Content check — University Physics 1, Chapter 9: Linear Momentum and Collisions

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 9: Linear Momentum and Collisions

73 problems · 15 figures

Check Your Understanding 9.1 §9.2 source ↗

The U.S. Air Force uses “10gs” (an acceleration equal to

10 \times 9.8 {m/s}^{2}

) as the maximum acceleration a human can withstand (but only for several seconds) and survive. How much time must the Enterprise spend accelerating if the humans on board are to experience an average of at most 10gs of acceleration? (Assume the inertial dampeners are offline.)

Answer

To reach a final speed of

v_{f} =

\frac{1}{4} (3.0 \times 10^{8} m/s)

at an acceleration of 10g, the time
required is

\begin{matrix} 10 g & = & \frac{v_{f}}{Δ t} \\ Δ t & = & \frac{v_{f}}{10 g} \frac{\frac{1}{4} (3.0 \times 10^{8} m/s)}{10 g} = 7.7 \times 10^{5} s = 8.9 d \end{matrix}

Check Your Understanding 9.2 §9.2 source ↗

What if we had assumed the phone did bounce on impact? Would this have increased the force on the iPhone, decreased it, or made no difference?

Answer

If the phone bounces up with approximately the same initial speed as its impact speed, the change in momentum of the phone will be

Δ \vec{p} =

m Δ \vec{v} -

(− m Δ \vec{v}) =

2 m Δ \vec{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.

Check Your Understanding 9.3 §9.3 source ↗

Suppose the second, smaller cart had been initially moving to the left. What would the sign of the final velocity have been in this case?

Answer

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

\begin{matrix} (m_{1} + m_{2}) {\vec{v}}_{f} & = & m_{1} v_{1} \hat{i} - m_{2} v_{2} \hat{i} \\ {\vec{v}}_{f} & = & (\frac{m_{1} v_{1} - m_{2} v_{2}}{m_{1} + m_{2}}) \hat{i} \\ = & [\frac{(0.675 kg) (0.75 m/s) - (0.500 kg) (1.33 m/s)}{1.175 kg}] \hat{i} \\ = & − (0.135 m/s) \hat{i} \end{matrix}

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

Check Your Understanding 9.4 §9.3 source ↗

Would the ball’s change of momentum have been larger, smaller, or the same, if it had collided with the floor and stopped (without bouncing)?

Answer

If the ball does not bounce, its final momentum

{\vec{p}}_{2}

is zero, so

\begin{matrix} Δ \vec{p} & = {\vec{p}}_{2} - {\vec{p}}_{1} \\ = (0) \hat{j} - (−1.4 kg \cdot m/s) \hat{j} \\ = + (1.4 kg \cdot m/s) \hat{j} \end{matrix}

Check Your Understanding 9.5 §9.3 source ↗

Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you would need to impose an additional condition on the problem. What is that additional condition?

Answer

Consider the impulse momentum theory, which is

\vec{J} = Δ \vec{p}

. If

\vec{J} = 0

, we have the situation described in the example. If a force acts on the system, then

\vec{J} = {\vec{F}}_{ave} Δ t

. Thus, instead of

{\vec{p}}_{f} = {\vec{p}}_{i}

, we have

{\vec{F}}_{ave} Δ t =

Δ \vec{p} =

{\vec{p}}_{f} -

{\vec{p}}_{i}

where

{\vec{F}}_{ave}

is the force due to friction.

Check Your Understanding 9.6 §9.3 source ↗

The changes of momentum for Philae and for Comet 67/P were equal (in magnitude). Were the impulses experienced by Philae and the comet equal? How about the forces? How about the changes of kinetic energies?

Answer

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.

Check Your Understanding 9.7 §9.4 source ↗

There is a possible mathematical second to the system of equations solved in this example (because the energy equation is quadratic):

v_{1,f} =

0,

v_{2,f} =

−2.5 m/s

. This solution is unacceptable on physical grounds; what’s wrong with it?

