A-Level Physics October/November 2024 Q6(c)(i): Use the principle of conservation of momentum to determine θ.

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  By the principle of conservation of momentum, the total momentum of the system is constant. Assuming the initial momentum is zero, the total final momentum must also be zero. We can resolve the momentum vectors into components perpendicular to an arbitrary axis (e.g., the initial line of flight if there was one, or any line if starting from rest).

  The total momentum in the perpendicular direction is conserved, and was initially zero. Therefore, the sum of the final momentum components in this direction must be zero. This means the magnitudes of the perpendicular components must be equal.

  Let $p_1$ be the momentum of the 212u fragment and $p_2$ be the momentum of the 4u fragment.

  Component of $p_1$ perpendicular to axis: $p_{1,\perp} = m_1 v_1 \sin 68^{\circ}$ Component of $p_2$ perpendicular to axis: $p_{2,\perp} = m_2 v_2 \sin \theta$

  $p_{1,\perp} = p_{2,\perp}$ $m_1 v_1 \sin 68^{\circ} = m_2 v_2 \sin \theta$

  Substituting the given values: $212 \times (1.3 \times 10^5) \times \sin 68^{\circ} = 4 \times (150 \times 10^5) \times \sin \theta$

  The factors of atomic mass unit (u) and $10^5$ can be cancelled from both sides: $212 \times 1.3 \times \sin 68^{\circ} = 4 \times 150 \times \sin \theta$ $275.6 \times \sin 68^{\circ} = 600 \times \sin \theta$ $255.55 = 600 \sin \theta$

  $\sin \theta = \frac{255.55}{600} = 0.4259...$

  $\theta = \arcsin(0.4259...)$ $\theta = 25.2^{\circ}$

  To 2 significant figures, $\theta = 25^{\circ}$.

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  How the marks are awarded

  • C1 — Correctly stating the expression for the perpendicular component of momentum for either fragment, e.g., $212 \times 1.3 \times 10^5 \sin 68^{\circ}$ or $4 \times 150 \times 10^5 \sin \theta$.
  • C1 — Applying the principle of conservation of momentum by equating the two perpendicular components: $212 \times 1.3 \times 10^5 \sin 68^{\circ} = 4 \times 150 \times 10^5 \sin \theta$.
  • A1 — Calculating the final correct answer for θ as 25°.

  Common mistakes

  • Using cosine instead of sine for resolving the momentum components.
  • Attempting to conserve momentum without resolving into components (e.g., setting $m_1v_1 = m_2v_2$).
  • Making a calculation error when rearranging the equation to solve for sin(θ) or forgetting to use the inverse sine function at the end.
  • Using a calculator in radians mode instead of degrees.

  Examiner tip: For any 2D conservation law problem, always resolve the vectors into two perpendicular components and apply the principle to each component separately.

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