Question 1

A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s. What is the magnitude of the angular momentum of the stick?

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Question 2

In a certain cyclotron, a proton of mass 1.67 × 10^-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 10⁶ rad/s. What is the angular momentum of the proton?

Accepted Answer

The angular momentum of a particle is given by L = mvR, where m is the mass of the particle, v is its velocity, and R is the radius of its circular path. Using this formula, we can calculate the angular momentum of the proton as:

L = (1.67 × 10^-27 kg)(1.6 m/2)(2.0 × 10^16 rad/s)
L = 2.13 × 10^-11 kg∙m^2/s

Therefore, the closest answer choice is C, 2.1 × 10^-21 kg∙m^2/s.

Question 3

A car heading north suddenly collides at an intersection with a truck of the same mass heading east. If they lock together and travel at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision, how fast was the car initially traveling?

Accepted Answer

The momentum of the system is conserved in both the x and y directions. The initial momentum of the car is $p_c = m_c v_c$, where $m_c$ is the mass of the car and $v_c$ is its initial velocity. The initial momentum of the truck is $p_t = m_t v_t$, where $m_t$ is the mass of the truck and $v_t$ is its initial velocity. The final momentum of the system is $(m_c + m_t) v_f$, where $v_f$ is the final velocity of the system.In the x direction, we have:$$p_{ci} + p_{ti} = (m_c + m_t) v_f \cos 15^\circ$$$$m_c v_c + m_t 0 = (m_c + m_t) 28 \cos 15^\circ$$$$v_c = \frac{(m_c + m_t) 28 \cos 15^\circ}{m_c}$$$$v_c = \frac{28 \cos 15^\circ}{1}$$$$v_c = 26.6 \mathrm{~m/s}$$Therefore, the car was initially traveling at 26.6 m/s.

Question 4

A uniform, solid, cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at $2.7 \mathrm { rad } / \mathrm { s } \text {. }$ What is the magnitude of the flywheel's angular momentum?

Accepted Answer

The angular momentum $ L $ of a solid cylinder is given by $ L = I \omega $, where $ I = \frac{1}{2} m r^2 $ is the moment of inertia. Substituting the given values, $ I = \frac{1}{2} \times 15 \, \text{kg} \times (1.4 \, \text{m})^2 = 14.7 \, \text{kg} \cdot \text{m}^2 $. Thus, $ L = 14.7 \, \text{kg} \cdot \text{m}^2 \times 2.7 \, \text{rad/s} = 39.69 \, \text{kg} \cdot \text{m}^2/\text{s} \approx 40 \, \text{kg} \cdot \text{m}^2/\text{s} $.

Question 5

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg ∙ m² is rotating freely with an angular speed of 1.0 rad/s. Two people, each of mass 60 kg, are standing right outside the edge of the merry-go-round and suddenly step onto the edge with negligible speed relative to the ground. What is the angular speed of the merry-go-round right after the two people have stepped on?

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Question 6

A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m² and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?

Accepted Answer

First, we need to find the initial angular momentum of the child before she jumps onto the merry-go-round:
$L_i = mrv$
$L_i = (40.0	ext{ kg})(1.50	ext{ m})(3.00	ext{ m/s})$
$L_i = 180	ext{ kg m}^2/	ext{s}$
Next, we need to find the final moment of inertia of the system after the child jumps onto the merry-go-round. We can use the parallel axis theorem:
$I_{final} = I_{merry-go-round} + I_{child} + 2md^2$
$I_{final} = 600	ext{ kg m}^2/	ext{s} + (40.0	ext{ kg})(1.50	ext{ m})^2 + 2(40.0	ext{ kg})(1.50	ext{ m})^2$
$I_{final} = 1500	ext{ kg m}^2/	ext{s}$
Conservation of angular momentum tells us that the initial angular momentum of the child must equal the final angular momentum of the system:
$L_i = L_{final}$
$mrv = I_{final}\omega_{final}$
$(40.0	ext{ kg})(3.00	ext{ m/s})(1.50	ext{ m}) = (1500	ext{ kg m}^2/	ext{s})\omega_{final}$
$\omega_{final} = 0.26	ext{ rad/s}$

Therefore, the answer is B, 0.261 rad/s.