Answer

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.

Check Your Understanding 9.8 §9.4 source ↗

Suppose there had been no friction (the collision happened on ice); that would make

μ_{k}

zero, and thus

v_{c,f} =

\sqrt{2 μ_{k} g d} =

0

, which is obviously wrong. What is the mistake in this conclusion?

Answer

If zero friction acts on the car, then it will continue to slide indefinitely (

d \to \infty

), so we cannot use the work-energy theorem as is done in the example. Thus, we could not solve the problem from the information given.

Check Your Understanding 9.9 §9.5 source ↗

Suppose the initial velocities were not at right angles to each other. How would this change both the physical result and the mathematical analysis of the collision?

Answer

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.

Check Your Understanding 9.10 §9.5 source ↗

Notice that the mass of the air in the tank was neglected in the analysis and solution. How would the solution method changed if the air was included? How large a difference do you think it would make in the final answer?

Answer

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{matrix} P V & = & N R T \\ N & = & \frac{P V}{R T} = \frac{(2500 psi) (0.011 m^{3})}{(8.31 J/mol \cdot K) (300 K)} (\frac{6894.8 Pa}{1 psi}) \\ = & 7.59 \times 10^{1} mol \end{matrix}

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.

Check Your Understanding 9.11 §9.6 source ↗

Suppose we included the sun in the system. Approximately where would the center of mass of the Earth-moon-sun system be located? (Feel free to actually calculate it.)

Answer

The average radius of Earth’s orbit around the Sun is

1.496 \times 10^{11} 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{matrix} R_{CM} & = \frac{m_{Sun} R_{Sun} + m_{em} R_{em}}{m_{Sun}} \\ = \frac{(1.989 \times 10^{30} kg) (0) + (5.97 \times 10^{24} kg + 7.36 \times 10^{22} kg) (1.496 \times 10^{11} m)}{1.989 \times 10^{30} kg} \\ = 455 km \end{matrix}

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

Check Your Understanding 9.12 §9.6 source ↗

Suppose you have a macroscopic salt crystal (that is, a crystal that is large enough to be visible with your unaided eye). It is made up of a huge number of unit cells. Is the center of mass of this crystal necessarily at the geometric center of the crystal?

Answer

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,

{\vec{r}}_{CM} =

\frac{1}{M} \sum_{j = 1}^{N} m_{j} {\vec{r}}_{j} =

\frac{1}{M} \sum_{j = 1}^{N} m {\vec{r}}_{j} =

\frac{m}{M} \sum_{j = 1}^{N} {\vec{r}}_{j} =

\frac{N m}{M} \frac{\sum_{j = 1}^{N} {\vec{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

{\vec{r}}_{CM} =

\frac{m}{M} \sum_{j = 1}^{N} {\vec{r}}_{j} =

\frac{N m}{M} \frac{\sum_{j = 1}^{N} {\vec{r}}_{j}}{N} =

\frac{1}{N} \sum_{j = 1}^{N} {\vec{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.

Check Your Understanding 9.13 §9.6 source ↗

How would the firework display change in deep space, far away from any source of gravity?

Answer

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

Check Your Understanding 9.14 §9.7 source ↗

What is the physical difference (or relationship) between

\frac{d m}{d t}

and

\frac{d m_{g}}{d t}

in this example?

Answer

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} (t)

. 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} (t)

. Differentiation with respect to time gives

\frac{d m}{d t} =

\frac{d m_{R}}{d t} +

\frac{d m_{g} (t)}{d t} =

\frac{d m_{g} (t)}{d t}

where we used

\frac{d m_{R}}{d t} = 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.

9-1 §9.1 source ↗

An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?

Answer

Since

K = p^{2} / 2 m

, then if the momentum is fixed, the object with smaller mass has more kinetic energy.

9-3 §9.2 source ↗

Is it possible for a small force to produce a larger impulse on a given object than a large force? Explain.

