AP Physics 1 · Free-response skill

Connect a derivation to an explanation

The move. Explain a physical situation in words, derive an equation for it from energy and force principles, and show that the equation and the reasoning tell the same story.

How to do it

  1. For a words-only part, reason from principles rather than formulas: name the forms of energy and say where the energy goes, or name the forces and say which way they point.
  2. For a derivation, start from a principle written in symbols, such as Ki+Ui=Kf+UfK_i + U_i = K_f + U_f or ∑F=ma\sum F = ma, and substitute an expression for each term before you solve. Keep every symbol until the end.
  3. To show that an equation agrees with your reasoning, point to a specific feature of it, a term, a factor or a variable, and say how that feature matches what you argued.
  4. When a part asks for a number, use your derived expression; getting a sensible value is a quick check on the algebra.

Practice prompt Try it

A designer plans a toy track with a vertical circular loop of radius R. A small car of mass m is released from rest on the track at a height h above the bottom of the loop, rolls down, and enters the loop at its lowest point. Treat the car as a point, and ignore friction and air resistance. The car stays on the track at the top of the loop as long as the track still pushes on it or just barely touches it.

Answer each part about the car and the loop.

  1. Without using equations, explain why releasing the car from a height of exactly 2R, the height of the top of the loop, would not get it around the loop.
  2. Write an expression for the car’s speed at the top of the loop in terms of h, R and g.
  3. At the smallest release height that works, the track exerts no force on the car at the top. Use this fact and your part (b) result to find that smallest release height in terms of R.
  4. Does a heavier car need a greater release height to get around the loop? Point to the feature of your work that shows it.
  5. The designer builds a loop with R = 0.30 m and releases the car from h = 0.90 m. Using g = 9.8 m/s², find the car’s speed at the top of the loop, and say whether the car makes it around.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) Released from 2R, the car could reach the top of the loop only by turning all of its kinetic energy back into potential energy, so it would have no speed to spare up there. But to follow the circle near the top, the car must still be moving fast enough that the track keeps pushing on it. Without that speed it cannot keep following the curve: it loses contact and leaves the track before it gets around. So it must start higher than 2R.

  • Point 1 Energy argument in words: from 2R there is not enough speed near the top to stay on the circle, so the car leaves the track.

(b) Energy is conserved from the release point to the top of the loop, which is at height 2R: mgh=mg(2R)+12mv2mgh = mg(2R) + \tfrac{1}{2}mv^2. So v2=2g(h−2R)v^2 = 2g(h - 2R) and v=2g(h−2R)v = \sqrt{2g(h - 2R)}.

  • Point 2 Starts from conservation of energy and solves for v.

(c) At the top, the center of the loop is straight down. If the track exerts no force, gravity alone supplies the inward force: mg=mv2/Rmg = mv^2/R, so v2=gRv^2 = gR. Setting this equal to the part (b) result: gR=2g(h−2R)gR = 2g(h - 2R), so h−2R=R/2h - 2R = R/2 and h=2.5Rh = 2.5R.

  • Point 3 Newton’s second law at the top with gravity as the only force.
  • Point 4 Combines the two equations to get h = 2.5R.

(d) No. The mass m appears in every term of the energy equation and on both sides of mg=mv2/Rmg = mv^2/R, so it cancels, and the result h=2.5Rh = 2.5R contains no m. A heavier car needs the same release height.

  • Point 5 Answers no and points to the feature of the work: m cancels, so h has no m.

(e) v=2(9.8)(0.90−0.60)=5.88≈2.4v = \sqrt{2(9.8)(0.90 - 0.60)} = \sqrt{5.88} \approx 2.4 m/s. The smallest height that works is 2.5R=0.752.5R = 0.75 m, and 0.90 m is higher, so the car makes it around. As a check, v2/R=19.6v^2/R = 19.6 m/s², more than g, so the track still pushes on the car at the top.

  • Point 6 Uses the part (b) expression with R = 0.30 m.
  • Point 7 Concludes the car makes it, compared against 2.5R = 0.75 m.