A-Level Mathematics May/June 2025 Q5(b): P is now projected on a smooth horizontal surface with speed vms⁻¹ directly towards a p…

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  Let the mass of particle P be $m_P$. From part (a), $m_P = 0.5$ kg. The mass of particle Q is $m_Q = 1.25$ kg.

  Conservation of Linear Momentum (CoM): The total momentum before the collision is equal to the total momentum after the collision.

  Initial momentum = $m_P u_P + m_Q u_Q = (0.5)v + (1.25)(0) = 0.5v$ Final momentum = $m_P v_P + m_Q v_Q = (0.5)w + (1.25)(2w) = 0.5w + 2.5w = 3w$

  By CoM: $0.5v = 3w$ $v = 6w$ (1)

  Loss of Kinetic Energy (KE): The loss of KE is the initial total KE minus the final total KE.

  Initial KE = $\frac{1}{2}m_P u_P^2 + \frac{1}{2}m_Q u_Q^2 = \frac{1}{2}(0.5)v^2 + 0 = 0.25v^2$ Final KE = $\frac{1}{2}m_P v_P^2 + \frac{1}{2}m_Q v_Q^2 = \frac{1}{2}(0.5)w^2 + \frac{1}{2}(1.25)(2w)^2$ Final KE = $0.25w^2 + 0.625(4w^2) = 0.25w^2 + 2.5w^2 = 2.75w^2$

  Loss of KE = Initial KE - Final KE $25 = 0.25v^2 - 2.75w^2$ (2)

  Solving the simultaneous equations: Substitute equation (1) into equation (2): $25 = 0.25(6w)^2 - 2.75w^2$ $25 = 0.25(36w^2) - 2.75w^2$ $25 = 9w^2 - 2.75w^2$ $25 = 6.25w^2$ $w^2 = \frac{25}{6.25} = 4$ $w = 2$ (since speed must be positive)

  Now substitute the value of $w$ back into equation (1) to find $v$: $v = 6w = 6(2)$ $v = 12$

  Final Answer: $v = 12$ and $w = 2$.

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  How the marks are awarded

  • M1 — For correctly applying the principle of conservation of momentum to form an equation relating v and w. The model answer shows this with $0.5v = 0.5w + 1.25(2w)$.
  • M1 — For correctly forming an equation for the loss of kinetic energy. The model answer shows this by setting the initial KE minus the final KE equal to 25: $0.25v^2 - 2.75w^2 = 25$.
  • DM1 — For using the two previously formed equations to create a single equation in one variable. This is dependent on both M marks and is shown by substituting $v=6w$ into the energy equation.
  • A1 — For obtaining the correct final values for both variables, $v=12$ and $w=2$.

  Common mistakes

  • A sign error in the conservation of momentum equation, such as assuming P rebounds without information, leading to an incorrect relationship like $v=4w$.
  • An error in the kinetic energy formula for particle Q, often by failing to square the entire velocity term $(2w)$, for example writing $\frac{1}{2}(1.25)(2w^2)$ instead of $\frac{1}{2}(1.25)(2w)^2$.
  • Confusing loss of KE with gain of KE, by writing Final KE - Initial KE = 25, which results in an equation with no real solutions.
  • Arithmetic errors when solving the simultaneous equations, for example calculating $(6w)^2$ as $6w^2$ instead of $36w^2$.

  Examiner tip: This question rewards the ability to form and solve simultaneous equations derived from two separate physical principles: conservation of momentum and change in kinetic energy.

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  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

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