Answer

Yes; impulse is the force applied multiplied by the time during which it is applied (

J = F Δ 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.

9-5 §9.2 source ↗

What external force is responsible for changing the momentum of a car moving along a horizontal road?

Answer

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

9-7 §9.3 source ↗

Under what circumstances is momentum conserved?

Answer

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.

9-9 §9.3 source ↗

Explain in terms of momentum and Newton’s laws how a car’s air resistance is due in part to the fact that it pushes air in its direction of motion.

Answer

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

\vec{F} = d \vec{p} / d t

. 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.

9-11 §9.3 source ↗

A sprinter accelerates out of the starting blocks. Can you consider him as a closed system? Explain.

Answer

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.

9-13 §9.4 source ↗

Two objects of equal mass are moving with equal and opposite velocities when they collide. Can all the kinetic energy be lost in the collision?

Answer

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

9-15 §9.5 source ↗

Momentum for a system can be conserved in one direction while not being conserved in another. What is the angle between the directions? Give an example.

Answer

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.

9-17 §9.7 source ↗

It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases?

Answer

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.

9-19 §9.1 source ↗

A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding without any friction on a smooth surface.

a. Find the momentum of the box with respect to the floor.

b. Find the momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface.

Answer

a. magnitude:

25 kg \cdot m/s;

b. same as a.

9-21 §9.1 source ↗

The mass of Earth is $5.97 \times 10^{24} kg$ and its orbital radius is an average of $1.50 \times 10^{11} m$ . Calculate the magnitude of its linear momentum at the location in the diagram.

An illustration of the earth orbiting the sun. The mass of the earth is given as 5.97 times 10 to the 24 kilograms and the radius of the orbit is labeled R earth = 1.5 times 10 to the 11 meters.

Answer

1.78 \times 10^{29} kg \cdot m/s

9-23 §9.1 source ↗

What is the average momentum of an avalanche that moves a 40-cm-thick layer of snow over an area of 100 m by 500 m over a distance of 1 km down a hill in 5.5 s? Assume a density of 350 kg/m³ for the snow.

Answer

1.3 \times 10^{9} kg \cdot m/s

9-25 §9.2 source ↗

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment (see the following figure).

A drawing of a car on a bridge. The car is labeled as having velocity v sub i equals 20 meters per second i hat to the right.

a. Calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm.

b. Calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 cm.

Answer

1.50 \times 10^{6} N

; b.

1.00 \times 10^{5} N

9-27 §9.2 source ↗

A cruise ship with a mass of $1.00 \times 10^{7} kg$ strikes a pier at a speed of 0.750 m/s. It comes to rest after traveling 6.00 m, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest, assuming a constant force.)

A drawing of a ship hitting a pier. The ship is moving to the right with v sub i equals 0.750 meters per second.

Answer

4.69 \times 10^{5} N

9-29 §9.2 source ↗

Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero.

Answer

2.10 \times 10^{3} N

9-31 §9.2 source ↗

What is the momentum (as a function of time) of a 5.0-kg particle moving with a velocity

\vec{v} (t) =

(2.0 \hat{i} + 4.0 t \hat{j}) m/s

? What is the net force acting on this particle?

Answer

\vec{p} (t) =

(10 \hat{i} + 20 t \hat{j}) kg \cdot m / s

;

\vec{F} = (20 N) \hat{j}

9-33 §9.2 source ↗

A hockey puck of mass 150 g is sliding due east on a frictionless table with a speed of 10 m/s. Suddenly, a constant force of magnitude 5 N and direction due north is applied to the puck for 1.5 s. Find the north and east components of the momentum at the end of the 1.5-s interval.

A puck is shown with force F equals 5.0 N north and v sub I = 10 meters per second east.

