| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Coalescence or perfectly inelastic collision |
| Difficulty | Moderate -0.3 This is a standard M2 mechanics question testing routine application of momentum conservation, energy methods, and forces on an incline. All parts follow textbook procedures with no novel problem-solving required, though the multi-step nature and combination of topics makes it slightly more substantial than the most basic exercises. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| WD against resistance \(=\) KE lost | M1 | |
| \(\frac{1}{2}\times0.04\times50^2=0.2F\) | A1 | |
| \(F=250\) and resistive force is 250 N | A1 | Accept \(-250\) |
| [3] | ||
| OR: Use suvat: \(v^2=u^2+2as\): \(a=-\frac{2500}{0.4}\) | M1 | Complete method: suvat and N2L |
| Use N2L: \(F=0.04\times a=-250\) | A1 | Correct \(a\) |
| Resistive force \(=250\) N | A1 | Correct \(F\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PCLM: \(0.04\times50=(3.96+0.04)V\) | M1 | |
| \(V=0.5\) so 0.5 m s\(^{-1}\) | A1 | cao |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Energy lost is \(\frac{1}{2}\times0.04\times50^2-\frac{1}{2}\times4\times0.5^2\) | M1 | Correct masses |
| \(=49.5\) J; equating WD against resistance to energy lost | A1ft | ft their 0.5 from (ii). May be implied |
| \(250x=49.5\) | M1 | ft their \(250\); \(x=\) a difference in non-zero KEs |
| \(x=0.198\) and distance is 0.198 m | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using W-E equation. Friction is \(F\) N: \(\frac{1}{2}\times6\times(7^2-1^2)=(91.5-F-6g\sin30)\times8\) | M1 | o.e. All 5 terms present, no extras. Allow sign errors |
| B1 | KE terms (both) | |
| B1 | Resolved weight or GPE term | |
| M1 | WD is Force \(\times\) distance | |
| \(F=44.1\) | A1 | cao |
| (As slipping) \(F=\mu R\) | M1 | Used |
| \(R=6g\cos30=6g\times\frac{\sqrt{3}}{2}\) | B1 | Does not need to be evaluated |
| \(\mu=\frac{44.1}{3g\times\sqrt{3}}=\frac{1.5}{\sqrt{3}}=\frac{\sqrt{3}}{2}\) (0.8660...) | A1 | cao. Any form. 0.87 or better |
| [8] | Using suvat and N2L: Max possible is B1 for resolved weight, then last 3 marks. Award SC(4) if N2L and suvat used, and \(\mu\) correct, www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Power is \(T\times v\), so \(91.5\times7=640.5\) W | M1 | |
| A1 | cao accept 640 or 641. Must be their final answer | |
| [2] |
# Question 2:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| WD against resistance $=$ KE lost | M1 | |
| $\frac{1}{2}\times0.04\times50^2=0.2F$ | A1 | |
| $F=250$ and resistive force is 250 N | A1 | Accept $-250$ |
| **[3]** | | |
| OR: Use suvat: $v^2=u^2+2as$: $a=-\frac{2500}{0.4}$ | M1 | Complete method: suvat and N2L |
| Use N2L: $F=0.04\times a=-250$ | A1 | Correct $a$ |
| Resistive force $=250$ N | A1 | Correct $F$ |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| PCLM: $0.04\times50=(3.96+0.04)V$ | M1 | |
| $V=0.5$ so 0.5 m s$^{-1}$ | A1 | cao |
| **[2]** | | |
## Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Energy lost is $\frac{1}{2}\times0.04\times50^2-\frac{1}{2}\times4\times0.5^2$ | M1 | Correct masses |
| $=49.5$ J; equating WD against resistance to energy lost | A1ft | ft their 0.5 from (ii). May be implied |
| $250x=49.5$ | M1 | ft their $250$; $x=$ a difference in non-zero KEs |
| $x=0.198$ and distance is 0.198 m | A1 | cao |
| **[4]** | | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using W-E equation. Friction is $F$ N: $\frac{1}{2}\times6\times(7^2-1^2)=(91.5-F-6g\sin30)\times8$ | M1 | o.e. All 5 terms present, no extras. Allow sign errors |
| | B1 | KE terms (both) |
| | B1 | Resolved weight or GPE term |
| | M1 | WD is Force $\times$ distance |
| $F=44.1$ | A1 | cao |
| (As slipping) $F=\mu R$ | M1 | Used |
| $R=6g\cos30=6g\times\frac{\sqrt{3}}{2}$ | B1 | Does not need to be evaluated |
| $\mu=\frac{44.1}{3g\times\sqrt{3}}=\frac{1.5}{\sqrt{3}}=\frac{\sqrt{3}}{2}$ (0.8660...) | A1 | cao. Any form. 0.87 or better |
| **[8]** | | Using suvat and N2L: Max possible is B1 for resolved weight, then last 3 marks. Award SC(4) if N2L and suvat used, and $\mu$ correct, www |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Power is $T\times v$, so $91.5\times7=640.5$ W | M1 | |
| | A1 | cao accept 640 or 641. Must be their final answer |
| **[2]** | | |
---2
\begin{enumerate}[label=(\alph*)]
\item A bullet of mass 0.04 kg is fired into a fixed uniform rectangular block along a line through the centres of opposite parallel faces, as shown in Fig. 2.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-3_209_1287_342_388}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
The bullet enters the block at $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and comes to rest after travelling 0.2 m into the block.
\begin{enumerate}[label=(\roman*)]
\item Calculate the resistive force on the bullet, assuming that this force is constant.
Another bullet of the same mass is fired, as before, with the same speed into a similar block of mass 3.96 kg . The block is initially at rest and is free to slide on a smooth horizontal plane.
\item By considering linear momentum, find the speed of the block with the bullet embedded in it and at rest relative to the block.
\item By considering mechanical energy, find the distance the bullet penetrates the block, given the resistance of the block to the motion of the bullet is the same as in part (i).
\end{enumerate}\item Fig. 2.2 shows a block of mass 6 kg on a uniformly rough plane that is inclined at $30 ^ { \circ }$ to the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-3_348_636_1382_712}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
A string with a constant tension of 91.5 N parallel to the plane pulls the block up a line of greatest slope. The speed of the block increases from $1 \mathrm {~ms} ^ { - 1 }$ to $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a distance of 8 m .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2016 Q2 [19]}}