AP Physics 1: Algebra-Based Practice Quiz: Circular Motion

Correct Answer: C

The content explicitly states, 'Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.'

Correct Answer: C

The magnitude of centripetal acceleration is given by $a_c = v^2/r$. If the radius *r* is doubled to *2r*, the new acceleration becomes $a_{c, new} = v^2/(2r) = (1/2)(v^2/r) = (1/2)a_{c, old}$.

Correct Answer: B

The content defines tangential acceleration as 'the rate at which an object's speed changes.'

Correct Answer: C

The text states, 'The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.'

Correct Answer: B

The provided content specifies, 'For a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction.'

Correct Answer: C

The text states, 'Centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.'

Correct Answer: C

Because the object is moving in a circle, it must have centripetal acceleration to change its direction. Because its speed is changing, it must also have tangential acceleration, which is defined as the rate of change of speed.

Correct Answer: B

In the equation $T^2 = \frac{4\pi^2}{GM}R^3$, the mass of the central body, *M*, is in the denominator. Therefore, if *M* increases, the value of the entire fraction decreases, meaning $T^2$ decreases, and thus the period *T* decreases.

Correct Answer: C

'Uniform circular motion' means the speed is constant. Since tangential acceleration is the rate of change of speed, it is zero. However, the object's direction is constantly changing, so it must have centripetal acceleration directed toward the center of the path.

Correct Answer: D

Centripetal acceleration is given by $a_c = v^2/r$. For the first car, $a_1 = v^2/r$. For the second car, $a_2 = (2v)^2/r = 4v^2/r$. The ratio $a_2/a_1$ is $(4v^2/r) / (v^2/r) = 4$. Therefore, the ratio is 4:1.

Correct Answer: C

The relationship is $T^2 \propto R^3$. Therefore, $(T_A/T_B)^2 = (R_A/R_B)^3$. Given $R_A = 4R_B$, we have $(T_A/T_B)^2 = (4R_B/R_B)^3 = 4^3 = 64$. Taking the square root of both sides gives $T_A/T_B = \sqrt{64} = 8$, so $T_A = 8 T_B$.

Correct Answer: D

The content explicitly states that tangential acceleration 'is directed tangent to the object's circular path.'

Correct Answer: B

The provided text states that the period and radius of the orbit are related to 'the mass of the central body,' which is represented by *M* in the formula.

Correct Answer: C

In uniform circular motion, the speed is constant by definition. Period and frequency, which describe the revolution, are also constant. However, the acceleration (centripetal acceleration) is always directed toward the center of the circle. As the object moves, the direction of the center relative to the object changes, so the direction of the acceleration vector is constantly changing. A vector is only constant if both its magnitude and direction are constant.

Correct Answer: B

The content states, 'The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path: $a_c = \frac{v^2}{r}$.'

Correct Answer: A

The equation shows that $T^2$ is directly proportional to $R^3$. The term $\frac{4\pi^2}{GM}$ is the constant of proportionality. Therefore, the square of the period is directly related to the cube of the orbital radius.