| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Moderate -0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. Part (a) is routine application of formulas, while part (b) requires the insight that reversed direction means final velocity is negative, leading to an inequality—slightly above pure recall but still a textbook exercise type. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(3m.2u - 4mu = 3mv_1 + 4mv_2\) | M1 | CLM. Need all terms. Condone sign slips |
| A1 | Correct but check their directions for \(v_1 \& v_2\). Impact law. Must be used the right way round, but condone sign slips | |
| \(e(2u + u) = -v_1 + v_2\) | M1 | Directions of \(v_1 \& v_2\) must be consistent between the two equations. (Ignore the diagram if necessary) |
| A1 | Eliminate \(v_1\) to produce an equation in \(v_2\) only. Dependent on both previous M marks – must be using both equations | |
| \(\frac{u(2+9e)}{7} = v_2\) | DM1 | |
| A1 | DO NOT accept the negative. The question asks for speed | |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_1 = \frac{2u(1-6e)}{7}\) | M1 | Use the work from (a) or restart to find \(v_1\) or \(\lambda v_1\) for a constant \(\lambda\). If using work from (a) this mark is dependent on the first 2 M marks |
| A1 | a.e.f. Correct for their direction. Allow for \(\lambda v_1\) | |
| DM1 | An appropriate inequality for their \(v_1\) (seen or implied) – requires previous M1 scored. Work on \(v_1=0\) scores M0 until the inequality is formed | |
| \(v_1 < 0 \Rightarrow e > \frac{1}{6}\) | A1 | Accept \(\frac{2}{12}\). Answer must follow from correct work for \(v_1\) |
| \(l \ge e > \frac{1}{6}\) | B1 | For (their value) \(< e \le 1\) |
| SR: from \(v_1 \le 0\) could score M1A0B1 | ||
| (5) | ||
| 11 |
**(a)**
| $3m.2u - 4mu = 3mv_1 + 4mv_2$ | M1 | CLM. Need all terms. Condone sign slips |
| | A1 | Correct but check their directions for $v_1 \& v_2$. Impact law. Must be used the right way round, but condone sign slips |
| $e(2u + u) = -v_1 + v_2$ | M1 | Directions of $v_1 \& v_2$ must be consistent between the two equations. (Ignore the diagram if necessary) |
| | A1 | Eliminate $v_1$ to produce an equation in $v_2$ only. Dependent on both previous M marks – must be using both equations |
| $\frac{u(2+9e)}{7} = v_2$ | DM1 | |
| | A1 | DO NOT accept the negative. The question asks for speed |
| | (6) | |
**(b)**
| $v_1 = \frac{2u(1-6e)}{7}$ | M1 | Use the work from (a) or restart to find $v_1$ or $\lambda v_1$ for a constant $\lambda$. If using work from (a) this mark is dependent on the first 2 M marks |
| | A1 | a.e.f. Correct for their direction. Allow for $\lambda v_1$ |
| | DM1 | An appropriate inequality for their $v_1$ (seen or implied) – requires previous M1 scored. Work on $v_1=0$ scores M0 until the inequality is formed |
| $v_1 < 0 \Rightarrow e > \frac{1}{6}$ | A1 | Accept $\frac{2}{12}$. Answer must follow from correct work for $v_1$ |
| $l \ge e > \frac{1}{6}$ | B1 | For (their value) $< e \le 1$ |
| | | SR: from $v_1 \le 0$ could score M1A0B1 |
| | (5) |
| | 11 | |
---2. A particle $P$ of mass $3 m$ is moving with speed $2 u$ in a straight line on a smooth horizontal plane. The particle $P$ collides directly with a particle $Q$ of mass $4 m$ moving on the plane with speed $u$ in the opposite direction to $P$. The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $Q$ immediately after the collision.
Given that the direction of motion of $P$ is reversed by the collision,
\item find the range of possible values of $e$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q2 [11]}}