A-Level Mathematics October/November 2024 Q5(c): Find the time from A being projected until C returns to O.

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  This question requires finding the total time from the start of the motion until the combined particle returns to the origin. This involves three stages: motion before collision, the collision itself, and motion after the collision.

  Let the mass of particles A and B be m. The collision occurs at $t = 3.5$ s after A is projected. The height of the collision is $h = 218.75$ m (from part b).

  1. Find velocities immediately before collision:

  Particle A is projected with $u_A = 80$ m s⁻¹ and travels for $t_A = 3.5$ s. Using $v = u + at$ with $a = -g = -10$ m s⁻²: $v_A = 80 - 10(3.5) = 80 - 35 = 45$ m s⁻¹

  Particle B is projected with $u_B = 100$ m s⁻¹ and travels for $t_B = 2.5$ s (as it was projected 1 s after A, and collision is at $t=3.5$s from A's projection). $v_B = 100 - 10(2.5) = 100 - 25 = 75$ m s⁻¹

  2. Apply Conservation of Momentum for the collision:

  The two particles coalesce to form a single particle C of mass 2m. Let the velocity of C immediately after the collision be $v_C$. $m_A v_A + m_B v_B = (m_A + m_B) v_C$ $m(45) + m(75) = (m + m) v_C$ $120m = 2m v_C$ $v_C = \frac{120}{2} = 60$ m s⁻¹

  The initial velocity of the combined particle C is 60 m s⁻¹ upwards.

  3. Find the time for C to return to O:

  Particle C starts at a height of $h = 218.75$ m with an initial upward velocity of $u_C = 60$ m s⁻¹. We need to find the time, $t_{fall}$, for it to reach the ground ($s=0$).

  Using the equation of motion $s = ut + \frac{1}{2}at^2$, with displacement $s = -218.75$ m: $-218.75 = 60t_{fall} + \frac{1}{2}(-10)t_{fall}^2$ $-218.75 = 60t_{fall} - 5t_{fall}^2$ $5t_{fall}^2 - 60t_{fall} - 218.75 = 0$

  Solving the quadratic equation for $t_{fall}$: $t_{fall} = \frac{-(-60) \pm \sqrt{(-60)^2 - 4(5)(-218.75)}}{2(5)}$ $t_{fall} = \frac{60 \pm \sqrt{3600 + 4375}}{10}$ $t_{fall} = \frac{60 \pm \sqrt{7975}}{10}$ $t_{fall} = \frac{60 \pm 89.302...}{10}$

  Since time must be positive, we take the positive root: $t_{fall} = \frac{60 + 89.302...}{10} = \frac{149.302...}{10} = 14.930...$ s

  4. Calculate total time:

  The total time is the time until collision plus the time for C to fall to the ground. Total time = $t_{collision} + t_{fall}$ Total time = 3.5 + 14.930... = 18.430...

  Total time = 18.4 s (3 s.f.)

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  How the marks are awarded

  • M1 — For using the equation of motion $v=u+at$ to find the velocity of at least one particle just before collision. For example, calculating $v_A = 80 - 10(3.5)$ or $v_B = 100 - 10(2.5)$.
  • DM1 — For applying the principle of conservation of momentum. This involves setting up the equation $m(45) + m(75) = (2m)v_C$, with three non-zero terms. This mark is dependent on the first M1.
  • A1 — For correctly calculating the velocity of the combined particle C as $v_C = 60$ m s⁻¹.
  • DM1 — For a complete method to find the time for particle C to fall to the ground. This involves setting up the quadratic equation of motion for C, $5t^2 - 60t - 218.75 = 0$, using the collision height from part (b) and their calculated $v_C$. This mark is dependent on both previous M marks.
  • A1 — For the correct final answer of 18.4 s, obtained by adding the time to collision (3.5 s) to the time calculated for C's subsequent motion (14.9 s).

  Common mistakes

  • Using an incorrect sign for displacement in the final stage of motion, for example setting up the equation as $+218.75 = 60t - 5t^2$.
  • Forgetting to add the initial time before collision (3.5 s) to the time calculated for particle C's motion, giving an answer of 14.9 s instead of 18.4 s.
  • Making an error in the conservation of momentum calculation, such as forgetting to use the combined mass (2m) for the particle after the collision.
  • Using incorrect time values when finding velocities before collision, for example using $t=3.5$ for both particles A and B.

  Examiner tip: Break down multi-stage problems into distinct phases (e.g., before collision, collision, after collision) and apply the relevant physical principles to each.

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