Answer

Let the positive x-axis be in the direction of the original momentum. Then

p_{x} =

1.5 kg \cdot m/s

and

p_{y} =

7.5 kg \cdot m/s

9-35 §9.3 source ↗

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of $1.50 \times 10^{5} kg$ and a velocity of $(0.30 m/s) \hat{i}$ , and the second having a mass of $1.10 \times 10^{5} kg$ and a velocity of $− (0.12 m/s) \hat{i}$ . What is their final velocity?

Answer

(0.122 m/s) \hat{i}

9-37 §9.3 source ↗

The figure below shows a bullet of mass 200 g traveling horizontally towards the east with speed 400 m/s, which strikes a block of mass 1.5 kg that is initially at rest on a frictionless table.

A drawing of a block on a table, and a bullet headed toward it.

After striking the block, the bullet is embedded in the block and the block and the bullet move together as one unit.

a. What is the magnitude and direction of the velocity of the block/bullet combination immediately after the impact?

b. What is the magnitude and direction of the impulse by the block on the bullet?

c. What is the magnitude and direction of the impulse from the bullet on the block?

d. If it took 3 ms for the bullet to change the speed from 400 m/s to the final speed after impact, what is the average force between the block and the bullet during this time?

Answer

a. 47 m/s in the bullet to block direction; b.

70.6 N \cdot s

, toward the bullet; c.

70.6 N \cdot s

, toward the block; d. magnitude is

2.35 \times 10^{4} N

9-39 §9.3 source ↗

A 4.5 kg puffer fish expands to 40% of its mass by taking in water. When the puffer fish is threatened, it releases the water toward the threat to move quickly forward. What is the ratio of the speed of the puffer fish forward to the speed of the expelled water backwards?

Answer

2:5

9-41 §9.3 source ↗

Two figure skaters are coasting in the same direction, with the leading skater moving at 5.5 m/s and the trailing skating moving at 6.2 m/s. When the trailing skater catches up with the leading skater, he picks her up without applying any horizontal forces on his skates. If the trailing skater is 50% heavier than the 50-kg leading skater, what is their speed after he picks her up?

Answer

5.9 m/s

9-43 §9.4 source ↗

A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 15.8° to the initial direction of the bowling ball and with a speed of 15.0 m/s.

a. Calculate the final velocity (magnitude and direction) of the bowling ball.

b. Is the collision elastic?

Answer

a. 6.80 m/s, –5.33°; b. yes (calculate the ratio of the initial and final kinetic energies)

9-45 §9.4 source ↗

A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away?

Answer

2.5 cm

9-47 §9.4 source ↗

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.60 m/s and that of the trailing car is 6.00 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds?

Answer

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

9-49 §9.4 source ↗

An alpha particle (⁴He) undergoes an elastic collision with a stationary uranium nucleus (²³⁵U). What percent of the kinetic energy of the alpha particle is transferred to the uranium nucleus? Assume the collision is one-dimensional.

Answer

6.6%

9-51 §9.4 source ↗

A 35-kg child rides a relatively massless sled down a hill and then coasts along the flat section at the bottom, where a second 35-kg child jumps on the sled as it passes by her. If the speed of the sled is 3.5 m/s before the second child jumps on, what is its speed after she jumps on?

Answer

1.8 m/s

9-53 §9.5 source ↗

A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of

35 °

. It catches a 0.325-kg pigeon from behind in midair. What is their combined velocity after impact if the pigeon’s initial velocity was 7.00 m/s directed horizontally? Note that

{\hat{v}}_{1, i}

is a unit vector pointing in the direction in which the falcon is initially flying.

A hawk is flying toward a dove. The hawk is moving in a direction that is 35 degrees down from the horizontal at v 1 i = 28.0 meters per second v 1 i hat. The dove is moving to the right at 7.00 meters per second i hat. — Figure 9.34 in the book ↗

Answer

22.1 m/s at

32.2 °

below the horizontal

9-55 §9.5 source ↗

A projectile of mass 2.0 kg is ﬁred in the air at an angle of 40.0 $°$ to the horizon at a speed of 50.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0-kg part falls straight down after breakup with an initial speed of 10.0 m/s, the 0.7-kg part moves in the original forward direction, and the 0.3-kg part goes straight up.

