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AP Physics 1- 2.9 Circular Motion - Exam Style questions - FRQs- New Syllabus

Vectors and Motion in Two Dimensions AP  Physics 1 FRQ

Unit: 2. Force and Translational  Dynamics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

A block of mass \( M \) is released from rest at Point \( A \), a height \( 6R \) above the horizontal. After being released, the block slides on a track, as shown. Released from Point \( A \), the block does not lose contact with the track at any point. Points \( B \) and \( C \) are located at the highest points of their respective circular loops, both of radius \( R \). All frictional forces are negligible.

Diagram \( \mathrm{A} \) shows an energy bar chart that represents the gravitational potential energy \( U_g \) of the block-Earth system and the kinetic energy \( K \) of the block at Point \( A \), when the block is released from rest at height \( 6R \).
(a) Draw shaded regions in Diagram \( \mathrm{B} \) that represent the gravitational potential energy \( U_g \) and kinetic energy \( K \) of the block-Earth system when the block is located at Point \( B \), a height \( 2R \) above the horizontal.
• Shaded regions should start at the dashed line that represents zero energy.
• Represent any energy that is equal to zero with a distinct line on the zero-energy line.
• The relative height of each shaded region should reflect the magnitude of the respective energy consistent with the scale used in Diagram \( \mathrm{A} \).
 
(b) Starting with conservation of energy, derive an expression for the speed of the block at Point \( B \). Express your answer in terms of \( R \) and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference book.
(c)
(i) On the following dot that represents the block, draw and label the forces \( (\text{not components}) \) that are exerted on the block at the instant the block slides through Point \( C \). Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
(ii) A student claims that \( 4R \) is the minimum height of Point \( A \), such that the block can slide through Point \( C \) without losing contact with the track after the block is released from rest. Briefly explain why this claim is incorrect.

Most-appropriate topic codes (AP Physics 1):

• Topic \( 3.1 \) — Translational Kinetic Energy (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
• Topic \( 3.3 \) — Potential Energy (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
• Topic \( 3.4 \) — Conservation of Energy (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)(ii)} \))
• Topic \( 2.2 \) — Forces and Free-Body Diagrams (Part \( \mathrm{(c)(i)} \))
• Topic \( 2.9 \) — Circular Motion (Part \( \mathrm{(c)(i)} \), Part \( \mathrm{(c)(ii)} \))
▶️ Answer/Explanation

(a)
At Point \( A \), the block is released from rest, so all of the mechanical energy is gravitational potential energy.

Since Point \( A \) is at height \( 6R \), the total energy is proportional to \( 6R \). At Point \( B \), the block is at height \( 2R \), so

\( U_g = Mg(2R) \)

and the remaining energy must be kinetic energy:

\( K = Mg(6R) – Mg(2R) = 4MgR \)

So in Diagram \( \mathrm{B} \), the \( U_g \) bar should have height \( 2 \) units and the \( K \) bar should have height \( 4 \) units, for a total of \( 6 \) units.

This keeps the total mechanical energy the same as in Diagram \( \mathrm{A} \).

(b)
Start with conservation of mechanical energy:

\( E_i = E_f \)

At Point \( A \):

\( E_i = U_{g,A} = Mg(6R) \)

At Point \( B \):

\( E_f = U_{g,B} + K_B = Mg(2R) + \dfrac{1}{2}Mv_B^2 \)

Therefore,

\( Mg(6R) = Mg(2R) + \dfrac{1}{2}Mv_B^2 \)

\( 6gR = 2gR + \dfrac{1}{2}v_B^2 \)

\( 4gR = \dfrac{1}{2}v_B^2 \)

\( v_B^2 = 8gR \)

\( \boxed{v_B = \sqrt{8gR}} \)

The mass cancels, which is what we expect for frictionless motion under gravity.

(c)(i)
At Point \( C \), the forces on the block are:

• A downward gravitational force \( F_g \)
• A downward normal force \( F_N \)

Both arrows should point downward from the dot because at the top of the loop, the center of the circular path is below the block, so the normal force points toward the center, which is downward.

(c)(ii)
The claim is incorrect because if the block starts from height \( 4R \), then by energy conservation it would reach Point \( C \) with zero kinetic energy.

Point \( C \) is also at height \( 4R \), so starting from \( 4R \) means

\( Mg(4R) = Mg(4R) + \dfrac{1}{2}Mv_C^2 \)

which gives

\( v_C = 0 \)

But to remain in contact with the track at the top of the loop, the block must still have some speed so that a centripetal force can be provided. If the speed is zero, the block does not have enough momentum to stay on the track and will lose contact.

So the minimum starting height must be greater than \( 4R \), not equal to \( 4R \).

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