)\widehat{j}\hfill \\ & =+\left(1.4\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}\right)\widehat{j}\hfill \end{array}$

Consider the impulse momentum theory, which is $\overrightarrow{J}=\text{\Delta}\overrightarrow{p}$. If $\overrightarrow{J}=0$, we have the situation described in the example. If a force acts on the system, then $\overrightarrow{J}={\overrightarrow{F}}_{\text{ave}}\text{\Delta}t$. Thus, instead of ${\overrightarrow{p}}_{f}={\overrightarrow{p}}_{i}$, we have

${\overrightarrow{F}}_{\text{ave}}\text{\Delta}t=\text{\Delta}\overrightarrow{p}={\overrightarrow{p}}_{\text{f}}-{\overrightarrow{p}}_{\text{i}}$

where ${\overrightarrow{F}}_{\text{ave}}$ is the force due to friction.

The impulse is the change in momentum multiplied by the time required for the change to occur. By conservation of momentum, the changes in momentum of the probe and the comment are of the same magnitude, but in opposite directions, and the interaction time for each is also the same. Therefore, the impulse each receives is of the same magnitude, but in opposite directions. Because they act in opposite directions, the impulses are not the same. As for the impulse, the force on each body acts in opposite directions, so the forces on each are not equal. However, the change in kinetic energy differs for each, because the collision is not elastic.

This solution represents the case in which no interaction takes place: the first puck misses the second puck and continues on with a velocity of 2.5 m/s to the left. This case offers no meaningful physical insights.

If zero friction acts on the car, then it will continue to slide indefinitely ($d\to \infty $), so we cannot use the work-kinetic-energy theorem as is done in the example. Thus, we could not solve the problem from the information given.

Were the initial velocities not at right angles, then one or both of the velocities would have to be expressed in component form. The mathematical analysis of the problem would be slightly more involved, but the physical result would not change.

The volume of a scuba tank is about 11 L. Assuming air is an ideal gas, the number of gas molecules in the tank is

$\begin{array}{ccc}\hfill PV& =\hfill & NRT\hfill \\ \hfill N& =\hfill & \frac{PV}{RT}=\frac{\left(2500\phantom{\rule{0.2em}{0ex}}\text{psi}\right)\left(0.011\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\right)}{\left(8.31\phantom{\rule{0.2em}{0ex}}\text{J/mol}\xb7\text{K}\right)\left(300\phantom{\rule{0.2em}{0ex}}\text{K}\right)}\left(\frac{6894.8\phantom{\rule{0.2em}{0ex}}\text{Pa}}{1\phantom{\rule{0.2em}{0ex}}\text{psi}}\right)\hfill \\ & =\hfill & 7.59\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{1}\phantom{\rule{0.2em}{0ex}}\text{mol}\hfill \end{array}$

The average molecular mass of air is 29 g/mol, so the mass of air contained in the tank is about 2.2 kg. This is about 10 times less than the mass of the tank, so it is safe to neglect it. Also, the initial force of the air pressure is roughly proportional to the surface area of each piece, which is in turn proportional to the mass of each piece (assuming uniform thickness). Thus, the initial acceleration of each piece would change very little if we explicitly consider the air.

The average radius of Earth’s orbit around the Sun is $1.496\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{m}$. Taking the Sun to be the origin, and noting that the mass of the Sun is approximately the same as the masses of the Sun, Earth, and Moon combined, the center of mass of the Earth + Moon system and the Sun is

$\begin{array}{cc}\hfill {R}_{\text{CM}}& =\frac{{m}_{\mathrm{Sun}}{R}_{\mathrm{Sun}}+{m}_{\text{em}}{R}_{\text{em}}}{{m}_{\mathrm{Sun}}}\hfill \\ & =\frac{\left(1.989\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(0\right)+\left(5.97\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{24}\phantom{\rule{0.2em}{0ex}}\text{kg}+7.36\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{22}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(1.496\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}{1.989\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}}\hfill \\ & =4.6\phantom{\rule{0.2em}{0ex}}\text{km}\hfill \end{array}$

Thus, the center of mass of the Sun, Earth, Moon system is 4.6 km from the center of the Sun.

