A-Level Mathematics May/June 2025 Q6(b): It is given instead that the plane BC is rough. The work done against the frictional fo…

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  We apply the work-energy principle to the system of two particles. The initial state is at rest (KE = 0) and the final state is when the particles are moving with speed $v$.

  Work-Energy Principle: Gain in Kinetic Energy = Loss in Gravitational Potential Energy - Work Done against friction.

  1. Change in Gravitational Potential Energy (GPE): Particle P moves horizontally, so its change in GPE is 0. Particle Q moves 2 m down the plane inclined at $30^\circ$. The vertical distance it descends is $h = 2 \sin 30^\circ = 2 \times 0.5 = 1$ m. Loss in GPE of the system = $m_Q g h = 0.6 \times g \times 1 = 0.6g$ J.

  2. Change in Kinetic Energy (KE): Both particles are connected by an inextensible string, so they move at the same speed $v$. Gain in KE of the system = $\frac{1}{2} m_P v^2 + \frac{1}{2} m_Q v^2 = \frac{1}{2} (0.3 + 0.6) v^2 = 0.45 v^2$ J.

  3. Work Done (WD) against Friction: The work done against friction for particle Q is given as $1.8$ J. From the context of the problem (implied from the mark scheme), the plane P is on is rough with a coefficient of friction $\mu = 0.4$. The normal reaction on P is $R = m_P g = 0.3g$. The work done against friction for P is $WD_P = F_f \times d = (\mu R) \times d = (0.4 \times 0.3g) \times 2 = 0.24g$ J.

  4. Form and solve the energy equation: Gain in KE = Loss in GPE - Total WD against friction $0.45 v^2 = (0.6g \times 2 \sin 30^\circ) - (0.3g \times 2 \times 0.4) - 1.8$

     Using $g = 10 \text{ ms}^{-2}$: $0.45 v^2 = (0.6 \times 10 \times 1) - (0.3 \times 10 \times 2 \times 0.4) - 1.8$ $0.45 v^2 = 6 - 2.4 - 1.8$ $0.45 v^2 = 1.8$ $v^2 = \frac{1.8}{0.45} = 4$ $v = 2 \text{ ms}^{-1}$ (since speed $v > 0$)

  The speed of Q when it has moved 2 m down the plane is $2 \text{ ms}^{-1}$.

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  How the marks are awarded

  • B1 — Correctly calculating the work done against friction for particle P. This is the term 0.3g × 2 × 0.4 in the energy equation, which evaluates to 2.4 J (using g=10).
  • B1 — Correctly calculating the loss in gravitational potential energy for particle Q. This is the term 0.6g × 2 sin 30° in the energy equation, which evaluates to 6 J (using g=10).
  • M1 — Setting up a valid work-energy equation with four distinct terms: gain in kinetic energy for the system, loss in potential energy for Q, work done against friction for Q, and work done against friction for P. Sign errors are permitted at this stage.
  • A1 — A fully correct energy equation with all terms and signs correct, leading to the final accurate answer of 2 ms⁻¹.

  Common mistakes

  • Incorrectly setting up the work-energy equation, for example by adding the work done against friction instead of subtracting it from the energy input (GPE loss).
  • Forgetting to include a term in the energy equation, most commonly the gain in KE for particle P or the work done against friction for particle P.
  • Using the distance along the slope (2 m) instead of the vertical height (2 sin 30°) when calculating the change in gravitational potential energy for particle Q.
  • Calculating the frictional force on P but forgetting to multiply by the distance (2 m) to find the work done against friction.

  Examiner tip: For energy questions involving connected particles, always write down a single work-energy equation for the whole system, carefully accounting for every particle's change in KE, every particle's change in GPE, and all work done by external non-conservative forces like friction.

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