Book cover for College Physics for AP® Courses

College Physics for AP® Courses

Irina Lyublinskaya, Gregg Wolfe, Douglas Ingram , Liza Pujji

ISBN #9781938168932

2,282 Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
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Summary

This section emphasizes that linear momentum is a fundamental vector quantity defined by mass and velocity, and its conservation governs collision dynamics. Whether examining elastic collisions where both momentum and kinetic energy are conserved, or in inelastic collisions where objects stick together and some kinetic energy is lost, the conservation principles provide a powerful tool for analyzing the results. The concept of impulse connects force and the time over which it acts to yield changes in momentum, with practical applications ranging from sports to rocket propulsion.

Learning Objectives

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Key Concepts

CONCEPT

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Example Problems

Example 1

(a) Calculate the momentum of a $2000-\mathrm{kg}$ elephant charging a hunter at a speed of 7.50 $\mathrm{m} / \mathrm{s} .$ (b) Compare the elephant's momentum with the momentum of a $0.0400-\mathrm{kg}$ tranquilizer dart fired at a speed of 600 $\mathrm{m} / \mathrm{s}$ . (c) What is the momentum of the 90.0 -kg hunter running at 7.40 $\mathrm{m} / \mathrm{s}$ after missing the elephant?

Example 2

(a) What is the mass of a large ship that has a momentum of $1.60 \times 10^{9} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ , when the ship is moving at a speed of 48.0 $\mathrm{km} / \mathrm{h} ?$ (b) Compare the ship's momentum to the momentum of a $1100-\mathrm{kg}$ artillery shell fired at a speed of 1200 $\mathrm{m} / \mathrm{s} .$

Example 3

(a) At what speed would a $2.00 \times 10^{4}$ -kg airplane have to fly to have a momentum of $1.60 \times 10^{9} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ (the same as the ship's momentum in the problem above)? (b) What is the plane's momentum when it is taking off at a speed of 60.0 $\mathrm{m} / \mathrm{s} ?(\mathrm{c})$ lf the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

Example 4

(a) What is the momentum of a garbage truck that is $1.20 \times 10^{4} \mathrm{kg}$ and is moving at 10.0 $\mathrm{m} / \mathrm{s} ?$ (b) At what speed would an $8.00-\mathrm{kg}$ trash can have the same momentum as the truck?

Example 5

A runaway train car that has a mass of $15,000 \mathrm{kg}$ travels at a speed of 5.4 $\mathrm{m} / \mathrm{s}$ down a track. Compute the time required for a force of 1500 $\mathrm{N}$ to bring the car to rest.
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Step-by-Step Explanations

QUESTION

How do you calculate the momentum of a 110-kg football player running at 8.00 m/s?\nStep-by-step Answer:\nStep 1: Write down the formula for momentum: p = m * v.\nStep 2: Substitute the given mass (110 kg) and velocity (8.00 m/s) into the formula.\nStep 3: Multiply the mass by the velocity: p = 110 kg * 8.00 m/s = 880 kg\u00b7m/s.\nFinal Answer: The momentum of the player is 880 kg\u00b7m/s.\n\n- Topic: Determining average force from impulse \nQuestion: If a tennis ball of mass 0.057 kg is hit such that its velocity changes from 0 m/s to 58 m/s in 5.0 ms, how do you calculate the average force exerted by the racquet?\nStep-by-step Answer:\nStep 1: Calculate the change in momentum using \u0394p = m*(v_final - v_initial). Here, \u0394p = 0.057 kg * (58 m/s - 0 m/s) = 3.306 kg\u00b7m/s.\nStep 2: Convert the contact time into seconds: 5.0 ms = 0.005 s.\nStep 3: Use the impulse formula F = \u0394p/\u0394t, so F = 3.306 kg\u00b7m/s / 0.005 s = 661 N.\nFinal Answer: The average force is approximately 661 N.\n\n- Topic: Using momentum conservation in a collision \nQuestion: In a head-on elastic collision, a 0.500-kg ball moving at 4.00 m/s hits a stationary 3.50-kg ball. How can we apply momentum conservation to determine post-collision velocities?\nStep-by-step Answer:\nStep 1: Write the conservation of momentum equation: m1*v1 = m1*v1' + m2*v2' (since the second ball is initially at rest).\nStep 2: Write the conservation of kinetic energy equation: 0.5*m1*v1^2 = 0.5*m1*v1'^2 + 0.5*m2*v2'^2.\nStep 3: Solve the momentum equation for one of the unknowns (for example, express v2' in terms of v1' and the known values).\nStep 4: Substitute that expression into the energy equation and solve the resulting quadratic for v1'.\nStep 5: Discard the trivial solution (if v1' equals initial velocity, it represents the situation with no collision) and find the physically meaningful value.\nFinal Answer: Using the provided algebra (as illustrated in the textbook example), you arrive at v1' \u2248 -3.00 m/s (recoil) and v2' \u2248 1.00 m/s.\n\n"

STEP-BY-STEP ANSWER:

