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FRQ: Energy and Circular Motion in a Vertical Loop — The Science Fair Loop Track

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  CHALLENGE  |  8 points  |  20 min

▤ Scenario

For a science fair, a student builds the track shown in Figure 1. A small block of mass m is released from rest at a height H above the floor. It slides down a ramp and into a vertical circular loop of radius R, whose lowest point is at floor level. The top of the loop is therefore a height 2R above the floor.

Figure 1. The ramp and the vertical loop of radius R. The block is released from rest at height H.

Friction and air resistance are negligible, and the block is small compared with the loop. H_min is the smallest release height for which the block remains in contact with the track at the top of the loop.

✎ Free Response Questions

Part A

Indicate whether H_min is greater than, less than, or equal to 2R by writing one of the following.

• H_min > 2R

• H_min < 2R

• H_min = 2R

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Starting with conservation of energy, derive an expression for H_min. Express your final answer in terms of R, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Justify how your derived equation in part B is or is not consistent with your reasoning in part A.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board's published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

H_min > 2R

To follow the circular track over the top of the loop, the block must be moving there. A block with no speed at the top would simply fall straight down, away from the track. A moving block has kinetic energy. The mechanical energy of the block–Earth system is constant, so the potential energy at release must cover both the potential energy at the top of the loop and that kinetic energy. More potential energy at release than at the top means a release point higher than the top. The block must still have kinetic energy at the top of the loop, so it has to start from above the top.

Scoring (3 points):

▸ A1 — 1 point: For selecting H_min > 2R with an attempt at a relevant justification.

▸ A2 — 1 point: For indicating that the block must have a nonzero speed, and so kinetic energy, at the top of the loop to stay on the circular path.

▸ A3 — 1 point: For indicating that, because mechanical energy is constant, the potential energy at release must be greater than the potential energy at the top, so the release height is greater than 2R.

⚠︎ Common error (partial credit): selects H_min > 2R on the grounds that some energy is lost on the way down: the track is stated to be frictionless — earns 1 of 3 points (A1).

✗ Common error (no credit): selects H_min = 2R because the block then has “just enough energy to reach that height”: reaching a height of 2R with no speed is not the same as travelling around the top of a circle, and a block released from 2R leaves the track before it gets there.

Part B — Model Answer

Apply conservation of energy to the block–Earth system between the release point and the top of the loop, with the floor as the zero of potential energy. At the minimum release height the block only just stays in contact at the top, so the normal force there is zero and the gravitational force alone provides the centripetal acceleration.

mgH_min = ½mv_top² + mg(2R)
mg = mv_top²/R, so v_top² = gR
mgH_min = ½m(gR) + 2mgR
H_min = (5/2)R

The mass cancels, and the block needs an extra R/2 of height to supply the kinetic energy it must have at the top.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that starts with conservation of energy between the release point and the top of the loop.

▸ B2 — 1 point: For applying Newton's second law at the top of the loop with the gravitational force as the only force providing the centripetal acceleration: mg = mv²/R, or v² = gR.

▸ B3 — 1 point: For a correct final expression, H_min = (5/2)R or 2.5R.

Scoring Note: A correct, isolated, final expression for H_min earns points B2 and B3.

⚠︎ Common error (partial credit): takes the top of the loop to be at height R and obtains H_min = (3/2)R — earns 2 of 3 points (B1, B2).

⚠︎ Common error (partial credit): sets the speed at the top to zero and obtains H_min = 2R — earns 1 of 3 points (B1).

Part C — Model Answer

The derived expression, H_min = (5/2)R, is greater than 2R, so it agrees with the answer in part A. It can be written as 2R + R/2. The 2R is the height of the top of the loop. The extra R/2 comes from the term ½mv_top² = ½mgR, which is the kinetic energy that part A argued the block must still have at the top. If no speed were needed there, that term would vanish and the equation would give H_min = 2R. The amount by which H_min exceeds 2R is exactly the kinetic energy needed at the top.

Scoring (2 points):

▸ C1 — 1 point: For stating whether the expression derived in part B is consistent with the selection in part A by comparing it with 2R, in agreement with the student's own answers.

▸ C2 — 1 point: For relating a feature of the derived equation to the reasoning in part A: the amount by which H_min exceeds 2R comes from the kinetic energy (the speed) required at the top of the loop.

⚠︎ Common error (partial credit): states only that (5/2)R is greater than 2R, with no link to the reasoning in part A — earns 1 of 2 points (C1).

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