Free Pearson Edexcel International A Level Physics lesson

Measure impulses, collisions and circular motion

Synchronise sensors and video before interpreting a disagreement.

What you’ll learn

  • Plan and evaluate core practicals 9 and 10.
  • Calibrate video and integrate force records.
  • Test a circular-motion relationship with controlled geometry.

Core practical 9: compare a force pulse with momentum change

Use a force sensor aligned with a trolley’s impact direction and a calibrated method for velocity immediately before and after contact. Measure the total moving mass, including attachments. Define one positive axis and retain the sign of a rebound. Start logging before contact so the zero-force baseline and full pulse are included.

Choose sampling fast enough to resolve the pulse and a force range that avoids clipping. Subtract a measured baseline, then integrate the force–time curve by the logger or a justified trapezium sum. Compare the signed area with m(v − u). Vary incident speed or cushioning over a suitable range while keeping the force and motion measurements synchronised.

A persistent offset in the force sensor contributes offset multiplied by integration time to the impulse. Poorly chosen velocity windows include frictional slowing outside the contact interval. Repeat trials, inspect each trace and compare the discrepancy with force, timing, velocity and mass uncertainty. Secure the track and use a stop or catch to prevent an unexpected rebound leaving the apparatus.

J ≈ Σ½(Fᵢ + Fᵢ₊₁)Δt; J = m(v − u); baseline contribution = Foffset × duration

Symbols and units

  • J is impulse in N s
  • Fᵢ are sampled resultant forces in N
  • Δt is sampling interval in s
  • m is mass in kg
  • u and v are signed velocities in m s⁻¹.

How the equation explains the idea

  1. Approximate each interval’s force–time area by the trapezium ½(Fᵢ + Fᵢ₊₁)Δt and add signed areas across the pulse.
  2. Compare with m(v − u). A constant force zero error Foffset adds Foffset × duration to the integral; subtract a justified baseline before integration rather than calling every mismatch lost momentum.
A baseline offset contributes extra apparent impulse

Conditions. Illustrative triangular true resultant pulse: duration 0.40 s, peak 4 N. A second trace adds constant sensor offset +1 N throughout the integration window.

Time / s; Force reading / N.

True impulse is ½ × 0.40 × 4 = 0.80 N s. The offset adds rectangular area 1 × 0.40 = 0.40 N s, making the uncorrected integral 1.20 N s. More samples do not remove this baseline error.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Three force samples

  1. Force samples at 0, 0.010 and 0.020 s are 0, 20 and 0 N after zero correction.
  2. Two trapezia give J = ½(0 + 20)0.010 + ½(20 + 0)0.010 = 0.20 N s.
  3. A 0.10 kg object changing velocity by 2.0 m s⁻¹ gives the same momentum change.

A triangular pulse must not be treated as its peak force acting throughout.

Keep this idea. Synchronise the full corrected force pulse with the velocity change of the same moving mass.

Core practical 10: extract two-dimensional velocities from video

Position a camera perpendicular to the collision plane and include a calibration scale in that plane. Secure it so the image does not move. Use a frame rate and shutter setting that resolve the spheres without excessive blur, and verify the actual frame interval. Record masses and mark a coordinate system fixed to the table.

Track each sphere centre over several frames before and after contact. Fit x against t and y against t to obtain the two velocity components; a displacement over several frames reduces relative pixel uncertainty but must not include the collision itself. Calculate total momentum components before and after. Also compare translational kinetic energies if investigating elasticity, with rolling energy limitations stated.

Repeat across several approach directions or speeds. Check whether the table slopes or friction produces appreciable external impulse during the chosen interval. Misidentifying the same edge of a blurred sphere can systematically bias positions. Compare component discrepancies with estimated scale, frame and tracking uncertainties rather than using a single vector magnitude to conceal opposing component errors.

vₓ = Δx/Δt; vᵧ = Δy/Δt; Σpₓ,before = Σpₓ,after and Σpᵧ,before = Σpᵧ,after

Symbols and units

  • x and y are position components in m
  • Δt is frame interval in s
  • vₓ and vᵧ are velocity components in m s⁻¹
  • p components are kg m s⁻¹.

How the equation explains the idea

  1. Calibrate image distance into metres before dividing coordinate differences by the frame interval. Use intervals before and after the collision rather than one interval straddling it.
  2. For each body calculate mΔx/Δt and mΔy/Δt, then compare the system totals component by component. Camera perspective and uncertain frame timing can bias every reconstructed velocity.

Worked example

A tracked outgoing sphere

  1. Over 0.040 s, its centre moves +0.080 m in x and −0.060 m in y.
  2. vₓ = 2.0 m s⁻¹ and vᵧ = −1.5 m s⁻¹.
  3. Speed = √(2.0² + 1.5²) = 2.5 m s⁻¹.

Keep the component signs even though speed is positive.

Keep this idea. Calibrate space and time, then compare both signed momentum components.

Test a circular-force model without changing the radius accidentally

To test F = mv²/r, specify which quantity changes and which remain fixed. At fixed m and r, a force-against-v² graph should be linear. Alternatively express speed as 2πr/T to obtain F = 4π²mr/T². A period measurement can be more reliable than estimating the instantaneous tangential speed directly.

Measure the radius from the rotation centre to the object’s centre, not merely the free string length. A marker on the string can help detect changes in radius. Time many complete revolutions past a fixed reference, keeping speed steady. A hanging mass supplies approximate tension only while it remains stationary and friction through the guide is negligible.

Protect the swept area and use securely attached small masses under supervision. Record whether the orbit is truly horizontal and whether the hanging mass oscillates, because these change the force geometry. Uncertainty in radius enters differently depending on the chosen expression; in F = 4π²mr/T² it contributes once, while timing contributes twice.

F = mv²/r = 4π²mr/T²; for fixed m,r, gradient of F against 1/T² = 4π²mr

Symbols and units

  • F is inward resultant force in N
  • m is rotating mass in kg
  • r is radius in m
  • T is revolution period in s.

How the equation explains the idea

  1. Substitute v = 2πr/T into F = mv²/r to obtain F = 4π²mr/T².
  2. At fixed m and r, plot F against 1/T². The slope is 4π²mr with units N s². If radius drifts as speed changes, a straight-line interpretation with one fixed r is not justified.

Worked example

Predict a radial resultant

  1. m = 0.050 kg, r = 0.40 m and T = 0.80 s.
  2. F = 4π² × 0.050 × 0.40/0.80².
  3. F = 1.23 N approximately.

Changing period changes the required force quadratically.

Keep this idea. Control the actual orbit geometry and time complete turns before testing a force law.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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