A bullet at launch has v sub i = 50.0 meters per second directed at 40 degrees above the horizontal. At peak before explosion, the bullet is directed to the right with vector v sub i, x = v sub i x x hat. At leak after explosion, there are three pieces. M 1 = 1.0 k g has v 1 f = minus 10 meters per second j hat, downward. M 2 = 0.7 k g has vector v sub 2, f = v sub 2 f i hat to the right. M 3 = 0.3 k g has vector v sub 3, f = v sub 3 f j hat up..

a. Find the speeds of the 0.3-kg and 0.7-kg pieces immediately after the break-up.

b. How high from the break-up point does the 0.3-kg piece go before coming to rest?

c. Where does the 0.7-kg piece land relative to where it was ﬁred from?

Answer

a. 33 m/s and 110 m/s; b. 57 m; c. 480 m

9-57 §9.5 source ↗

A 200-kg rocket in deep space moves with a velocity of

(121 m/s) \hat{i} +

(38.0 m/s) \hat{j}

. Suddenly, it explodes into three pieces, with the first (78 kg) moving at

− (321 m/s) \hat{i} +

(228 m/s) \hat{j}

and the second (56 kg) moving at

(16.0 m/s) \hat{i} -

(88.0 m/s) \hat{j}

. Find the velocity of the third piece.

Answer

(732 m/s) \hat{i} +

(−79.6 m/s) \hat{j}

9-59 §9.5 source ↗

Three 70-kg deer are standing on a flat 200-kg rock that is on an ice-covered pond. A gunshot goes off and the deer scatter, with deer A running at

(15 m/s) \hat{i} +

(5.0 m/s) \hat{j}

, deer B running at

(−12 m/s) \hat{i} +

(8.0 m/s) \hat{j}

, and deer C running at

(1.2 m/s) \hat{i} -

(18.0 m/s) \hat{j}

. What is the velocity of the rock on which they were standing?

Answer

− (1.47 m/s) \hat{i} +

(1.75 m/s) \hat{j}

9-61 §9.5 source ↗

An oxygen atom (mass 16 u) moving at 733 m/s at 15.0° with respect to the

\hat{i}

direction collides and sticks to an oxygen molecule (mass 32 u) moving at 528 m/s at 128° with respect to the

\hat{i}

direction. The two stick together to form ozone. What is the final velocity of the ozone molecule?

Answer

341 m/s at 86.8° with respect to the

\hat{i}

axis.

9-63 §9.6 source ↗

Three point masses are placed at the corners of a triangle as shown in the figure below.

A right triangle with sides length 3 c m and 4 c m has masses of 100 g at the vertex between the hypotenuse and the 4 c m side, 75 g at the vertex between the hypotenuse and the 3 c m side, and 150 g at the vertex between the the 3 c m side and the 4 c m side.

Find the center of mass of the three-mass system.

Answer

With the origin defined to be at the position of the 150-g mass,

x_{CM} = −1.23 cm

and

y_{CM} = 0.69 cm

9-65 §9.6 source ↗

Two particles of masses

m_{1}

and

m_{2}

separated by a horizontal distance D are let go from the same height h at different times. Particle 1 starts at

t = 0

, and particle 2 is let go at

t = T

. Find the vertical position of the center of mass at a time before the first particle strikes the ground. Assume no air resistance.

