AP® Physics 1 Practice Test & Exam Question Bank (QBank)
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Question
A communications satellite in orbit around Earth spins about a central axis with a known angular speed, as shown above. The satellite has two solar panels that can fold toward or unfold away from the body of the satellite. When the panels are folded toward the body of the satellite, the angular speed increases from 2 rad/s to 4 rad/s. Which of the following could account for the increase in the satellite's angular speed?
| A. Folding the panels exerts a positive net torque that increased the rotational inertia of the satellite. | |
| B. Folding the panels exerts a positive net torque that decreased the rotational inertia of the satellite. | |
| C. The rotational inertia of the satellite decreased in the absence of a net torque. | |
| D. The angular momentum of the satellite increased as a result of the folding panels. |
Explanation
Rotational inertia (I) measures the resistance of an object to changes in its rotation. It is proportional to the object's mass (m) multiplied by the square of the radial distance r from the axis of rotation:
I ∝ mr2
The angular momentum (L) of a rotating object is the product of its rotational inertia and angular speed (ω):
L = Iω
According to Newton's second law of motion, a net external torque exerted on a system changes the angular momentum of the system. Conservation of angular momentum implies that the angular momentum at an initial time Li is equal to the angular momentum at a final time Lf when the net external torque is zero:
Li = Lf
(Iω)i = (Iω)f
In this question, angular momentum is conserved because the folding of the solar panels exerts zero external torque on the satellite. However, folding the solar panels inward moves some of the satellite's mass closer to the axis of rotation, decreasing both r and I:
rsatellite,f < rsatellite,i
Isatellite,f < Isatellite,i
Because the satellite's angular momentum is conserved, the decrease in rotational inertia must be compensated by an increase in angular speed (i.e., the satellite begins rotating faster):
ωf > ωi
Therefore, the increase in the satellite's angular speed accounts for the decrease in rotational inertia in the absence of net external torque.
(Choices A and B) Folding the panels will not exert a net external torque on the satellite. Regardless, torque cannot increase or decrease rotational inertia because rotational inertia is a property of the system.
(Choice D) No net external torque is exerted on the satellite during the folding of the panels. Therefore, angular momentum is conserved and does not increase.
Things to remember:
The angular momentum of a rotating object is conserved when the net external torque on the object is zero. When angular momentum is conserved, changes in rotational inertia cause changes in angular speed.
Question
A 90 kg basketball player jogs on the court with a speed of 2.5 m/s. The translational kinetic energy of the player is most nearly
| A. 0 J | |
| B. 110 J | |
| C. 280 J | |
| D. 560 J |
Explanation
An object possesses translational kinetic energy KE when it is in linear motion with respect to some reference frame. KE equals one half the product of the object's mass m and squared speed v2:
$$KE = \tfrac{1}{2}mv^2$$
In this question, the basketball player moves with a speed of 2.5 m/s relative to the court. Hence, the player's KE can be calculated by substituting m = 90 kg and v = 2.5 m/s into the above equation:
$$KE = \tfrac{1}{2}(90 \,\text{kg})(2.5 \,\text{m/s})^2$$
$$KE \approx 280 \,\text{J}$$
Therefore, the player's KE is most nearly 280 J.
(Choice A) 0 J is the player's KE if the player is at rest.
(Choice B) 110 J results from incorrectly multiplying the player's mass by their speed v instead of v2.
(Choice D) 560 J follows from neglecting the overall factor of one half in the formula for KE.
Things to remember:
Translational kinetic energy is equal to one half the product of mass and squared speed.
Question
A large tank of water, open to the air at the top, has a diameter of 5 m and a depth of 10 m. A hole with a diameter of 0.01 m is present 2 m above the bottom of the tank, which sits 1 m above the ground, as shown in the figure above. The speed of the water as it exits the hole is most nearly
| A. 0.4 m/s | |
| B. 7.7 m/s | |
| C. 12.6 m/s | |
| D. 14.1 m/s |
Explanation
In this question, the speed of the water exiting the hole in the tank can be found by applying Torricelli's theorem. Because the pressure at the top of the tank and at the hole are both equal to atmospheric pressure, conservation of energy can be used to relate the speed of the water v to the difference in height y between the top surface of the fluid and the exit hole.
The sum of the kinetic and gravitational potential energy densities is equal at all locations:
Furthermore, the speed of the water at the top surface is zero:
Solving for the exit speed yields:
In this case, Δy = 10 m − 2 m = 8 m. Substituting this result into the above equation and using g ≈ 10 m/s² gives:
(Choice A) 0.4 m/s results from incorrectly using the diameter of the hole (0.01 m) as the difference in height in Torricelli's theorem.
(Choice B) 7.7 m/s follows from incorrectly using the height of the hole above the ground (3 m) as the difference in height. The height of the hole above the ground does not affect the speed of the water out of the hole.
(Choice D) 14.1 m/s follows from using the depth of the water (10 m) as the difference in height. However, the exit hole's height of 2 m above the bottom of the tank must be subtracted from the water's depth.
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