Question 7

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m². If he pulls in his arms so his moment of inertia decreases to 1.80 kg ∙ m², what will be his new angular speed?

Accepted Answer

The conservation of angular momentum states that the initial angular momentum of a system remains constant if no external torque is applied. Therefore, we can use the equation:
I1ω1 = I2ω2
where I1 and ω1 are the initial moment of inertia and angular speed, respectively, and I2 and ω2 are the final moment of inertia and angular speed, respectively.
Plugging in the given values, we get:
(2.25 kg ∙ m2) (5.00 rad/s) = (1.80 kg ∙ m2) ω2
Solving for ω2, we get:
ω2 = (2.25 kg ∙ m2) (5.00 rad/s) / (1.80 kg ∙ m2) = 6.25 rad/s
Therefore, the new angular speed is 6.25 rad/s, which corresponds to choice C.

Question 8

A puck with a mass m₁ = 50 g moving at 1.0 m/s approaches a stationary puck with a mass m₂ = 100 g on a frictionless air table, and they undergo a sudden two-dimensional elastic collision. After the collision, both pucks have identical speeds, but travel in different directions. What is the angle between the original and final paths of puck with mass m₁?

Accepted Answer

Since both pucks have identical speeds after the collision, their paths must make an angle of 90° with each other. This is because in an elastic collision, the momentum and kinetic energy are conserved, which means that the final kinetic energy is the same as the initial kinetic energy, but the direction of the momentum can change. Therefore, the angle between the original and final paths is 90°.

Question 9

Two dogs, Rover and Fido, run on a level frictionless surface. Rover runs eastward with a momentum of 24 kg . m/s, and Fido runs northward with momentum 10 kg . m/s. They make a sudden perfectly inelastic collision. What is the magnitude of their combined momentum after the collision?

Accepted Answer

In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since momentum is a vector quantity, we calculate the resultant momentum using the Pythagorean theorem, given the eastward and northward components are perpendicular: √(24^2 + 10^2) = √(576 + 100) = √676 = 26 kg.m/s.

Question 10

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s. Calculate the magnitude of the net angular momentum of the system.

Accepted Answer

The net angular momentum is calculated by finding the angular momentum of each flywheel and then summing them, taking into account their directions. The angular momentum $L$ of a solid cylindrical flywheel is given by $L = I\omega$, where $I = \frac{1}{2}mr^2$ is the moment of inertia of the cylinder, $m$ is the mass, $r$ is the radius, and $\omega$ is the angular velocity. For the two flywheels rotating in the same direction, $L_{1+2} = 2 \times \frac{1}{2} \times 65.0 \times (1.47)^2 \times 8.94$. For the one rotating in the opposite direction, $L_{3} = \frac{1}{2} \times 65.0 \times (1.47)^2 \times (-3.42)$. The net angular momentum is the sum of these, which gives approximately 1020 kg·m^2/s.

Question 11

A 24.5-kg child is standing on the outer edge of a horizontal merry-go-round that has a moment of inertia of $989 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ about a vertical axis through its center and a radius of 2.40 m. The entire system (including the child) is initially rotating at 0.180 rev/s. Find the angular velocity if the child moves to a new position 1.10 m from the center of the merry-go-round.

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Question 12

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together after the collision. What is the speed of the wreckage just after the collision?

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Question 13

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south. The two cars stick together after the collision. What is the speed of the cars after collision?

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Question 14

An ice skater has a moment of inertia of 5.0 kg ∙ m² when her arms are outstretched, and at this time she is spinning at 3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to 2.0 kg ∙ m², how fast will she be spinning?

Accepted Answer

The law of conservation of angular momentum states that the total angular momentum of a system remains constant if no external torque acts on the system. Mathematically, it can be expressed as:

I1ω1 = I2ω2

where I1 and ω1 are the moment of inertia and angular velocity of the ice skater with arms outstretched, and I2 and ω2 are the moment of inertia and angular velocity of the ice skater with arms pulled in.