Answer

y_{CM} = {\begin{matrix} \frac{h}{2} - \frac{1}{4} g t^{2}, t < T \\ h - \frac{1}{2} g t^{2} - \frac{1}{4} g T^{2} + \frac{1}{2} g t T, t \geq T \end{matrix}

9-67 §9.6 source ↗

Two particles of masses $m_{1}$ and $m_{2}$ move uniformly in different circles of radii $R_{1}$ and $R_{2}$ about the origin in the x, y-plane. The coordinates of the two particles in meters are given as follows ( $z = 0$ for both). Here t is in seconds:
$\begin{matrix} x_{1} (t) & = & 4 cos (2 t) \\ y_{1} (t) & = & 4 sin (2 t) \\ x_{2} (t) & = & 2 cos (3 t - \frac{π}{2}) \\ y_{2} (t) & = & 2 sin (3 t - \frac{π}{2}) \end{matrix}$

a. Find the radii of the circles of motion of both particles.

b. Find the x- and y-coordinates of the center of mass.

c. Decide if the center of mass moves in a circle by plotting its trajectory.

Answer

R_{1} = 4 m

R_{2} = 2 m

; b.

X_{CM} =

\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}},

Y_{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}}

9-69 §9.6 source ↗

Find the center of mass of a rod of length L whose mass density changes from one end to the other quadratically. That is, if the rod is laid out along the x-axis with one end at the origin and the other end at

x = L

, the density is given by

ρ (x) =

ρ_{0} +

(ρ_{1} - ρ_{0}) {(\frac{x}{L})}^{2}

, where

ρ_{0}

and

ρ_{1}

are constant values.

Answer

x_{c m} =

\frac{3}{4} L (\frac{ρ_{1} + ρ_{0}}{ρ_{1} + 2 ρ_{0}})

9-71 §9.6 source ↗

Find the center of mass of a rectangular material of length a and width b made up of a material of nonuniform density. The density is such that when the rectangle is placed in the xy-plane, the density is given by

ρ (x, y) = ρ_{0} x y

Answer

(\frac{2 a}{3}, \frac{2 b}{3})

9-73 §9.6 source ↗

Find the center of mass of a cone of uniform density that has a radius R at the base, height h, and mass M. Let the origin be at the center of the base of the cone and have +z going through the cone vertex.

Answer

(x_{CM}, y_{CM}, z_{CM}) =

(0,0, h / 4)

9-75 §9.6 source ↗

Find the center of mass of a uniform thin semicircular plate of radius R. Let the origin be at the center of the semicircle, the plate arc from the +x axis to the −x axis, and the z axis be perpendicular to the plate.

Answer

(x_{CM}, y_{CM}, z_{CM}) =

(0, 4 R / (3 π), 0)

9-77 §9.7 source ↗

(a) A 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-N frictional force opposing the squid’s movement?

(b) How much energy is lost to work done against friction?

Answer

(a) 0.413 m/s, (b) about 0.2 J

9-79 §9.7 source ↗

Repeat the preceding problem but for a rocket that takes off from a space station, where there is no gravity other than the negligible gravity due to the space station.

Answer

1551 kg

9-81 §9.7 source ↗

What exhaust speed is required to accelerate a rocket in deep space from 800 m/s to 1000 m/s in 5.0 s if the total rocket mass is 1200 kg and the rocket only has 50 kg of fuel left?

Answer

4.9 km/s

9-84 §9.AP source ↗

Which has a larger magnitude of momentum: a 3000-kg elephant moving at 40 km/h or a 60-kg cheetah moving at 112 km/h?

Answer

the elephant has a higher momentum

9-86 §9.AP source ↗

You friend claims that momentum is mass multiplied by velocity, so things with more mass have more momentum. Do you agree? Explain.

Answer

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.

9-88 §9.AP source ↗

Your 1500-kg sports car accelerates from 0 to 30 m/s in 10 s. What average force is exerted on it during this acceleration?

Answer

4.5 \times 10^{3} N

9-90 §9.AP source ↗

Repeat the preceding problem, but including a drag force due to air of

f_{drag} = − b \vec{v} .

