| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with multiple routine parts: showing an object reaches a point using energy conservation, finding velocity with work done against friction, applying coefficient of restitution, and a straightforward power calculation. All techniques are textbook applications requiring no novel insight, though the multi-step nature and careful bookkeeping elevate it slightly above the easiest questions. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| KE at A is \(\frac{1}{2} \times 10 \times 16.6^2 = 1377.8\) J | ||
| GPE at B is \(10 \times 9.8 \times 14 = 1372\) J | M1 | Calculate relevant quantities (KE at A and PE at B or \(v=1.08\) at B or \(h=14.1\)) |
| KE at A \(>\) GPE at B so gets beyond B | E1 | Clear argued comparison (e.g. \(1377.8 > 1372\)) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| M1 | Use of \(WD = 14x\) | |
| B1 | \(x = 25\) | |
| A to D: \(\frac{1}{2}\times10\times v^2 - \frac{1}{2}\times10\times16.6^2\) | M1 | WE equation with at least one KE, \(\Delta\)GPE and WD by friction terms, all of correct form |
| \(= -10\times9.8\times7 - 25\times14\) | A1 | Allow only sign errors |
| \((v^2 = 68.36)\) | ||
| \(v = 8.2680\ldots\) so \(8.27\) (3 s.f.) | A1 | cao |
| *OR:* B to D: \(\frac{1}{2}\times10\times v^2 - (1377.8 - 1372)\) | M1 | WE equation with at least one KE, \(\Delta\)GPE and WD by friction terms, all of correct form |
| \(= 10\times9.8\times7 - 25\times14\) | M1 | Use of \(WD = 14x\) |
| B1 | \(x = 25\) | |
| A1 | Allow only sign errors | |
| \((v^2 = 68.36)\), \(v = 8.2680\ldots\) so \(8.27\) (3 s.f.) | A1 | cao |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V = \sqrt{2\times9.8\times7}\ (11.7)\) AND \(\frac{1}{2}V = \sqrt{2\times9.8\times h}\ (5.86)\) | M1 | Use of \(v^2 = 2gs\). Must be 7 in \(V\). Using '8.27' as \(u\) gives M0. \(e\) used appropriately: must use their attempt at a vertical velocity |
| \(h = \frac{1}{4}\times7 = 1.75\) | A1 | cao [Award SC 2 if 1.75 seen WWW] |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Driving force \((D) = \frac{P}{50}\) | B1 | Use of \(P =\) force \(\times\) velocity. May be implied e.g. by sight of \(0.8P/50\) or \(0.2P/50\) in N2L |
| \(P = 50F\ (D = F)\) | B1 | Accept any form |
| N2L along the road: \(\frac{0.8P}{50} - F = 1500\times -0.08\) | M1 | Use of N2L with all terms attempted and consistent with power reduction. Allow sign errors. |
| \(0.8P - 50F = -6000\) | A1 | Accept any form |
| Solving gives \(F = 600\quad P = 30\,000\) | M1 | Attempt to solve 2 equations each involving \(P\) and \(F\). Dependent on N2L equation attempted with 3 terms. |
| A1 | cao both [Taking 80% reduction in \(P\) gives \(P=7500\) and \(F=150\) for 5/6] | |
| [6] |
# Question 4(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| KE at A is $\frac{1}{2} \times 10 \times 16.6^2 = 1377.8$ J | | |
| GPE at B is $10 \times 9.8 \times 14 = 1372$ J | M1 | Calculate relevant quantities (KE at A and PE at B or $v=1.08$ at B or $h=14.1$) |
| KE at A $>$ GPE at B so gets beyond B | E1 | Clear argued comparison (e.g. $1377.8 > 1372$) |
| | **[2]** | |
---
# Question 4(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| | M1 | Use of $WD = 14x$ |
| | B1 | $x = 25$ |
| A to D: $\frac{1}{2}\times10\times v^2 - \frac{1}{2}\times10\times16.6^2$ | M1 | WE equation with at least one KE, $\Delta$GPE and WD by friction terms, all of correct form |
| $= -10\times9.8\times7 - 25\times14$ | A1 | Allow only sign errors |
| $(v^2 = 68.36)$ | | |
| $v = 8.2680\ldots$ so $8.27$ (3 s.f.) | A1 | cao |
| *OR:* B to D: $\frac{1}{2}\times10\times v^2 - (1377.8 - 1372)$ | M1 | WE equation with at least one KE, $\Delta$GPE and WD by friction terms, all of correct form |
| $= 10\times9.8\times7 - 25\times14$ | M1 | Use of $WD = 14x$ |
| | B1 | $x = 25$ |
| | A1 | Allow only sign errors |
| $(v^2 = 68.36)$, $v = 8.2680\ldots$ so $8.27$ (3 s.f.) | A1 | cao |
| | **[5]** | |
---
# Question 4(a)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = \sqrt{2\times9.8\times7}\ (11.7)$ AND $\frac{1}{2}V = \sqrt{2\times9.8\times h}\ (5.86)$ | M1 | Use of $v^2 = 2gs$. Must be 7 in $V$. Using '8.27' as $u$ gives M0. $e$ used appropriately: must use their attempt at a vertical velocity |
| $h = \frac{1}{4}\times7 = 1.75$ | A1 | cao [Award SC 2 if 1.75 seen WWW] |
| | **[3]** | |
---
# Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Driving force $(D) = \frac{P}{50}$ | B1 | Use of $P =$ force $\times$ velocity. May be implied e.g. by sight of $0.8P/50$ or $0.2P/50$ in N2L |
| $P = 50F\ (D = F)$ | B1 | Accept any form |
| N2L along the road: $\frac{0.8P}{50} - F = 1500\times -0.08$ | M1 | Use of N2L with all terms attempted and consistent with power reduction. Allow sign errors. |
| $0.8P - 50F = -6000$ | A1 | Accept any form |
| Solving gives $F = 600\quad P = 30\,000$ | M1 | Attempt to solve 2 equations each involving $P$ and $F$. Dependent on N2L equation attempted with 3 terms. |
| | A1 | cao both [Taking 80% reduction in $P$ gives $P=7500$ and $F=150$ for 5/6] |
| | **[6]** | |4
\begin{enumerate}[label=(\alph*)]
\item A small heavy object of mass 10 kg travels the path ABCD which is shown in Fig. 4. ABCD is in a vertical plane; CD and AEF are horizontal. The sections of the path AB and CD are smooth but section BC is rough.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-5_368_1323_402_338}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
You should assume that
\begin{itemize}
\item the object does not leave the path when travelling along ABCD and does not lose energy when changing direction
\item there is no air resistance.
\end{itemize}
Initially, the object is projected from A at a speed of $16.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up the slope.
\begin{enumerate}[label=(\roman*)]
\item Show that the object gets beyond B .
The section of the path BC produces a constant resistance of 14 N to the motion of the object.
\item Using an energy method, find the velocity of the object at D .
At D , the object leaves the path and bounces on the smooth horizontal ground between E and F , shown in Fig. 4. The coefficient of restitution in the collision of the object with the ground is $\frac { 1 } { 2 }$.
\item Calculate the greatest height above the ground reached by the object after its first bounce.
\end{enumerate}\item A car of mass 1500 kg travelling along a straight, horizontal road has a steady speed of $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when its driving force has power $P \mathrm {~W}$.
When at this speed, the power is suddenly reduced by $20 \%$. The resistance to the car's motion, $F \mathrm {~N}$, does not change and the car begins to decelerate at $0.08 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Calculate the values of $P$ and $F$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2014 Q4 [16]}}