A-Level Physics May/June 2024 Q2(b)(v): By considering energy, calculate the speed of the object just before it lands.

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  accepted ✓

  By the principle of conservation of energy, the total initial energy is equal to the total final energy. The object has initial kinetic energy (KE) and gravitational potential energy (GPE) which is converted to final KE as it lands.

  Initial Energy = Final Energy

  $E_{k, initial} + E_{p, initial} = E_{k, final} + E_{p, final}$

  Taking the landing ground as the zero potential energy level ($h=0$):

  $\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv^2 + 0$

  The mass, $m$, cancels from every term:

  $\frac{1}{2}u^2 + gh = \frac{1}{2}v^2$

  Substitute the known values (initial speed $u = 6.0\ \text{m s}^{-1}$, height $h = 2.5\ \text{m}$, and $g = 9.81\ \text{m s}^{-2}$):

  $\frac{1}{2}(6.0)^2 + (9.81)(2.5) = \frac{1}{2}v^2$

  $18 + 24.525 = 0.5v^2$

  $42.525 = 0.5v^2$

  $v^2 = 85.05$

  $v = \sqrt{85.05}$

  $v = 9.222... \approx 9.2\ \text{m s}^{-1}$ (to 2 s.f.)

  Final speed, $v = 9.2\ \text{m s}^{-1}$

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  How the marks are awarded

  • C1 — Stating the energy conservation principle by writing an equation involving kinetic energy and gravitational potential energy, such as ½mu² + mgh = ½mv².
  • C1 — Correctly substituting the given values for initial speed (6.0), height (2.5) and g (9.81) into a valid energy conservation equation, as shown in the line ½(6.0)² + (9.81)(2.5) = ½v².
  • A1 — Calculating the final correct answer of v = 9.2 m s⁻¹. This mark is for the final numerical value being correct.

  Common mistakes

  • Forgetting the initial kinetic energy and only calculating the speed gained from the loss in GPE (i.e., setting mgh = ½mv²), which gives an incorrect answer of 7.0 m s⁻¹.
  • Ignoring the instruction 'By considering energy' and incorrectly attempting to use kinematics (suvat) equations, which would score no marks and is prone to error.
  • Making a calculation error, such as forgetting to square the initial speed (using ½ × 6.0 instead of ½ × 6.0²) or forgetting to take the square root at the final step.
  • Incorrectly combining the energy terms, for example by subtracting the initial kinetic energy from the potential energy.

  Examiner tip: When a question involves a change in both height and speed, applying the principle of conservation of energy is often the most direct method as it avoids the complexities of vector calculations.

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