| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Session | Specimen |
| Marks | 13 |
| 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 multi-part energy conservation question with standard mechanics techniques. Parts (i)-(ii) are routine calculations (mgh and work-energy theorem). Part (iii) requires setting up an energy equation with friction work on an incline, which is standard A-level Further Mechanics. Part (iv) involves finding a range and making a practical comment, but the mathematics remains straightforward. Slightly above average difficulty due to the multi-step nature and friction on an incline, but all techniques are standard textbook applications. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | Change in GPE is 40 × 9.8 × 12 = 4704 J |
| [1] | 1.1 | n |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (ii) | 1 |
| Answer | Marks |
|---|---|
| = 5508 J | M1 |
| Answer | Marks |
|---|---|
| [3] | m3.3 |
| Answer | Marks |
|---|---|
| 1.1 | e |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iii) | e |
| Answer | Marks |
|---|---|
| so α = 30.227599… so 30.2° ( 3 s.f.) | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Use of F (cid:32)(cid:80)R |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iv) | (A) |
| 29.3° < α < 30.2°. | B1 | |
| [1] | 3.4 | |
| 6 | (iv) | (B) |
| [1] | 3.5a |
Question 6:
6 | (i) | Change in GPE is 40 × 9.8 × 12 = 4704 J | B1
[1] | 1.1 | n
4 s.f. not required
6 | (ii) | 1
4704(cid:14) (cid:117)40(cid:117)42 + 484
2
= 5508 J | M1
B1
c
A1
[3] | m3.3
i
1.1
1.1 | e
Use of GPE and KE and WD.
Accept wrong
sign
KE term correct
FT only from (i). 4 s.f. not
required
6 | (iii) | e
12
Distance travelled is p
sin(cid:68)
Frictional force is 0.6(cid:117)40gcos(cid:68)
S
Using the WE equation
1 1 12
(cid:117)40(cid:117)32 (cid:16) (cid:117)40(cid:117)42 (cid:32)4704(cid:16)0.6(cid:117)40gcos(cid:68)(cid:117)
2 2 sin(cid:68)
tan(cid:68)(cid:32)0.5826589...
so α = 30.227599… so 30.2° ( 3 s.f.) | B1
M1
B1
M1
B1
B1
A1
[7] | 1.1
3.4
1.1
1.1
3.4
1.1
1.1 | Use of F (cid:32)(cid:80)R
Value of R
All terms present. Allow sign
errors
KE terms. Allow sign errors
Friction term. Allow sign errors
FT only from (i)
Must show ° or rad (0.52757…)
6 | (iv) | (A) | For speed 0 at D, α = 29.3°
29.3° < α < 30.2°. | B1
[1] | 3.4
6 | (iv) | (B) | Not practical to keep angle within such small margins. | B1
[1] | 3.5a6 A sack of beans of mass 40 kg is pulled from rest at point A up a non-uniform slope onto and along a horizontal platform. Fig. 6 shows this slope AB and the platform BC , which is a vertical distance of 12 m above A.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-6_253_1203_504_477}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the gain in the gravitational potential energy of the sack when it is moved from A to the platform.
The sack has a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ by the time it reaches C at the far end of the platform. The total work done against friction in moving the sack from A to C is 484 J . There are no other resistances to the sack's motion.
\item Calculate the total work done in moving the sack between the points A and C .
At point C , travelling at $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the sack starts to slide down a straight chute inclined at $\alpha$ to the horizontal. Point D at the bottom of the chute is at the same vertical height as A , as shown in Fig. 6. The chute is rough and the coefficient of friction between the chute and the sack is 0.6 . During this part of the motion, again the only resistance to the motion of the sack is friction.
\item Use an energy method to calculate the value of $\alpha$ given that the sack is travelling at $3 \mathrm {~ms} ^ { - 1 }$ when it reaches D .
For safety reasons the sack needs to arrive at D with a speed of less than $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The value of $\alpha$ can be adjusted to try to achieve this.
\item (A) Find the range of values of $\alpha$ which achieve a safe speed at D .\\
(B) Comment on whether adjusting $\alpha$ is a practical way of achieving a safe speed at D .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS Q6 [13]}}