A-Level Mathematics May/June 2025 Q5(c): When the system has been in motion for 1.5s, the string attached to A breaks. Find the…

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  This question involves two distinct stages of motion for particle A.

  Stage 1: Motion from rest until the string breaks (for 1.5 s)

  The system is released from rest, so the initial velocity is $u = 0$. From a previous part of the question, the acceleration of the system is $a = 2 \text{ m s}^{-2}$. The time of motion is $t = 1.5 \text{ s}$.

  First, we find the distance A travels up the plane in this stage, $s_1$, using $s = ut + \frac{1}{2}at^2$. $s_1 = (0)(1.5) + \frac{1}{2}(2)(1.5)^2$ $s_1 = 1.5^2 = 2.25 \text{ m}$

  Next, we find the speed of A at the moment the string breaks. This will be the initial speed for the second stage of motion. We use $v = u + at$. $v = 0 + (2)(1.5)$ $v = 3 \text{ m s}^{-1}$

  Stage 2: Motion after the string breaks until A comes to rest

  After the string breaks, the tension force is removed. The forces acting on A down the plane are the frictional force, $F$, and the component of its weight parallel to the plane, $m_A g \sin \theta$.

  From previous parts of the question, we have:

  • Mass of A, $m_A = 3$ kg
  • Angle of the plane, $\theta = 30^{\circ}$
  • Constant frictional force on A, $F = 11$ N

  We apply Newton's Second Law to A, taking the direction up the plane as positive. The net force is $-F - m_A g \sin \theta$. $F_{net} = m_A a_2$ $-F - m_A g \sin 30^{\circ} = m_A a_2$ $-11 - (3)(10) \sin 30^{\circ} = 3a_2$ $-11 - (3)(10)(0.5) = 3a_2$ $-11 - 15 = 3a_2$ $-26 = 3a_2$ $a_2 = -\frac{26}{3} \text{ m s}^{-2}$

  Now we find the additional distance, $s_2$, that A travels up the plane. The initial speed for this stage is $u_2 = 3 \text{ m s}^{-1}$, the final speed is $v_2 = 0$ (as A comes to rest), and the acceleration is $a_2 = -\frac{26}{3} \text{ m s}^{-2}$. We use $v^2 = u^2 + 2as$. $0^2 = 3^2 + 2(-\frac{26}{3})s_2$ $0 = 9 - \frac{52}{3}s_2$ $\frac{52}{3}s_2 = 9$ $s_2 = \frac{9 \times 3}{52} = \frac{27}{52} \text{ m}$

  Total Distance

  The total distance travelled by A is the sum of the distances from both stages. Total Distance $= s_1 + s_2$ Total Distance $= 2.25 + \frac{27}{52}$ Total Distance $= \frac{9}{4} + \frac{27}{52} = \frac{9 \times 13}{52} + \frac{27}{52} = \frac{117 + 27}{52} = \frac{144}{52} = \frac{36}{13}$

  Total Distance $\approx 2.76923...$

  Total distance travelled by A is $2.77 \text{ m}$ (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly calculating the distance travelled by A in the first 1.5 seconds, shown in the model answer as $s_1 = 2.25 \text{ m}$.
  • B1 — Correctly calculating the speed of A at the instant the string breaks, shown as $v = 3 \text{ m s}^{-1}$.
  • M1 — A correct attempt to apply Newton's Second Law to particle A after the string breaks. This involves identifying the two forces acting down the plane (friction and weight component) and equating their sum to $ma$. This is shown by the equation $-11 - (3)(10) \sin 30^{\circ} = 3a_2$.
  • DM1 — Using a complete kinematic method to find the further distance travelled. This mark is dependent on the previous M1. It is awarded for using $v^2 = u^2 + 2as$ with $v=0$, their calculated initial speed ($u=3$), and their new negative acceleration ($a_2 = -26/3$), as seen in the calculation $0^2 = 3^2 + 2(-\frac{26}{3})s_2$.
  • A1 — Obtaining the correct final answer by summing the two distances ($s_1$ and $s_2$) and rounding correctly to 3 significant figures, $2.77 \text{ m}$.

  Common mistakes

  • Calculating only the distance travelled after the string breaks ($"s_2 = 27/52$") and forgetting to add the initial distance ($s_1=2.25$) to find the total distance.
  • Using the initial acceleration ($a=2$) for the second stage of motion, instead of recalculating the new acceleration after the string breaks.
  • Making an error in the forces for the second stage, for example, forgetting the friction force or the weight component when applying Newton's Second Law.
  • Using an initial velocity of $u=0$ for the second stage of motion, failing to recognise that the final velocity of the first stage is the initial velocity for the second.

  Examiner tip: For multi-stage dynamics problems, clearly define each stage, identify the initial and final conditions for each, and remember that the final velocity of one stage becomes the initial velocity for the next.

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