| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Explosion or separation of particles |
| Difficulty | Standard +0.3 This is a multi-part mechanics question involving standard momentum conservation, impulse calculation, and collision analysis. While it has several steps and requires careful bookkeeping of velocities and directions, each individual component uses routine A-level mechanics techniques (conservation of momentum, equations of motion with friction, coefficient of restitution formula). The 'show that' part in (ii) guides students to the answer, reducing problem-solving demand. Slightly easier than average due to its structured, procedural nature. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Take j north and i east; velocity before \(5\mathbf{i} - 5\sqrt{3}\mathbf{j}\), after \(3\mathbf{i}\) | B1 | Resolving initial velocity (may be implied). Allow \(5\mathbf{i} + 5\sqrt{3}\mathbf{j}\) or \(5\mathbf{i} - 5\sqrt{3}\mathbf{j}\) |
| \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\) | M1 | May be implied. Allow if only one direction considered or both combined without vectors. Must include attempt to resolve 10 |
| \(\mathbf{I} = 120\,000\,000(-2\mathbf{i} + 5\sqrt{3}\mathbf{j})\); Modulus is \(120\,000\,000 \times 8.888194...= 1.0665... \times 10^9\) N s | A1 | Accept mass of 120 000 |
| \(1.07 \times 10^9\) N s (to 3 s.f.) | A1 | cao; Alternative method using diagram, cos and sine rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PCLM: \(0.4 \times 6 = 0.5V\) | M1 | Implied by 4.8 or -4.8 |
| \(V = 4.8\) ms\(^{-1}\) direction opposite to P | A1 | Allow -4.8 as the speed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| P travels \(6 \times \frac{2}{3} = 4\) m before collision | B1 | Or find \(t = \frac{13}{24}\) for time from edge to collision AND \(d = 3.25\) |
| Q travels \(4 - 2 \times 0.75 = 2.5\) m in \(\frac{2}{3}\) s | B1 | \(3.25 - 0.75 = 2.5\) |
| \(2.5 = \frac{(4.8 + v_Q)}{2} \times \frac{2}{3}\) | M1 | Using appropriate suvat, FT their 2.5 |
| \(v_Q = 2.7\) ms\(^{-1}\) | E1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Suppose friction on Q is \(F\); \(-F \times \frac{2}{3} = 0.5(2.7 - 4.8)\) so \(F = 1.575\) | B1 | Using \(Ft = m(v-u)\) or find \(a = -3.15\) and use \(F = ma\). FT their 2.7 |
| \(1.575 = \mu \times 0.5 \times 9.8\) | M1 | \(F = \mu R\) |
| \(R\) correct (4.9) | A1 | |
| \(\mu = 0.321\) (3 s.f.) | A1 | cao; Note: \(F=ma\) and \(R=mg\) give \(\mu = \frac{a}{g}\) (M1A1). Find \(a=-3.15\)(B1) gives 0.321 (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PCLM: \(0.4 \times 6 + 0.5 \times 2.7 = 0.4V_P + 0.5V_Q\) | M1 | PCLM. FT their 2.7 from (ii). Award M1A0 for use of 4.8 from (i) instead of 2.7 |
| \(4V_P + 5V_Q = 37.5\) | A1 | FT their 2.7 from (ii). Accept any form |
| NEL: \(\frac{V_Q - V_P}{2.7 - 6} = -\frac{1}{8}\) | M1 | NEL. FT their 2.7 from (ii). Award M1A0 for use of 4.8 from (i) instead of 2.7 |
| \(V_Q - V_P = 0.4125\) | A1 | FT their 2.7 from (ii). Accept any form |
| \(V_Q = 4.35\) ms\(^{-1}\) | A1 | cao |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Take **j** north and **i** east; velocity before $5\mathbf{i} - 5\sqrt{3}\mathbf{j}$, after $3\mathbf{i}$ | B1 | Resolving initial velocity (may be implied). Allow $5\mathbf{i} + 5\sqrt{3}\mathbf{j}$ or $5\mathbf{i} - 5\sqrt{3}\mathbf{j}$ |
| $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ | M1 | May be implied. Allow if only one direction considered or both combined without vectors. Must include attempt to resolve 10 |