Substituting the given values, we get:

5.0 kg ∙ m2 ∙ s-1 × 2π × 3.0 rev/s = 2.0 kg ∙ m2 ∙ s-1 × 2π × ω2

Simplifying the equation, we get:

ω2 = (5.0 × 3.0) / 2.0 = 7.5 rev/s

Therefore, the ice skater will be spinning at 7.5 rev/s when she pulls in her arms and decreases her moment of inertia. The correct choice is C.

Question 15

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south. The two cars stick together after the collision. What is the direction of motion of the cars after collision?

Accepted Answer

We can use conservation of momentum in two dimensions to solve this problem. 
Before the collision, the 1500-kg car has a momentum of 1500 kg * 25 m/s = 37500 kg*m/s toward the east (since 90 km/h is 25 m/s) and the 3000-kg car has a momentum of 3000 kg * 16.67 m/s = 50010 kg*m/s toward the south (since 60 km/h is 16.67 m/s). 
After the collision, the two cars stick together and move with the same velocity. Let's call this velocity v. We can set up an equation for conservation of momentum in the x-direction: 
1500 kg * 25 m/s + 3000 kg * 0 m/s = (1500 kg + 3000 kg) * v_x 
where v_x is the x-component of the final velocity. Solving for v_x, we get v_x = 12.5 m/s. 
We can set up another equation for conservation of momentum in the y-direction: 
1500 kg * 0 m/s + 3000 kg * 16.67 m/s = (1500 kg + 3000 kg) * v_y 
where v_y is the y-component of the final velocity. Solving for v_y, we get v_y = 11.11 m/s. 
Finally, we can use trigonometry to find the direction of motion of the cars: 
tan(theta) = v_y / v_x 
theta = atan(v_y / v_x) 
theta = 53.1° south of east.

Question 16

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together after the collision. In what direction does the wreckage move just after the collision?

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Question 17

A solid wood door, 90.0 cm wide by 2.00 m tall has a mass of 35.0 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20.0 m/s and hits the door 60.0 cm from the hinged side. The ball rebounds with a speed of 16.0 m/s along the same line. What is the angular speed of the door just after the collision with the ball?

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Question 18

A light-weight potter's wheel, having a moment of inertia of $24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ is spinning freely at 40.0 rpm. The potter drops a small but dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?

Accepted Answer

The correct answer is A.Explanation:The angular momentum of the system is conserved, so:$$I_i\omega_i = I_f\omega_f$$where $I_i$ and $\omega_i$ are the initial moment of inertia and angular speed, and $I_f$ and $\omega_f$ are the final moment of inertia and angular speed.The moment of inertia of the wheel and clay is:$$I_f = I_i + mr^2$$where $m$ is the mass of the clay and $r$ is the distance from the rotational axis.Substituting into the conservation of angular momentum equation, we get:$$I_i\omega_i = (I_i + mr^2)\omega_f$$Solving for $m$, we get:$$m = \frac{I_i(\omega_i - \omega_f)}{r^2\omega_f}$$Plugging in the given values, we get:$$m = \frac{(24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 })(40.0 \mathrm {~rpm} - 32 \mathrm {~rpm})}{(1.2 \mathrm {~m} )^2(32 \mathrm {~rpm})}$$$$m = 4.2 \mathrm {~kg}$$

Question 19

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s. The cars stick together after colliding. What is the speed of the wreckage just after the collision?

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Question 20

A ballerina spins initially at 1.5 rev/s when her arms are extended. She then draws in her arms to her body and her moment of inertia becomes 0.88 kg ∙ m², and her angular speed increases to 4.0 rev/s. What was her initial moment of inertia?

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Quiz 9: Momentum

A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s. What is the magnitude of the angular momentum of the stick?

In a certain cyclotron, a proton of mass 1.67 × 10-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 106 rad/s. What is the angular momentum of the proton?

A car heading north suddenly collides at an intersection with a truck of the same mass heading east. If they lock together and travel at 28 m/s28 \mathrm {~m} / \mathrm { s }28 m/s at 15° north of east just after the collision, how fast was the car initially traveling?

A uniform, solid, cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at 2.7rad/s. 2.7 \mathrm { rad } / \mathrm { s } \text {. }2.7rad/s. What is the magnitude of the flywheel's angular momentum?

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m2. If he pulls in his arms so his moment of inertia decreases to 1.80 kg ∙ m2, what will be his new angular speed?