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Stroboscopic pictures: worked exercises

Five stroboscopic pictures to work with, with questions, detailed answers and figures. Print with or without answers.

Measure distances on the figures with a ruler and use the drawn scale (bottom left) to get real distances. Answers are computed from the exact positions; your measurements may differ slightly. “Open in the tool” loads the stroboscopic picture into the video tracker, which recomputes everything and draws the vectors.

  1. Puck on an air table (Year 10–11)
  2. A falling marble (Year 10–12)
  3. A thrown ball (Year 12–13)
  4. A point on a spinning disc (Year 12–13)
  5. A toy car braking (Year 10–12)

Exercise 1 — Puck on an air table Year 10–11

A puck slides on a horizontal air table. Its positions are shown every τ = 40 ms.

M1M2M3M4M5M6M7M8M9M10M11M12M1310 cmτ = 40 ms
  1. Describe the trajectory and how the puck's speed changes.
  2. Measure the distance M1M13 using the scale, then compute the puck's speed.
  3. What can be said about the forces acting on the puck? Justify.

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Answers
M1M2M3M4M5M6M7M8M9M10M11M12M1310 cmτ = 40 msv3v7v110.2 m/s
  1. The points are in a line and evenly spaced (1.6 cm apart): the motion is uniform and straight.
  2. M1M13 = 19.2 cm, covered in 12τ = 0.48 s: v = 0.192 / 0.48 = 0.40 m/s.
  3. The velocity vector is constant: by Newton's first law, the forces on the puck (weight and the push of the air) balance.

Exercise 2 — A falling marble Year 10–12

A marble falls vertically; its positions are recorded every τ = 40 ms (it is already moving at M1).

M1M2M3M4M5M6M7M8M920 cmτ = 40 ms
  1. How does the marble's speed change? Justify without calculation.
  2. Compute the speeds v3 and v5 at points M3 and M5.
  3. Draw the velocity vectors at M3 and M5 (velocity scale: 1 cm for 1 m/s), then the change-in-velocity vector Δv4 = v5 − v3.
  4. Deduce the marble's acceleration and compare with g = 9.81 m/s².

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Answers
M1M2M3M4M5M6M7M8M920 cmτ = 40 msv3v5Δv41.0 m/s
  1. The gap between successive positions grows: the speed increases (straight, accelerated motion).
  2. v3 = M2M4 / 2τ = 0.143 / 0.08 = 1.78 m/s; v5 = M4M6 / 2τ = 0.206 / 0.08 = 2.57 m/s.
  3. Both vectors are vertical and downwards; Δv4 is therefore vertical and downwards, of size 0.78 m/s (see figure).
  4. a4 = Δv4 / (t5 − t3) = 0.78 / 0.08 = 9.81 m/s², equal to g: the marble is in free fall (negligible friction).

Exercise 3 — A thrown ball Year 12–13

A ball is thrown upwards and forwards. Its positions are marked every τ = 60 ms; air resistance is neglected.

M1M2M3M4M5M6M7M8M9M10M11M12M13M14M1550 cmτ = 60 ms
  1. What do you notice about the horizontal spacing of the positions? What does it tell you about the horizontal velocity vx?
  2. Compute and draw the velocity vectors at M4 and M6.
  3. Construct Δv5 = v6 − v4. What are its direction and sense?
  4. Compute the acceleration at M5. Which force explains this result?

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Answers
M1M2M3M4M5M6M7M8M9M10M11M12M13M14M1550 cmτ = 60 msv4v6Δv51.0 m/s
  1. The positions are evenly spaced horizontally (19.3 cm): vx is constant, vx = 3.21 m/s.
  2. v4 = M3M5 / 2τ = 3.82 m/s; v6 = M5M7 / 2τ = 3.33 m/s. Each vector is parallel to the matching chord, in the direction of motion.
  3. Δv5 is vertical and downwards (Δvx = 0.00 m/s), of size 1.18 m/s.
  4. a5 = Δv5 / 2τ = 1.18 / 0.12 = 9.81 m/s² ≈ g, vertical and downwards: only the weight acts (Newton's second law).

Exercise 4 — A point on a spinning disc Year 12–13

A mark stuck to the edge of a disc is photographed every τ = 100 ms. The disc spins at a steady rate about O.

OM1M2M3M4M5M6M7M8M9M10M11M1210 cmτ = 100 ms
  1. What kind of motion is it? Does the speed change? Does the velocity vector change?
  2. Compute the speed at M3 and M5, then draw the velocity vectors.
  3. Construct Δv4 = v5 − v3. Where does it point?
  4. Compute the acceleration at M4 and compare it with v²/R.

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Answers
OM1M2M3M4M5M6M7M8M9M10M11M1210 cmτ = 100 msv3v5Δv40.5 m/s
  1. The points lie on a circle and are evenly spaced: uniform circular motion. The speed is constant, but the direction of the velocity vector keeps changing.
  2. v3 = M2M4 / 2τ = 0.75 m/s = v5. The vectors are tangent to the circle. (Exact value: 2πR/T = 0.79 m/s; the chord is slightly shorter than the arc.)
  3. Δv4 points towards the centre O: the acceleration is centripetal.
  4. a4 = Δv4 / 2τ = 0.75 / 0.2 = 3.75 m/s²; v²/R = 3.75 m/s² with the measured speed: good agreement, within the method's error.

Exercise 5 — A toy car braking Year 10–12

A toy car brakes on a horizontal track; its positions are recorded every τ = 100 ms until it stops.

M1M2M3M4M5M6M7M810 cmτ = 100 ms
  1. Describe the car's motion.
  2. Compute the speeds v2, v4 and v6.
  3. Does the speed drop steadily? Compute the mean acceleration between M2 and M6.
  4. Which way does the net force on the car point?

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Answers
M1M2M3M4M5M6M7M810 cmτ = 100 msv2v4v60.2 m/s
  1. The points are in a line and closer and closer together: straight motion that is slowing down.
  2. v2 = 1.05 m/s; v4 = 0.75 m/s; v6 = 0.45 m/s.
  3. It drops by 0.30 m/s every 0.2 s: yes. a = (v6 − v2) / (t6 − t2) = (0.45 − 1.05) / 0.4 = −0.75 m/s².
  4. Δv is opposite to the motion: the net force points against the direction of motion (braking forces).

Going further

The method is explained in the method sheet. To make your own worksheets, film a motion and track it with the tool: “Print a student sheet” gives a stroboscopic picture with questions, “Print the full document” gives the answers.