| $\mathbf{I} = 120\,000\,000(-2\mathbf{i} + 5\sqrt{3}\mathbf{j})$; Modulus is $120\,000\,000 \times 8.888194...= 1.0665... \times 10^9$ N s | A1 | Accept mass of 120 000 |
| $1.07 \times 10^9$ N s (to 3 s.f.) | A1 | cao; Alternative method using diagram, cos and sine rules |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| PCLM: $0.4 \times 6 = 0.5V$ | M1 | Implied by 4.8 or -4.8 |
| $V = 4.8$ ms$^{-1}$ direction opposite to P | A1 | Allow -4.8 as the speed |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| P travels $6 \times \frac{2}{3} = 4$ m before collision | B1 | Or find $t = \frac{13}{24}$ for time from edge to collision AND $d = 3.25$ |
| Q travels $4 - 2 \times 0.75 = 2.5$ m in $\frac{2}{3}$ s | B1 | $3.25 - 0.75 = 2.5$ |
| $2.5 = \frac{(4.8 + v_Q)}{2} \times \frac{2}{3}$ | M1 | Using appropriate suvat, FT their 2.5 |
| $v_Q = 2.7$ ms$^{-1}$ | E1 | Answer given |
## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Suppose friction on Q is $F$; $-F \times \frac{2}{3} = 0.5(2.7 - 4.8)$ so $F = 1.575$ | B1 | Using $Ft = m(v-u)$ or find $a = -3.15$ and use $F = ma$. FT their 2.7 |
| $1.575 = \mu \times 0.5 \times 9.8$ | M1 | $F = \mu R$ |
| $R$ correct (4.9) | A1 | |
| $\mu = 0.321$ (3 s.f.) | A1 | cao; Note: $F=ma$ and $R=mg$ give $\mu = \frac{a}{g}$ (M1A1). Find $a=-3.15$(B1) gives 0.321 (A1) |
## Part (b)(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| PCLM: $0.4 \times 6 + 0.5 \times 2.7 = 0.4V_P + 0.5V_Q$ | M1 | PCLM. FT their 2.7 from (ii). Award M1A0 for use of 4.8 from (i) instead of 2.7 |
| $4V_P + 5V_Q = 37.5$ | A1 | FT their 2.7 from (ii). Accept any form |
| NEL: $\frac{V_Q - V_P}{2.7 - 6} = -\frac{1}{8}$ | M1 | NEL. FT their 2.7 from (ii). Award M1A0 for use of 4.8 from (i) instead of 2.7 |
| $V_Q - V_P = 0.4125$ | A1 | FT their 2.7 from (ii). Accept any form |
| $V_Q = 4.35$ ms$^{-1}$ | A1 | cao |
---1
\begin{enumerate}[label=(\alph*)]
\item Fig. 1.1 shows the velocities of a tanker of mass 120000 tonnes before and after it changed speed and direction.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-2_237_917_360_577}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
Calculate the magnitude of the impulse that acted on the tanker.
\item An object of negligible size is at rest on a horizontal surface. It explodes into two parts, P and Q , which then slide along the surface.
Part P has mass 0.4 kg and speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Part Q has mass 0.5 kg .
\begin{enumerate}[label=(\roman*)]
\item Calculate the speed of Q immediately after the explosion. State how the directions of motion of P and Q are related.
The explosion takes place at a distance of 0.75 m from a raised vertical edge, as shown in Fig. 1.2. P travels along a line perpendicular to this edge.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-2_238_1205_1366_429}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
After the explosion, P has a perfectly elastic direct collision with the raised edge and then collides again directly with Q . The collision between P and Q occurs $\frac { 2 } { 3 } \mathrm {~s}$ after the explosion. Both collisions are instantaneous.
The contact between P and the surface is smooth but there is a constant frictional force between Q and the surface.
\item Show that Q has speed $2.7 \mathrm {~ms} ^ { - 1 }$ just before P collides with it.
\item Calculate the coefficient of friction between Q and the surface.
\item Given that the coefficient of restitution between P and Q is $\frac { 1 } { 8 }$, calculate the speed of Q immediately after its collision with P .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2013 Q1 [19]}}