A-Level Physics October/November 2024 Q9(b): Determine an expression, in terms of m, q and V, for the momentum p of an electron in t…

1 answer

• ▲0
  accepted ✓

  The work done on the electron by the accelerating potential difference $V$ is converted into kinetic energy ($E_k$).

  Work done = $qV$ $E_k = \frac{1}{2}mv^2$

  Equating these gives: $qV = \frac{1}{2}mv^2$ [1]

  The momentum $p$ of the electron is given by: $p = mv$ [2]

  From equation [1], we can rearrange for $v^2$: $2qV = mv^2$

  From equation [2], we can square it to relate $p^2$ to $v^2$: $p^2 = m^2v^2$

  Substitute the expression for $mv^2$ from the energy equation into the momentum equation: $p^2 = m(mv^2)$ $p^2 = m(2qV)$ $p^2 = 2qVm$

  Finally, take the square root to find the expression for momentum $p$: $p = \sqrt{2qVm}$

  ---

  How the marks are awarded

  • C1 — For correctly stating or using the principle that the work done on the electron ($qV$) is equal to its gain in kinetic energy ($\frac{1}{2}mv^2$).
  • C1 — For correctly stating or using the definition of momentum, $p = mv$.
  • A1 — For the correct algebraic combination of the two initial equations to derive the final, correct expression $p = \sqrt{2qVm}$.

  Common mistakes

  • Incorrectly equating energy and momentum, for example by starting with $p = qV$.
  • Making an algebraic error during rearrangement, such as forgetting to take the square root to get the final answer, leaving it as $p^2 = 2qVm$.
  • Omitting the factor of $\frac{1}{2}$ in the kinetic energy formula, leading to an incorrect final answer of $p = \sqrt{qVm}$.
  • Incorrectly combining the equations, for example by substituting $v$ from one into the other and making a mistake with the square roots, leading to an incorrect power of $m$ in the final answer.

  Examiner tip: Practice combining standard equations, such as those for kinetic energy and momentum, to eliminate a variable and derive a new relationship, as this is a common problem-solving technique.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.