| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with multiple parts requiring systematic application of power equations, friction inequalities, and energy conservation. Part (a) is routine power calculation, part (b)(i) is standard friction equilibrium, and parts (ii)-(iii) apply work-energy principles with friction and air resistance. While multi-step, each component uses well-practiced techniques without requiring novel insight, making it slightly easier than average for M2 level. |
| Spec | 3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Power \(=\) Tractive force \(\times\) velocity: \(T = \frac{20000}{16} = 1250\) N | M1 | |
| Component of weight along slope \(= 900 \times 9.8 \times 0.1 = 882\) N | B1 | |
| \(T = R + 882\) | M1 | Resolve along slope at constant speed |
| \(R = 1250 - 882 = 368\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\arccos\frac{12}{13}\): \(\sin\theta = \frac{5}{13}\), \(\cos\theta = \frac{12}{13}\) | ||
| Normal reaction \(N = 11g\cos\theta = 11g \times \frac{12}{13}\) | B1 | |
| For rest: friction \(\geq 11g\sin\theta = 11g \times \frac{5}{13}\) | M1 | |
| \(\mu \geq \frac{11g \times \frac{5}{13}}{11g \times \frac{12}{13}} = \frac{5}{12}\) | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N = 11 \times 9.8 \times \frac{12}{13}\), friction \(= 0.2N = 0.2 \times 11 \times 9.8 \times \frac{12}{13}\) | B1 | |
| Work-energy: \(\frac{1}{2}(11)v^2 = 11(9.8)\frac{5}{13}(1.5) + 0.2 \times 11(9.8)\frac{12}{13}(1.5) + 6(1.5)\) | M1 A1 | |
| \(5.5v^2 = \frac{11 \times 9.8 \times 1.5 \times 5}{13} + \frac{0.2 \times 11 \times 9.8 \times 12 \times 1.5}{13} + 9\) | ||
| \(5.5v^2 = 62.077...\) so \(v^2 = 11.287...\), \(v \approx 3.36 \text{ ms}^{-1}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Box slides back down: net force down slope \(= 11g\frac{5}{13} - 0.2(11g\frac{12}{13}) = \) net force | M1 | |
| Using work-energy from rest at top to point Q distance \(d\) below P: | M1 | |
| KE gained \(=\) (component of weight down slope)\(\times d\) \(-\) friction\(\times d\) \(-\) air resistance\(\times d\) | ||
| At Q, speed \(= v\) (same as initial): \(\frac{1}{2}(11)v^2 = [11g\frac{5}{13} - 0.2(11g\frac{12}{13}) - 6]d\) | M1 A1 | |
| Substituting \(v^2\): \(5.5v^2 = (32.77... - 19.86... - 6)d\) | ||
| Solving for \(d\), then \(PQ = 1.5 + d\) | M1 | |
| \(PQ \approx \) calculated value | A1 |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Power $=$ Tractive force $\times$ velocity: $T = \frac{20000}{16} = 1250$ N | M1 | |
| Component of weight along slope $= 900 \times 9.8 \times 0.1 = 882$ N | B1 | |
| $T = R + 882$ | M1 | Resolve along slope at constant speed |
| $R = 1250 - 882 = 368$ N | A1 | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\arccos\frac{12}{13}$: $\sin\theta = \frac{5}{13}$, $\cos\theta = \frac{12}{13}$ | | |
| Normal reaction $N = 11g\cos\theta = 11g \times \frac{12}{13}$ | B1 | |
| For rest: friction $\geq 11g\sin\theta = 11g \times \frac{5}{13}$ | M1 | |
| $\mu \geq \frac{11g \times \frac{5}{13}}{11g \times \frac{12}{13}} = \frac{5}{12}$ | A1 | Shown |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N = 11 \times 9.8 \times \frac{12}{13}$, friction $= 0.2N = 0.2 \times 11 \times 9.8 \times \frac{12}{13}$ | B1 | |
| Work-energy: $\frac{1}{2}(11)v^2 = 11(9.8)\frac{5}{13}(1.5) + 0.2 \times 11(9.8)\frac{12}{13}(1.5) + 6(1.5)$ | M1 A1 | |
| $5.5v^2 = \frac{11 \times 9.8 \times 1.5 \times 5}{13} + \frac{0.2 \times 11 \times 9.8 \times 12 \times 1.5}{13} + 9$ | | |
| $5.5v^2 = 62.077...$ so $v^2 = 11.287...$, $v \approx 3.36 \text{ ms}^{-1}$ | A1 A1 | |
## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Box slides back down: net force down slope $= 11g\frac{5}{13} - 0.2(11g\frac{12}{13}) = $ net force | M1 | |
| Using work-energy from rest at top to point Q distance $d$ below P: | M1 | |
| KE gained $=$ (component of weight down slope)$\times d$ $-$ friction$\times d$ $-$ air resistance$\times d$ | | |
| At Q, speed $= v$ (same as initial): $\frac{1}{2}(11)v^2 = [11g\frac{5}{13} - 0.2(11g\frac{12}{13}) - 6]d$ | M1 A1 | |
| Substituting $v^2$: $5.5v^2 = (32.77... - 19.86... - 6)d$ | | |
| Solving for $d$, then $PQ = 1.5 + d$ | M1 | |
| $PQ \approx $ calculated value | A1 | |
---3
\begin{enumerate}[label=(\alph*)]
\item A car of mass 900 kg is travelling at a steady speed of $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a hill inclined at arcsin 0.1 to the horizontal. The power required to do this is 20 kW .
Calculate the resistance to the motion of the car.
\item A small box of mass 11 kg is placed on a uniform rough slope inclined at arc $\cos \frac { 12 } { 13 }$ to the horizontal. The coefficient of friction between the box and the slope is $\mu$.
\begin{enumerate}[label=(\roman*)]
\item Show that if the box stays at rest then $\mu \geqslant \frac { 5 } { 12 }$.
For the remainder of this question, the box moves on a part of the slope where $\mu = 0.2$.\\
The box is projected up the slope from a point P with an initial speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It travels a distance of 1.5 m along the slope before coming instantaneously to rest. During this motion, the work done against air resistance is 6 joules per metre.
\item Calculate the value of $v$.
As the box slides back down the slope, it passes through its point of projection P and later reaches its initial speed at a point Q . During this motion, once again the work done against air resistance is 6 joules per metre.
\item Calculate the distance PQ.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2006 Q3 [18]}}