| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Standard +0.3 This is a standard M2 question testing routine applications of conservation of momentum, kinetic energy equations, and work-energy principles. Part (a) requires simultaneous equations from momentum and energy conservation—a textbook exercise. Parts (b)(i) and (b)(ii) are straightforward energy method calculations with clear given values and standard formulas. All parts follow predictable solution paths with no novel insight required, making this slightly easier than average. |
| 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 = Fv6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = -15\sqrt{\frac{5}{3}} (= -\sqrt{375i})\) | M1, A1, M1, A1, M1, A1, F1, A1 | Use of KE in two terms in an equation. Any form. PCLM. Accept sign errors. Any form. Valid method for elimination of \(v\) or \(V\) from a linear and a quadratic. Accept 1.29099...i. Accept no direction. Accept – 19.3649...i. Accept no direction. Second answer follows from first (Relative) directions indicated - accept diagram. Both speeds correct |
| Answer | Marks | Guidance |
|---|---|---|
| so 145 600 J done by car against resistances | M1, B1, A1, A1 | Use of WE. Must have KE, W and GPE. Allow -W. Both KE terms. Accept sign error. All correct with W or -W. cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(800 \times 9.8 \times \frac{1}{22.5} + 750\right) \times 18 = 19,772\) W | B1, M1, M1, A1, A1, B1, M1, M1, A1, A1 | Use of \(P = \frac{\text{Work done}}{(\text{elapsed time})}\) used for at least one work done term. WD is force × distance used for at least one force. Allow only sign errors both terms. cao. Use of \(P = Fv\) used for at least one term. Attempt at weight component. Allow only sign errors both terms. cao |
## Part (a)
Energy: $\frac{1}{2} \times 0.004 \times v^2 + \frac{1}{2} \times 0.060 \times V^2 = 0.8$
$v^2 + 15V^2 = 400$
PCLM in $i$ direction: $0.06V - 0.004v = 0$
$v = 15V$
Solving
$(15V)^2 + 15V^2 = 400$
so $V^2 = \frac{400}{240} = \frac{5}{3}$ and $V = \sqrt{\frac{5}{3}}$
$v = -15\sqrt{\frac{5}{3}} (= -\sqrt{375i})$ | M1, A1, M1, A1, M1, A1, F1, A1 | Use of KE in two terms in an equation. Any form. PCLM. Accept sign errors. Any form. Valid method for elimination of $v$ or $V$ from a linear and a quadratic. Accept 1.29099...i. Accept no direction. Accept – 19.3649...i. Accept no direction. Second answer follows from first (Relative) directions indicated - accept diagram. Both speeds correct | 8 marks
## Part (b)(i)
W is work done by resistances on car
$\frac{1}{2} \times 800 \times (12^2 - 30^2) = -800 \times 9.8 \times 20 + W$
$W = - 145,600$
so 145 600 J done by car against resistances | M1, B1, A1, A1 | Use of WE. Must have KE, W and GPE. Allow -W. Both KE terms. Accept sign error. All correct with W or -W. cao | 4 marks
## Part (b)(ii)
**either**
The slope is 18 × 25 = 450 m long
$\frac{800 \times 9.8 \times 20 + 750 \times 450}{25} = 19,772$ W
**or**
The angle of the slope is arcsin (1/22.5)
$\left(800 \times 9.8 \times \frac{1}{22.5} + 750\right) \times 18 = 19,772$ W | B1, M1, M1, A1, A1, B1, M1, M1, A1, A1 | Use of $P = \frac{\text{Work done}}{(\text{elapsed time})}$ used for at least one work done term. WD is force × distance used for at least one force. Allow only sign errors both terms. cao. Use of $P = Fv$ used for at least one term. Attempt at weight component. Allow only sign errors both terms. cao | 5 marks
---\begin{enumerate}[label=(\alph*)]
\item A firework is instantaneously at rest in the air when it explodes into two parts. One part is the body B of mass 0.06 kg and the other a cap C of mass 0.004 kg. The total kinetic energy given to B and C is 0.8 J. B moves off horizontally in the $\mathbf{i}$ direction.
By considering both kinetic energy and linear momentum, calculate the velocities of B and C immediately after the explosion. [8]
\item A car of mass 800 kg is travelling up some hills.
In one situation the car climbs a vertical height of 20 m while its speed decreases from 30 m s$^{-1}$ to 12 m s$^{-1}$. The car is subject to a resistance to its motion but there is no driving force and the brakes are not being applied.
\begin{enumerate}[label=(\roman*)]
\item Using an energy method, calculate the work done by the car against the resistance to its motion. [4]
\end{enumerate}
In another situation the car is travelling at a constant speed of 18 m s$^{-1}$ and climbs a vertical height of 20 m in 25 s up a uniform slope. The resistance to its motion is now 750 N.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item Calculate the power of the driving force required. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2011 Q2 [17]}}