Answer

\vec{J} =

\int_{0}^{τ} [m \vec{g} - m \vec{g} (1 - e^{− b t / m})] d t =

\frac{m^{2}}{b} \vec{g} (e^{− b τ / m} - 1)

9-92 §9.AP source ↗

A car crashes into a large tree that does not move. The car goes from 30 m/s to 0 in 1.3 m. (a) What impulse is applied to the 70 kg driver by the seatbelt, assuming he follows the same motion as the car? (b) What is the average force applied to the driver by the seatbelt?

Answer

− (2.1 \times 10^{3} kg \cdot m/s) \hat{i}

, b.

− (24 \times 10^{3} N) \hat{i}

9-94 §9.AP source ↗

You are coasting on your 10-kg bicycle at 15 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 2.0 m/s in the same direction as you. If your mass is 60 kg, (a) what is the initial momentum of you plus your bicycle? (b) What is the initial momentum of the bug? (c) What is your change in velocity due to the collision with the bug? (d) What would the change in velocity have been if the bug were traveling in the opposite direction?

Answer

(1.1 \times 10^{3} kg \cdot m/s) \hat{i}

, b.

(0.010 kg \cdot m/s) \hat{i}

, c.

− (0.00093 m/s) \hat{i}

, d.

− (0.0012 m/s) \hat{i}

9-96 §9.AP source ↗

Two carts on a straight track collide head on. The first cart was moving at 3.6 m/s in the positive x direction and the second was moving at 2.4 m/s in the opposite direction. After the collision, the second car continues moving in its initial direction of motion at 0.24 m/s. If the mass of the second car is 5.0 times that of the first, what is the final velocity of the first car?

Answer

− (7.2 m/s) \hat{i}

9-98 §9.AP source ↗

Derive the equations giving the final speeds for two objects that collide elastically, with the mass of the objects being

m_{1}

and

m_{2}

and the initial speeds being

v_{1,i}

and

v_{2,i} = 0

(i.e., second object is initially stationary).

Answer

v_{1,f} =

v_{1,i} \frac{m_{1} - m_{2}}{m_{1} + m_{2}},

v_{2,f} =

v_{1,i} \frac{2 m_{1}}{m_{1} + m_{2}}

9-100 §9.AP source ↗

A child sleds down a hill and collides at 5.6 m/s into a stationary sled that is identical to his. The child is launched forward at the same speed, leaving behind the two sleds that lock together and slide forward more slowly. What is the speed of the two sleds after this collision?

Answer

2.8 m/s

9-102 §9.AP source ↗

A 90-kg football player jumps vertically into the air to catch a 0.50-kg football that is thrown essentially horizontally at him at 17 m/s. What is his horizontal speed after catching the ball?

Answer

0.094 m/s

9-104 §9.AP source ↗

Two billiard balls are at rest and touching each other on a pool table. The cue ball travels at 3.8 m/s along the line of symmetry between these balls and strikes them simultaneously. If the collision is elastic, what is the velocity of the three balls after the collision?

Answer

final velocity of cue ball is

− (0.76 m/s) \hat{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

9-106 §9.AP source ↗

Two identical billiard balls collide. The first one is initially traveling at

(2.2 m/s) \hat{i} -

(0.4 m/s) \hat{j}

and the second one at

− (1.4 m/s) \hat{i} +

(2.4 m/s) \hat{j}

. Suppose they collide when the center of ball 1 is at the origin and the center of ball 2 is at the point

(2 R, 0)

where R is the radius of the balls. What is the final velocity of each ball?

Answer

ball 1:

− (1.4 m/s) \hat{i} -

(0.4 m/s) \hat{j}

, ball 2:

(2.2 m/s) \hat{i} +

(2.4 m/s) \hat{j}

9-108 §9.AP source ↗

Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point

(\sqrt{3} R / 2, R / 2)

Answer

ball 1:

(1.4 m/s) \hat{i} -

(1.7 m/s) \hat{j}

, ball 2:

− (2.8 m/s) \hat{i} +

(0.012 m/s) \hat{j}

9-110 §9.AP source ↗

Where is the center of mass of a slice of pizza that was cut into eight equal slices? Assume the origin is at the apex of the slice and measure angles with respect to an edge of the slice. The radius of the pizza is R.

