A-Level Physics October/November 2024 Q2(c)(i): Determine the magnitude and direction of the force exerted on ball X by ball Y during t…

1 answer

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  accepted ✓

  Force is the rate of change of momentum.

  $F = \frac{\Delta p}{\Delta t}$

  Let the initial direction of ball X be positive. Initial velocity of X, $u = +8.0$ m/s Final velocity of X, $v = -8.0$ m/s Mass of X, $m = 0.24$ kg Time of collision, $\Delta t = 2.0 \text{ ms} = 2.0 \times 10^{-3}$ s

  Change in momentum of X, $\Delta p = m(v-u)$ $\Delta p = 0.24 \times (-8.0 - 8.0) = 0.24 \times (-16) = -3.84$ Ns

  The force on X is: $F = \frac{-3.84}{2.0 \times 10^{-3}} = -1920$ N

  The magnitude of the force is $1920$ N. Rounding to 2 significant figures gives $1900$ N.

  The negative sign indicates the force is in the opposite direction to the initial motion, i.e. to the left.

  Alternatively, calculating the magnitude directly: $F = \frac{m \Delta v}{\Delta t} = \frac{0.24 \times 16}{2.0 \times 10^{-3}}$ $F = 1920 \text{ N} \approx 1900 \text{ N}$ (2 s.f.)

  Magnitude = 1900 N Direction = to the left

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  How the marks are awarded

  • C1 — Correctly substituting values into the formula for force as the rate of change of momentum. The working line '$F = \frac{0.24 \times 16}{2.0 \times 10^{-3}}$' shows the correct change in momentum (or impulse) divided by the correct time.
  • A1 — Calculating the final magnitude of the force as 1900 N. An answer of 1920 N would also be accepted.
  • B1 — Stating the correct direction of the force as 'to the left'.

  Common mistakes

  • Calculating the change in velocity incorrectly, for example as 8.0 - 8.0 = 0, by failing to treat velocity as a vector quantity. This would incorrectly lead to a force of zero.
  • Forgetting to convert the collision time from milliseconds (ms) to seconds (s), leading to a calculation of 3.84 / 2.0 and an answer that is 1000 times too small.
  • Calculating the change in momentum (Δp = 3.84 Ns) and giving this as the final answer for the force, thereby confusing impulse with force.
  • Stating the direction as 'to the right'. This would be the direction of the force on ball Y, not ball X, showing a misunderstanding of Newton's third law or a misreading of the question.

  Examiner tip: Always define a positive direction at the start of any mechanics problem and use it consistently to assign signs to vector quantities like velocity, momentum and force.

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