| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Moderate -0.3 This is a standard M2 collision problem requiring routine application of conservation of momentum and Newton's law of restitution. Part (a) involves algebraic manipulation to find a constraint on e, while part (b) is direct impulse calculation. The multi-step nature and algebraic work make it slightly easier than average, but it follows a well-practiced template with no novel insight required. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLM: \(0.7 \times 6 = 0.7 \times v + 1.2w\) | M1 | Requires all terms & dimensionally correct |
| \(42 = 7v + 12w\) | A1 | Correct unsimplified |
| Impact: \(w - v = 6e\) | M1, A1 | Used the right way round, condone sign errors |
| Equation in \(e\) and \(v\) only: \(42 - 72e = 19v\) | DM1 | Dependent on the two previous M marks |
| Use direction to form an inequality | M1 | Independent. Applied correctly for their \(v\) |
| \(42 - 72e > 0 \Rightarrow e < \dfrac{7}{12}\) | A1 | Given answer |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse on \(Q\): \(I = w \times 1.2\) | M1 | |
| Solve for \(w\): \(w = v + 6e = \dfrac{42 - 72 \times \frac{1}{4}}{19} + 6 \times \dfrac{1}{4}\) | B1 | Accept unsimplified with \(e\) substituted. Have to be using \(w\) in part (b). \(w = \dfrac{105}{38} = 2.763...\) seen or implied |
| \(I = 1.2 \times \dfrac{42}{19} \times \dfrac{5}{4} = \dfrac{63}{19} (= 3.32)\) (N s) | A1 | 3.3 or better |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse on \(Q = -\) impulse on \(P\) \(= -0.7(v-6)\) | M1 | Accept negative here |
| \(= -0.7\left(\dfrac{42 - \frac{1}{4} \times 72}{19} - 6\right)\) | B1 | Substitute for \(e\) in their \(v\). \(v = \dfrac{24}{19} = 1.263...\) seen or implied. Accept negative here. |
| \(= \dfrac{63}{19}\) | A1 | Final answer must be positive. 3.3 or better |
| (3) |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM: $0.7 \times 6 = 0.7 \times v + 1.2w$ | M1 | Requires all terms & dimensionally correct |
| $42 = 7v + 12w$ | A1 | Correct unsimplified |
| Impact: $w - v = 6e$ | M1, A1 | Used the right way round, condone sign errors |
| Equation in $e$ and $v$ only: $42 - 72e = 19v$ | DM1 | Dependent on the two previous M marks |
| Use direction to form an inequality | M1 | Independent. Applied correctly for their $v$ |
| $42 - 72e > 0 \Rightarrow e < \dfrac{7}{12}$ | A1 | **Given answer** |
| | **(7)** | |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse on $Q$: $I = w \times 1.2$ | M1 | |
| Solve for $w$: $w = v + 6e = \dfrac{42 - 72 \times \frac{1}{4}}{19} + 6 \times \dfrac{1}{4}$ | B1 | Accept unsimplified with $e$ substituted. Have to be using $w$ in part (b). $w = \dfrac{105}{38} = 2.763...$ seen or implied |
| $I = 1.2 \times \dfrac{42}{19} \times \dfrac{5}{4} = \dfrac{63}{19} (= 3.32)$ (N s) | A1 | 3.3 or better |
| | **(3)** | |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse on $Q = -$ impulse on $P$ $= -0.7(v-6)$ | M1 | Accept negative here |
| $= -0.7\left(\dfrac{42 - \frac{1}{4} \times 72}{19} - 6\right)$ | B1 | Substitute for $e$ in their $v$. $v = \dfrac{24}{19} = 1.263...$ seen or implied. Accept negative here. |
| $= \dfrac{63}{19}$ | A1 | Final answer must be positive. 3.3 or better |
| | **(3)** | |
---2. A particle $P$ of mass 0.7 kg is moving in a straight line on a smooth horizontal surface. The particle $P$ collides with a particle $Q$ of mass 1.2 kg which is at rest on the surface. Immediately before the collision the speed of $P$ is $6 \mathrm {~ms} ^ { - 1 }$. Immediately after the collision both particles are moving in the same direction. The coefficient of restitution between the particles is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that $e < \frac { 7 } { 12 }$
Given that $e = \frac { 1 } { 4 }$
\item find the magnitude of the impulse exerted on $Q$ in the collision.\\
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\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q2 [10]}}