Exam 3 Numbers 11-20 Ucm AP

1. 20. The two spheres pictured have equal densities and are subject only to their mutual gravitational attraction. Which of the following quantities must have the same magnitude for both spheres?

A. Acceleration

B. Velocity

C. Inertia

D. Gravitational force

The gravitational force between two objects depends on the masses of the objects and the distance between them. Since the spheres have equal densities, their masses will be proportional to their volumes, and therefore their masses will be equal. Since the masses are equal, the gravitational force between the spheres will also be equal. Therefore, the magnitude of the gravitational force must be the same for both spheres.

Explanation

The gravitational force between two objects depends on the masses of the objects and the distance between them. Since the spheres have equal densities, their masses will be proportional to their volumes, and therefore their masses will be equal. Since the masses are equal, the gravitational force between the spheres will also be equal. Therefore, the magnitude of the gravitational force must be the same for both spheres.

2. 19. If the string were suddenly cut when the ball is at position II as shown in the figure, the subsequent motion of the ball would be

A. To move to the right

B. To move to the left

C. To move to the top of the page

D. To move down and to the right

Based on the given information and the figure, it can be inferred that if the string is suddenly cut when the ball is at position II, the ball will continue its motion in the upward direction. This is because the ball is already moving upwards due to the tension in the string, and when the string is cut, there will be no force acting on the ball to change its direction. Therefore, the ball will continue to move upwards.

Explanation

Based on the given information and the figure, it can be inferred that if the string is suddenly cut when the ball is at position II, the ball will continue its motion in the upward direction. This is because the ball is already moving upwards due to the tension in the string, and when the string is cut, there will be no force acting on the ball to change its direction. Therefore, the ball will continue to move upwards.

3. 16. Most commercial satellites operate in geostationary orbits. A geostationary orbit is an orbit whose position in the sky remains the same for a stationary observer on Earth. However, some satellites operate at medium Earth orbits or low Earth orbits, and have periods that are less than 24 hours. Four satellites are in a circular orbit around Earth, well outside the atmosphere. The mass (m) andorbital radius (r) of each satellite are given in the provided data table. Which satellite has the greatest speed?

A. A

B. B

C. C

D. D

Satellite B has the greatest speed because it has the smallest orbital radius compared to the other satellites. According to Kepler's second law, satellites closer to the center of the orbit will have a higher orbital speed. Therefore, since satellite B has the smallest radius, it will have the highest speed.

Explanation

Satellite B has the greatest speed because it has the smallest orbital radius compared to the other satellites. According to Kepler's second law, satellites closer to the center of the orbit will have a higher orbital speed. Therefore, since satellite B has the smallest radius, it will have the highest speed.

4. Number 14

A

B

C

D

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Explanation

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5. 18. The ball has a mass m and a speed v as it moves around the vertical circle of radius r. Which of the following expressions can be used to find the minimum speed of the ball at position I such that the circular path is maintained?

A

B

C

D

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Explanation

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6. 15. An object weighing 4 N swings back and forth on the end of a string. At the bottom of the swing, the tension in the string is 6 N. What is the magnitude of the centripetal acceleration of the object at the bottom of the swing?

A. 0

B. ½ g

C. g

D. 3/2 g

The centripetal acceleration of an object moving in a circular path is given by the equation a = v^2/r, where v is the velocity of the object and r is the radius of the circular path. At the bottom of the swing, the tension in the string is equal to the weight of the object plus the centripetal force required to keep it moving in a circle. Therefore, the tension in the string is 6 N, which is greater than the weight of the object (4 N). Since the tension in the string is greater than the weight, there must be an upward net force acting on the object at the bottom of the swing. This upward net force provides the centripetal force required for circular motion. Since the net force is upward, and the centripetal force is inward, the centripetal acceleration is also upward. Therefore, the magnitude of the centripetal acceleration at the bottom of the swing is equal to ½ g, where g is the acceleration due to gravity.

Explanation

The centripetal acceleration of an object moving in a circular path is given by the equation a = v^2/r, where v is the velocity of the object and r is the radius of the circular path. At the bottom of the swing, the tension in the string is equal to the weight of the object plus the centripetal force required to keep it moving in a circle. Therefore, the tension in the string is 6 N, which is greater than the weight of the object (4 N). Since the tension in the string is greater than the weight, there must be an upward net force acting on the object at the bottom of the swing. This upward net force provides the centripetal force required for circular motion. Since the net force is upward, and the centripetal force is inward, the centripetal acceleration is also upward. Therefore, the magnitude of the centripetal acceleration at the bottom of the swing is equal to ½ g, where g is the acceleration due to gravity.

7. Number 11

A

B

C

D

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Explanation

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8. Number 12

A

B

C

D

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Explanation

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9. 17. A ball on the end of a string is being swung in a vertical circle, rotating clockwise as shown in the diagram. Position I is at the top of the circle and position III is at the bottom of the circle.

A. I ˃ II ˃ III

B. II ˃ I = III

C. III ˃ II ˃ I

D. I = II = III

The correct answer is c. III > II > I. This is because when the ball is at position III at the bottom of the circle, it has the greatest gravitational potential energy and the least kinetic energy. As the ball moves up to position II, it gains kinetic energy and loses gravitational potential energy. Finally, at position I at the top of the circle, the ball has the least gravitational potential energy and the greatest kinetic energy. Therefore, the correct order is III > II > I.

Explanation

The correct answer is c. III > II > I. This is because when the ball is at position III at the bottom of the circle, it has the greatest gravitational potential energy and the least kinetic energy. As the ball moves up to position II, it gains kinetic energy and loses gravitational potential energy. Finally, at position I at the top of the circle, the ball has the least gravitational potential energy and the greatest kinetic energy. Therefore, the correct order is III > II > I.

10. Number 13

A

B

C

D

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Explanation

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