| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle moving through liquid or resistance |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic SUVAT application, reading values from a speed-time graph, calculating areas under graphs, and simple integration. All parts are routine textbook exercises with no problem-solving insight required—easier than average A-level maths. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Area under curve \(0.5 \times 2 \times 20 + 0.5 \times (20 + 10) \times 4 + 0.5 \times 10 \times 1 = 85\) m | M1 | Attempt to find any area under curve or use const accn results |
| B1 | Any area correct (Accept 20 or 60 or 5 without explanation) | |
| A1 | cao | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{20 - 10}{4} = 2.5\) upwards | M1 | \(\Delta v / \Delta t\) |
| A1 | accept \(\pm 2.5\) | |
| B1 | Accept – 2.5 downwards (allow direction specified by diagram etc). Accept 'opposite direction to motion'. | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = -2.5t + c\) | M1 | Allow their \(a\) in the form \(v = \pm at + c\) or \(v = \pm a(t - 2) + c\) |
| \(v = 20\) when \(t = 2\) | M1 | |
| \(v = -2.5t + 25\) | A1 | cao [Allow \(v = 20 - 2.5(t - 2)\)] [Allow 2/3 for different variable to \(t\) used, e.g. \(x\). Allow any variable name for speed] |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Falling with negligible resistance | E1 | Accept 'zero resistance', or 'no resistance' seen. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1.5 \times 4 + 9.5 \times 2 + 7 = 20\) | E1 | One of the results shown |
| \(-1.5 \times 36 + 9.5 \times 6 + 7 = 10\) | E1 | |
| \(-1.5 \times 49 + 9.5 \times 7 + 7 = 0\) | E1 | All three shown. Be generous about the 'show'. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_2^7 (-1.5t^2 + 9.5t + 7) dt\) | M1 | Limits not required |
| \(= [-0.5t^3 + 4.75t^2 + 7t]_2^7\) | A1 | A1 for each term. Limits not required. Condone \(+ c\) |
| A1 | ||
| \(= \left( -\frac{343}{2} - \frac{19 \times 49}{4} + 49 \right) - (-4 + 19 + 14)\) | M1 | Attempt to use both limits on an integrated expression |
| A1 | Correct substitution in their expression including subtraction (may be left as an expression). | |
| \(= 81.25\) m | A1 | cao. |
| Total: 7 |
**Part (i)**
| Area under curve $0.5 \times 2 \times 20 + 0.5 \times (20 + 10) \times 4 + 0.5 \times 10 \times 1 = 85$ m | M1 | Attempt to find any area under curve or use const accn results |
| | B1 | Any area correct (Accept 20 or 60 or 5 without explanation) |
| | A1 | cao |
| | | **Total: 3** |
**Part (ii)**
| $\frac{20 - 10}{4} = 2.5$ upwards | M1 | $\Delta v / \Delta t$ |
| | A1 | accept $\pm 2.5$ |
| | B1 | Accept – 2.5 downwards (allow direction specified by diagram etc). Accept 'opposite direction to motion'. |
| | | **Total: 3** |
**Part (iii)**
| $v = -2.5t + c$ | M1 | Allow their $a$ in the form $v = \pm at + c$ or $v = \pm a(t - 2) + c$ |
| $v = 20$ when $t = 2$ | M1 | |
| $v = -2.5t + 25$ | A1 | cao [Allow $v = 20 - 2.5(t - 2)$] [Allow 2/3 for different variable to $t$ used, e.g. $x$. Allow any variable name for speed] |
| | | **Total: 3** |
**Part (iv)**
| Falling with negligible resistance | E1 | Accept 'zero resistance', or 'no resistance' seen. |
| | | **Total: 1** |
**Part (v)**
| $-1.5 \times 4 + 9.5 \times 2 + 7 = 20$ | E1 | One of the results shown |
| $-1.5 \times 36 + 9.5 \times 6 + 7 = 10$ | E1 | |
| $-1.5 \times 49 + 9.5 \times 7 + 7 = 0$ | E1 | All three shown. Be generous about the 'show'. |
| | | **Total: 2** |
**Part (vi)**
| $\int_2^7 (-1.5t^2 + 9.5t + 7) dt$ | M1 | Limits not required |
| $= [-0.5t^3 + 4.75t^2 + 7t]_2^7$ | A1 | A1 for each term. Limits not required. Condone $+ c$ |
| | A1 | |
| $= \left( -\frac{343}{2} - \frac{19 \times 49}{4} + 49 \right) - (-4 + 19 + 14)$ | M1 | Attempt to use both limits on an integrated expression |
| | A1 | Correct substitution in their expression including subtraction (may be left as an expression). |
| $= 81.25$ m | A1 | cao. |
| | | **Total: 7** |
---6 In this question take $g$ as $10 \mathrm {~m \mathrm {~s} ^ { - 2 }$.}
A small ball is released from rest. It falls for 2 seconds and is then brought to rest over the next 5 seconds. This motion is modelled in the speed-time graph Fig. 6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c84a748a-a6f4-48c5-b864-fe543569bdf5-5_659_1105_578_493}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
For this model,\\
(i) calculate the distance fallen from $t = 0$ to $t = 7$,\\
(ii) find the acceleration of the ball from $t = 2$ to $t = 6$, specifying the direction,\\
(iii) obtain an expression in terms of $t$ for the downward speed of the ball from $t = 2$ to $t = 6$,\\
(iv) state the assumption that has been made about the resistance to motion from $t = 0$ to $t = 2$.
The part of the motion from $t = 2$ to $t = 7$ is now modelled by $v = - \frac { 3 } { 2 } t ^ { 2 } + \frac { 19 } { 2 } t + 7$.\\
(v) Verify that $v$ agrees with the values given in Fig. 6 at $t = 2 , t = 6$ and $t = 7$.\\
(vi) Calculate the distance fallen from $t = 2$ to $t = 7$ according to this model.
\hfill \mbox{\textit{OCR MEI M1 2005 Q6 [19]}}