| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| 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 mechanics question involving friction on an inclined plane with multiple routine parts. Parts (i)-(iii) require straightforward application of friction formulas, work-energy principles, and potential energy calculations. Parts (iv)-(v) involve power calculations and energy methods with friction. While multi-step, each part follows textbook procedures without requiring novel insight or complex problem-solving, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = 80g\cos\theta\) or \(784\cos\theta\) | B1 | Seen |
| \(F_{\max} = \mu R\) | M1 | |
| so \(32g\cos\theta\) or \(313.6\cos\theta\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance is \(\frac{1.25}{\sin\theta}\) | B1 | |
| WD is \(F_{\max}d\) | M1 | |
| so \(32g\cos\theta \times \frac{1.25}{\sin\theta}\) | E1 | Award for this or equivalent seen |
| \(= \frac{392}{\tan\theta}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\Delta\)GPE is \(mgh\) | M1 | |
| so \(80 \times 9.8 \times 1.25 = 980\) J | A1 | Accept \(100g\) J |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| either \(P = Fv\) | M1 | |
| so \((80g\sin 35 + 32g\cos 35) \times 1.5\) | B1 | Weight term |
| A1 | All correct | |
| \(= 1059.85\ldots\) so \(1060\) W (3 s.f.) | A1 | cao |
| or \(P = \frac{\text{WD}}{\Delta t}\), so \(\frac{980 + \frac{392}{\tan 35}}{\left(\frac{1.25}{\sin 35}\right) \div 1.5}\) | M1, B1, B1 | Numerator FT their GPE; Denominator |
| \(= 1059.85\ldots\) so \(1060\) W (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| either Using W-E equation | M1 | Attempt speed at ground or dist to reach required speed. Allow only init KE omitted |
| \(0.5 \times 80 \times v^2 - 0.5 \times 80 \times \left(\frac{1}{2}\right)^2 = 980 - \frac{392}{\tan 35}\) | B1 | KE terms. Allow sign errors. FT from (iv). |
| B1 | Both WD against friction and GPE terms. Allow sign errors. FT from parts above. | |
| A1 | All correct | |
| \(v = 3.2793\ldots\) so yes | A1 | CWO |
| or N2L down slope; \(a = 2.409973\ldots\) | M1, A1 | All forces present |
| distance slid, using \(uvast\) is \(1.815372\ldots\) | A1 | |
| vertical distance is \(1.815372\ldots \times \sin 35 = 1.0412\ldots < 1.25\) so yes | M1, A1 | valid comparison; CWO |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 80g\cos\theta$ or $784\cos\theta$ | B1 | Seen |
| $F_{\max} = \mu R$ | M1 | |
| so $32g\cos\theta$ or $313.6\cos\theta$ N | A1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance is $\frac{1.25}{\sin\theta}$ | B1 | |
| WD is $F_{\max}d$ | M1 | |
| so $32g\cos\theta \times \frac{1.25}{\sin\theta}$ | E1 | Award for this or equivalent seen |
| $= \frac{392}{\tan\theta}$ | | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Delta$GPE is $mgh$ | M1 | |
| so $80 \times 9.8 \times 1.25 = 980$ J | A1 | Accept $100g$ J |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **either** $P = Fv$ | M1 | |
| so $(80g\sin 35 + 32g\cos 35) \times 1.5$ | B1 | Weight term |
| | A1 | All correct |
| $= 1059.85\ldots$ so $1060$ W (3 s.f.) | A1 | cao |
| **or** $P = \frac{\text{WD}}{\Delta t}$, so $\frac{980 + \frac{392}{\tan 35}}{\left(\frac{1.25}{\sin 35}\right) \div 1.5}$ | M1, B1, B1 | Numerator FT **their** GPE; Denominator |
| $= 1059.85\ldots$ so $1060$ W (3 s.f.) | A1 | cao |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **either** Using W-E equation | M1 | Attempt speed at ground or dist to reach required speed. Allow only init KE omitted |
| $0.5 \times 80 \times v^2 - 0.5 \times 80 \times \left(\frac{1}{2}\right)^2 = 980 - \frac{392}{\tan 35}$ | B1 | KE terms. Allow sign errors. FT from (iv). |
| | B1 | Both WD against friction and GPE terms. Allow sign errors. FT from parts above. |
| | A1 | All correct |
| $v = 3.2793\ldots$ so yes | A1 | CWO |
| **or** N2L down slope; $a = 2.409973\ldots$ | M1, A1 | All forces present |
| distance slid, using $uvast$ is $1.815372\ldots$ | A1 | |
| vertical distance is $1.815372\ldots \times \sin 35 = 1.0412\ldots < 1.25$ so yes | M1, A1 | valid comparison; CWO |
---2 One way to load a box into a van is to push the box so that it slides up a ramp. Some removal men are experimenting with the use of different ramps to load a box of mass 80 kg .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3865b4b3-97c7-412b-aabd-2705a954a847-3_345_1301_402_422}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Fig. 2 shows the general situation. The ramps are all uniformly rough with coefficient of friction 0.4 between the ramp and the box. The men push parallel to the ramp. As the box moves from one end of the ramp to the other it travels a vertical distance of 1.25 m .\\
(i) Find the limiting frictional force between the ramp and the box in terms of $\theta$.\\
(ii) From rest at the bottom, the box is pushed up the ramp and left at rest at the top. Show that the work done against friction is $\frac { 392 } { \tan \theta } \mathrm {~J}$.\\
(iii) Calculate the gain in the gravitational potential energy of the box when it is raised from the ground to the floor of the van.
For the rest of the question take $\theta = 35 ^ { \circ }$.\\
(iv) Calculate the power required to slide the box up the ramp at a steady speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(v) The box is given an initial speed of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the ramp and then slides down without anyone pushing it. Determine whether it reaches a speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ while it is on the ramp.
\hfill \mbox{\textit{OCR MEI M2 2009 Q2 [17]}}