Step 1: Write down the formula for momentum: p = m * v.\nStep 2: Substitute the given mass (110 kg) and velocity (8.00 m/s) into the formula.\nStep 3: Multiply the mass by the velocity: p = 110 kg * 8.00 m/s = 880 kg\u00b7m/s.\nFinal Answer: The momentum of the player is 880 kg\u00b7m/s.\n\n- Topic: Determining average force from impulse \nQuestion: If a tennis ball of mass 0.057 kg is hit such that its velocity changes from 0 m/s to 58 m/s in 5.0 ms, how do you calculate the average force exerted by the racquet?\nStep-by-step Answer:\nStep 1: Calculate the change in momentum using \u0394p = m*(v_final - v_initial). Here, \u0394p = 0.057 kg * (58 m/s - 0 m/s) = 3.306 kg\u00b7m/s.\nStep 2: Convert the contact time into seconds: 5.0 ms = 0.005 s.\nStep 3: Use the impulse formula F = \u0394p/\u0394t, so F = 3.306 kg\u00b7m/s / 0.005 s = 661 N.\nFinal Answer: The average force is approximately 661 N.\n\n- Topic: Using momentum conservation in a collision \nQuestion: In a head-on elastic collision, a 0.500-kg ball moving at 4.00 m/s hits a stationary 3.50-kg ball. How can we apply momentum conservation to determine post-collision velocities?\nStep-by-step Answer:\nStep 1: Write the conservation of momentum equation: m1*v1 = m1*v1' + m2*v2' (since the second ball is initially at rest).\nStep 2: Write the conservation of kinetic energy equation: 0.5*m1*v1^2 = 0.5*m1*v1'^2 + 0.5*m2*v2'^2.\nStep 3: Solve the momentum equation for one of the unknowns (for example, express v2' in terms of v1' and the known values).\nStep 4: Substitute that expression into the energy equation and solve the resulting quadratic for v1'.\nStep 5: Discard the trivial solution (if v1' equals initial velocity, it represents the situation with no collision) and find the physically meaningful value.\nFinal Answer: Using the provided algebra (as illustrated in the textbook example), you arrive at v1' \u2248 -3.00 m/s (recoil) and v2' \u2248 1.00 m/s.\n\n"
Final Answer: The momentum of the player is 880 kg\u00b7m/s.\n\n- Topic: Determining average force from impulse \nQuestion: If a tennis ball of mass 0.057 kg is hit such that its velocity changes from 0 m/s to 58 m/s in 5.0 ms, how do you calculate the average force exerted by the racquet?\nStep-by-step Answer:\nStep 1: Calculate the change in momentum using \u0394p = m*(v_final - v_initial). Here, \u0394p = 0.057 kg * (58 m/s - 0 m/s) = 3.306 kg\u00b7m/s.\nStep 2: Convert the contact time into seconds: 5.0 ms = 0.005 s.\nStep 3: Use the impulse formula F = \u0394p/\u0394t, so F = 3.306 kg\u00b7m/s / 0.005 s = 661 N.\nFinal Answer: The average force is approximately 661 N.\n\n- Topic: Using momentum conservation in a collision \nQuestion: In a head-on elastic collision, a 0.500-kg ball moving at 4.00 m/s hits a stationary 3.50-kg ball. How can we apply momentum conservation to determine post-collision velocities?\nStep-by-step Answer:\nStep 1: Write the conservation of momentum equation: m1*v1 = m1*v1' + m2*v2' (since the second ball is initially at rest).\nStep 2: Write the conservation of kinetic energy equation: 0.5*m1*v1^2 = 0.5*m1*v1'^2 + 0.5*m2*v2'^2.\nStep 3: Solve the momentum equation for one of the unknowns (for example, express v2' in terms of v1' and the known values).\nStep 4: Substitute that expression into the energy equation and solve the resulting quadratic for v1'.\nStep 5: Discard the trivial solution (if v1' equals initial velocity, it represents the situation with no collision) and find the physically meaningful value.\nFinal Answer: Using the provided algebra (as illustrated in the textbook example), you arrive at v1' \u2248 -3.00 m/s (recoil) and v2' \u2248 1.00 m/s.\n\n"

"- Topic: Calculating the momentum of an object \nQuestion: How do you calculate the momentum of a 110-kg football player running at 8.00 m/s?\nStep-by-step Answer:\nStep 1: Write down the formula for momentum: p = m * v.\nStep 2: Substitute the given mass (110 kg) and velocity (8.00 m/s) into the formula.\nStep 3: Multiply the mass by the velocity: p = 110 kg * 8.00 m/s = 880 kg\u00b7m/s.\nFinal Answer: The momentum of the player is 880 kg\u00b7m/s.\n\n- Topic: Determining average force from impulse \nQuestion: If a tennis ball of mass 0.057 kg is hit such that its velocity changes from 0 m/s to 58 m/s in 5.0 ms, how do you calculate the average force exerted by the racquet?\nStep-by-step Answer:\nStep 1: Calculate the change in momentum using \u0394p = m*(v_final - v_initial). Here, \u0394p = 0.057 kg * (58 m/s - 0 m/s) = 3.306 kg\u00b7m/s.\nStep 2: Convert the contact time into seconds: 5.0 ms = 0.005 s.\nStep 3: Use the impulse formula F = \u0394p/\u0394t, so F = 3.306 kg\u00b7m/s / 0.005 s = 661 N.\nFinal Answer: The average force is approximately 661 N.\n\n- Topic: Using momentum conservation in a collision \nQuestion: In a head-on elastic collision, a 0.500-kg ball moving at 4.00 m/s hits a stationary 3.50-kg ball. How can we apply momentum conservation to determine post-collision velocities?\nStep-by-step Answer:\nStep 1: Write the conservation of momentum equation: m1*v1 = m1*v1' + m2*v2' (since the second ball is initially at rest).\nStep 2: Write the conservation of kinetic energy equation: 0.5*m1*v1^2 = 0.5*m1*v1'^2 + 0.5*m2*v2'^2.\nStep 3: Solve the momentum equation for one of the unknowns (for example, express v2' in terms of v1' and the known values).\nStep 4: Substitute that expression into the energy equation and solve the resulting quadratic for v1'.\nStep 5: Discard the trivial solution (if v1' equals initial velocity, it represents the situation with no collision) and find the physically meaningful value.\nFinal Answer: Using the provided algebra (as illustrated in the textbook example), you arrive at v1' \u2248 -3.00 m/s (recoil) and v2' \u2248 1.00 m/s.\n\n"

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