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Forces explain changes in motion Motion graphs and constant-acceleration equations From a velocity graph to kinetic energy Free fall through motion and energy Resultant work changes kinetic energy Momentum is conserved in more collisions than kinetic energy Photoelectron energy determines its maximum speed Accelerating voltage becomes electron kinetic energy Motion3.1 Forces in action3.2 Kinematics3.1.1 Linear motion3.1.2 Work and conservation of energy3.3.1 Kinetic and potential energies3.3.2 Collisions3.5.2 E.m.f. and potential difference4.2.2 The photoelectric effect4.5.2
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Free fall through motion and energy

OCR A AS Physics · Two ideas, one explanation

Motion equations and conservation of mechanical energy give the same speed for a freely falling object. One describes the acceleration; the other tracks energy transfer.

v² − u² = 2gΔh ΔEk = −ΔEp

Follow the reasoning

  1. Define the initial and final heights and speeds.
  2. Use either the vertical motion equation or an energy balance.
  3. Check that the height change has the correct sign.
Specification outcomes (5)
  • 3.1.2(a)(i) Apply equations of motion for constant acceleration in a straight line, including motion in a uniform gravitational field without air resistance.
  • 3.1.2(b)(i) Understand acceleration of free fall g.
  • 3.3.2(a) Define and calculate kinetic energy, Ek = ½mv², and derive it from first principles.
  • 3.3.2(b) Define and calculate gravitational potential energy, Ep = mgh, and derive it from first principles.
  • 3.3.2(c) Explain the exchange between gravitational potential energy and kinetic energy.
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H156 v2.0 · May 2026

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PhysicsUK Solve15 marks

A technician calibrates the spring bumper of a small instrument trolley. Figure 1 shows the force F needed to compress the spring by x from its natural length. The graph is a straight line through (0, 0), (0.020 m, 1.2 N), (0.040 m, 2.4 N) and (0.060 m, 3.6 N). The trolley has mass 0.60 kg. It is released from rest against the spring at compression 0.060 m on a horizontal track. A constant resistive force of 0.60 N opposes the motion as soon as the trolley moves. The spring's mass is negligible and its deformation is elastic. The trolley loses contact with the spring at its natural length. There are no other energy losses. (a) Use the gradient of the force-compression graph to determine the spring constant. [2] (b) Use the graph to calculate the work needed to compress the spring from its natural length to 0.060 m. [2] (c) Calculate the acceleration just after release, stating its direction relative to the bumper. [3] (d) Calculate the speed when the trolley loses contact with the spring. Account for the work done against resistance. [3] (e) Explain why a single constant-acceleration equation cannot describe the whole interval while the trolley is in contact with the spring. [2] (f) Calculate the distance travelled after losing contact before the trolley stops. Use a constant-acceleration equation for this part. [3]

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