Answer

θ_{cm} =

\frac{1}{8}, r_{cm} =

\frac{8 R}{3 π} \sqrt{2 - \sqrt{2}}

9-112 §9.AP source ↗

You friend wonders how a rocket continues to climb into the sky once it is sufficiently high above the surface of Earth so that its expelled gasses no longer push on the surface. How do you respond?

Answer

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.

9-114 §9.CP source ↗

A 65-kg person jumps from the first floor window of a burning building and lands almost vertically on the ground with a horizontal velocity of 3 m/s and vertical velocity of $−9 m/s$ . Upon impact with the ground he is brought to rest in a short time. The force experienced by his feet depends on whether he keeps his knees stiff or bends them. Find the force on his feet in each case.

A drawing of a person near the ground. His velocity vector is directed down and slightly to the left and is given as 3.0 meters per second i hat minus 9.0 meters per second j hat. The x y directions are shown for reference, with x to the right and y upward.

a. First find the impulse on the person from the impact on the ground. Calculate both its magnitude and direction.

b. Find the average force on the feet if the person keeps his leg stiff and straight and his center of mass drops by only 1 cm vertically and 1 cm horizontally during the impact.

c. Find the average force on the feet if the person bends his legs throughout the impact so that his center of mass drops by 50 cm vertically and 5 cm horizontally during the impact.

d. Compare the results of part (b) and (c), and draw conclusions about which way is better.

You will need to find the time the impact lasts by making reasonable assumptions about the acceleration opposite to the motion. Although the force is not constant during the impact, working with constant average force for this problem is acceptable.

Answer

617 N \cdot s

, 108°; b.

F_{x} =

2.91 \times 10^{4} N

F_{y} = 2.6 \times 10^{5} N

; c.

F_{x} = 5850 N

F_{y} = 5265 N

9-116 §9.CP source ↗

Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.

Answer

Conservation of momentum demands

m_{1} v_{1,i} +

m_{2} v_{2,i} =

m_{1} v_{1,f} +

m_{2} v_{2,f}

. We are given that

m_{1} = m_{2}

v_{1,i} = v_{2,f}

, and

v_{2,i} = v_{1,f} = 0

. Combining these equations with the equation given by conservation of momentum gives

v_{1,i} = v_{1,i}

, which is true, so conservation of momentum is satisfied. Conservation of energy demands

\frac{1}{2} m_{1} v_{1,i}^{2} +

\frac{1}{2} m_{2} v_{2,i}^{2} =

\frac{1}{2} m_{1} v_{1,f}^{2} +

\frac{1}{2} m_{2} v_{2,f}^{2}

. Again combining this equation with the conditions given above give

v_{1,i} = v_{1,i}

, so conservation of energy is satisfied.

9-118 §9.CP source ↗

Find the center of mass of the structure given in the figure below. Assume a uniform thickness of 20 cm, and a uniform density of $1 {g/cm}^{3} .$

A diagram of several masses arranged to look like a model of a person is shown. At the top is a sphere, radius 8 cm. Centered below it is a rectangle 25 cm wide horizontally and 60 cm tall that looks like the body of the person. On either side of the rectangle are rectangles measuring 60 cm horizontally and 5 cm tall that look like the outstretched arms. The tops or the arms are aligned with the top of the body, and each arm extends out from the sides of the body horizontally. At the end of each arm is a 5 cm wide square. Below the body are the legs. Each leg is 70 cm tall and 8 cm wide. The tops of the legs are aligned with the bottom of the body. The outer sides of the legs are aligned with the sides o the body. Below each leg are the feet, which are 3 cm tall and 15 cm wide. The inner side of each foot is aligned with the inner side of the leg above it.

Answer

Assume origin on centerline and at floor, then

(x_{CM}, y_{CM}) =

(0,86 cm)

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Problems and answers are adapted from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.