On a macroscopic scale, the size of a unit cell is negligible and the crystal mass may be considered to be distributed homogeneously throughout the crystal. Thus,

${\overrightarrow{r}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^{N}{m}_{j}{\overrightarrow{r}}_{j}}=\frac{1}{M}{\displaystyle \sum _{j=1}^{N}m}{\overrightarrow{r}}_{j}=\frac{m}{M}{\displaystyle \sum _{j=1}^{N}{\overrightarrow{r}}_{j}}=\frac{Nm}{M}\phantom{\rule{0.2em}{0ex}}\frac{{\displaystyle \sum _{j=1}^{N}{\overrightarrow{r}}_{j}}}{N}$

where we sum over the number *N* of unit cells in the crystal and *m* is the mass of a unit cell. Because *Nm* = *M*, we can write

${\overrightarrow{r}}_{\text{CM}}=\frac{m}{M}{\displaystyle \sum _{j=1}^{N}{\overrightarrow{r}}_{j}}=\frac{Nm}{M}\phantom{\rule{0.2em}{0ex}}\frac{{\displaystyle \sum _{j=1}^{N}{\overrightarrow{r}}_{j}}}{N}=\frac{1}{N}{\displaystyle \sum _{j=1}^{N}{\overrightarrow{r}}_{j}}.$

This is the definition of the geometric center of the crystal, so the center of mass is at the same point as the geometric center.

The explosions would essentially be spherically symmetric, because gravity would not act to distort the trajectories of the expanding projectiles.

The notation ${m}_{g}$ stands for the mass of the fuel and *m* stands for the mass of the rocket plus the initial mass of the fuel. Note that ${m}_{g}$ changes with time, so we write it as ${m}_{g}\left(t\right)$. Using ${m}_{R}$ as the mass of the rocket with no fuel, the total mass of the rocket plus fuel is $m={m}_{R}+{m}_{g}\left(t\right)$. Differentiation with respect to time gives

$\frac{dm}{dt}=\frac{d{m}_{R}}{dt}+\frac{d{m}_{g}\left(t\right)}{dt}=\frac{d{m}_{g}\left(t\right)}{dt}$

where we used $\frac{d{m}_{R}}{dt}=0$ because the mass of the rocket does not change. Thus, time rate of change of the mass of the rocket is the same as that of the fuel.

### Conceptual Questions

Since $K={p}^{2}\text{/}2m$, then if the momentum is fixed, the object with smaller mass has more kinetic energy.

Yes; impulse is the force applied multiplied by the time during which it is applied ($J=F\text{\Delta}t$), so if a small force acts for a long time, it may result in a larger impulse than a large force acting for a small time.

By friction, the road exerts a horizontal force on the tires of the car, which changes the momentum of the car.

Momentum is conserved when the mass of the system of interest remains constant during the interaction in question and when no *net* external force acts on the system during the interaction.

To accelerate air molecules in the direction of motion of the car, the car must exert a force on these molecules by Newton’s second law $\overrightarrow{F}=d\overrightarrow{p}\text{/}dt$. By Newton’s third law, the air molecules exert a force of equal magnitude but in the opposite direction on the car. This force acts in the direction opposite the motion of the car and constitutes the force due to air resistance.

No, he is not a closed system because a net nonzero external force acts on him in the form of the starting blocks pushing on his feet.

Yes, all the kinetic energy can be lost if the two masses come to rest due to the collision (i.e., they stick together).

The angle between the directions must be 90°. Any system that has zero net external force in one direction and nonzero net external force in a perpendicular direction will satisfy these conditions.

Yes, the rocket speed can exceed the exhaust speed of the gases it ejects. The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.

### Problems

$1.78\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{29}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}$

$1.3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}$

a. $1.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{N}$; b. $1.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$

$4.69\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$

$2.10\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}$

$\overrightarrow{p}(t)=\left(10\widehat{i}+20t\widehat{j}\right)\text{kg}\xb7\text{m}\text{/}\text{s}$;$\overrightarrow{F}=\left(20\phantom{\rule{0.2em}{0ex}}\text{N}\right)\widehat{j}$

Let the positive *x*-axis be in the direction of the original momentum. Then ${p}_{x}=1.5\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}$ and ${p}_{y}=7.5\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}$

a. 47 m/s in the bullet to block direction; b.$70.6\phantom{\rule{0.2em}{0ex}}\text{N}\xb7\text{s}$, toward the bullet; c. $70.6\phantom{\rule{0.2em}{0ex}}\text{N}\xb7\text{s}$, toward the block; d. magnitude is $2.35\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}$

the speed of the leading bumper car is 6.00 m/s and that of the trailing bumper car is 5.60 m/s

$\left(732\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(\mathrm{-79.6}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$

$\text{\u2212}\left(0.21\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(0.25\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$

With the origin defined to be at the position of the 150-g mass, ${x}_{\text{CM}}=\mathrm{-1.23}\text{cm}$ and ${y}_{\text{CM}}=0.69\text{cm}$

