A-Level Mathematics May/June 2024 Q6(c)(i): (c) It is given that the distance XY is 36m and the distance YZ is 98 m. (i) Find the t…

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  Let $t_1$ be the time until the first collision (between A and B at Y). From previous parts, the speed of particle A is $6 \text{ m s}^{-1}$. Distance $XY = 36 \text{ m}$.

  Time for the first collision, $t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{36}{6} = 6 \text{ s}$.

  This is the time at which the second phase of motion begins. We need to find the positions of particles C and D at $t = 6 \text{ s}$. Particle C starts moving at $t=0$ with a speed of $3 \text{ m s}^{-1}$. In the first 6 seconds, the distance travelled by C is: Distance$_C = \text{Speed} \times \text{Time} = 3 \times 6 = 18 \text{ m}$.

  The initial distance between Y and Z is 98 m. At $t=6 \text{ s}$, particle D starts moving from Z towards Y, and particle C is already 18 m away from Y towards Z. The distance between C and D at the moment of the first collision is: Separation = $98 - 18 = 80 \text{ m}$.

  Now we consider the relative motion of C and D. Let $t_2$ be the time from the first collision to the second collision. From part (a), the speed of particle D is $v_D = 5 \text{ m s}^{-1}$. The speed of particle C is $v_C = 3 \text{ m s}^{-1}$. Since they are moving towards each other, their relative speed is the sum of their speeds: Relative Speed = $v_C + v_D = 3 + 5 = 8 \text{ m s}^{-1}$.

  The time taken for them to meet (the time between the two collisions) is: $t_2 = \frac{\text{Separation}}{\text{Relative Speed}} = \frac{80}{8} = 10 \text{ s}$.

  Therefore, the time between the two collisions is 10 s.

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  How the marks are awarded

  • B1 — Awarded for correctly calculating the time until the first collision as 6 s, by dividing the distance XY (36 m) by the speed of particle A (6 m/s).
  • B1FT — Awarded for correctly calculating the distance between particles C and D at the moment of the first collision. This involves subtracting the distance C has travelled in the first 6 s (18 m) from the total distance YZ (98 m) to get 80 m. This mark allows a follow-through from the time calculated for the first collision.
  • DM1 — Awarded for setting up a correct equation for the time to the second collision using the concept of relative motion. This involves dividing the remaining distance (80 m) by the sum of the speeds of C and D (3 m/s + 5 m/s). The speed of D (5 m/s) is from part (a).
  • A1 — Awarded for the correct final answer of 10 s. No incorrect subsequent working is allowed.

  Common mistakes

  • Using the full distance of 98 m for the relative motion calculation, forgetting that particle C has already moved for 6 seconds before particle D starts moving. This leads to an incorrect answer of 98 / 8 = 12.25 s.
  • Calculating the total time from the start (t=0) until the second collision (6 s + 10 s = 16 s) instead of the time between the two collisions as requested.
  • Incorrectly calculating the relative speed by subtracting the speeds (5 - 3 = 2) instead of adding them, which is required when particles move towards each other.
  • Confusing the start times of the particles, for example by assuming C and D both start moving at t=0 or both start moving after the first collision.

  Examiner tip: For multi-stage problems, carefully define the 't=0' point for each stage and determine the positions and velocities of all relevant particles at that specific instant.

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