${y}_{\text{CM}}=\{\begin{array}{c}\hfill \frac{h}{2}-\frac{1}{4}g{t}^{2},\phantom{\rule{0.5em}{0ex}}t<T\hfill \\ \hfill h-\frac{1}{2}g{t}^{2}-\frac{1}{4}g{T}^{2}+\frac{1}{2}gtT,\phantom{\rule{0.5em}{0ex}}t\ge T\hfill \end{array}$

a. ${R}_{1}=4\phantom{\rule{0.2em}{0ex}}\text{m}$, ${R}_{2}=2\phantom{\rule{0.2em}{0ex}}\text{m}$; b. ${X}_{\text{CM}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}},{Y}_{\text{CM}}=\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}}{{m}_{1}+{m}_{2}}$; c. yes, with $R=\frac{1}{{m}_{1}+{m}_{2}}\sqrt{16{m}_{1}^{2}+4{m}_{2}^{2}}$

${x}_{cm}=\frac{3}{4}\phantom{\rule{0.2em}{0ex}}L\left(\frac{{\rho}_{1}+{\rho}_{0}}{{\rho}_{1}+2{\rho}_{0}}\right)$

### Additional Problems

Answers may vary. The first clause is true, but the second clause is not true in general because the velocity of an object with small mass may be large enough so that the momentum of the object is greater than that of a larger-mass object with a smaller velocity.

$4.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}$

$\overrightarrow{J}={\displaystyle {\int}_{0}^{\tau}\left[m\overrightarrow{g}-m\overrightarrow{g}\left(1-{e}^{\text{\u2212}bt\text{/}m}\right)\right]dt}=\frac{{m}^{2}}{b}\overrightarrow{g}\left({e}^{\text{\u2212}b\tau \text{/}m}-1\right)$

a. $\text{\u2212}\left(2.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}\right)\widehat{i}$, b. $\text{\u2212}\left(24\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}\right)\widehat{i}$

a. $\left(1.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}\right)\widehat{i}$, b. $\left(0.010\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7\text{m/s}\right)\widehat{i}$, c. $\text{\u2212}\left(0.00093\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}$, d. $\text{\u2212}\left(0.0012\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}$

${v}_{\text{1,f}}={v}_{\text{1,i}}\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}},\phantom{\rule{0.2em}{0ex}}{v}_{\text{2,f}}={v}_{\text{1,i}}\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}$

final velocity of cue ball is $\text{\u2212}\left(0.76\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}$, final velocities of the other two balls are 2.6 m/s at ±30° with respect to the initial velocity of the cue ball

ball 1: $\text{\u2212}\left(1.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(0.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$, ball 2: $\left(2.2\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(2.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$

ball 1: $\left(1.4\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}-\left(1.7\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$, ball 2: $\text{\u2212}\left(2.8\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{i}+\left(0.012\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\widehat{j}$

Answers may vary. The rocket is propelled forward not by the gasses pushing against the surface of Earth, but by conservation of momentum. The momentum of the gas being expelled out the back of the rocket must be compensated by an increase in the forward momentum of the rocket.

### Challenge Problems

a. $617\phantom{\rule{0.2em}{0ex}}\text{N}\xb7\text{s}$, 108°; b. ${F}_{x}=2.91\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}$, ${F}_{y}=2.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$; c. ${F}_{x}=5850\phantom{\rule{0.2em}{0ex}}\text{N}$, ${F}_{y}=5265\phantom{\rule{0.2em}{0ex}}\text{N}$

Conservation of momentum demands ${m}_{1}{v}_{\text{1,i}}+{m}_{2}{v}_{\text{2,i}}={m}_{1}{v}_{\text{1,f}}+{m}_{2}{v}_{\text{2,f}}$. We are given that ${m}_{1}={m}_{2}$, ${v}_{\text{1,i}}={v}_{\text{2,f}}$, and ${v}_{\text{2,i}}={v}_{\text{1,f}}=0$. Combining these equations with the equation given by conservation of momentum gives ${v}_{\text{1,i}}={v}_{\text{1,i}}$, which is true, so conservation of momentum is satisfied. Conservation of energy demands ${\scriptscriptstyle \frac{1}{2}}{m}_{1}{v}_{\text{1,i}}^{2}+{\scriptscriptstyle \frac{1}{2}}{m}_{2}{v}_{\text{2,i}}^{2}={\scriptscriptstyle \frac{1}{2}}{m}_{1}{v}_{\text{1,f}}^{2}+{\scriptscriptstyle \frac{1}{2}}{m}_{2}{v}_{\text{2,f}}^{2}$. Again combining this equation with the conditions given above give ${v}_{\text{1,i}}={v}_{\text{1,i}}$, so conservation of energy is satisfied.

Assume origin on centerline and at floor, then $\left({x}_{\text{CM}},{y}_{\text{CM}}\right)=\left(\mathrm{0,86}\phantom{\rule{0.2em}{0ex}}\text{cm}\right)$