| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Free fall: time or distance |
| Difficulty | Moderate -0.3 This is a multi-part question requiring straightforward application of SUVAT equations and interpretation of speed-time graphs. While it has several parts (5 sub-questions), each involves routine calculations: substitution into d=5T², reading area under a graph, and integrating a simple quadratic. The conceptual demand is low—students just apply standard mechanics techniques without novel problem-solving. The commentary aspects require minimal insight. Slightly easier than average due to its structured, guided nature. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(d = 5 \times 8^2 = 320\), so 320 m | B1 | |
| This value is too great. It is not between 150 and 200 m. | B1 | Accept "inconsistent". Dependent on previous mark. |
| \(s = ut + \frac{1}{2}at^2\) with \(s = d\), \((u = 0)\), \(a = 10\) and \(t = 7\) | M1 | |
| Giving \(d = \frac{1}{2} \times 10 \times 7^2 = 5T^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Depth = Area under the graph | M1 | oe |
| \(= \frac{1}{2} \times 5 \times 50 + 3 \times 50\) | A1 | |
| \(= 275\) m | A1 | |
| Outside the 150 to 200 m interval so inconsistent | B1 | A numerical comparison is required for this mark but may refer to values for it stated in part (i). Dependent on previous mark. Special Case Allow up to M1 A0 A1 B1 for a response in which the time at which \(v\) becomes constant is near but not equal to 5 (eg 4 or 4.5). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The same: initial constant acceleration (of 10 m s\(^{-2}\)) | B1 | Do not allow statements about the initial speed or the time taken |
| Different: two part motion with constant speed at end | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(0 \leq t \leq 5\), the distance travelled is \(\int_0^{t} (10t - t^2) dt\) | M1 | Or equivalent using indefinite integration |
| \(= \left[5t^2 - \frac{t^3}{3}\right]_0^t\) | A1 | Limits not required for this mark |
| \(5 \times 5^2 - \frac{5^3}{3} = (83 \frac{1}{3})\) | A1 | \(A \Rightarrow M\) |
| For \(5 < t \leq 8\), the distance travelled is \(25 \times 3\) (= 75) | B1 | Seen or implied |
| \(d = 83\frac{1}{3} + 75 = 158\frac{1}{3}\) | A1 | CAO |
| This is within the given interval. | B1 | Dependent on previous mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Similar: constant speed for \(5 < t \leq 8\) | B1 | |
| Different: acceleration is not constant for \(0 \leq t \leq 5\). | B1 |
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $d = 5 \times 8^2 = 320$, so 320 m | B1 | |
| This value is too great. It is not between 150 and 200 m. | B1 | Accept "inconsistent". Dependent on previous mark. |
| $s = ut + \frac{1}{2}at^2$ with $s = d$, $(u = 0)$, $a = 10$ and $t = 7$ | M1 | |
| Giving $d = \frac{1}{2} \times 10 \times 7^2 = 5T^2$ | A1 | |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Depth = Area under the graph | M1 | oe |
| $= \frac{1}{2} \times 5 \times 50 + 3 \times 50$ | A1 | |
| $= 275$ m | A1 | |
| Outside the 150 to 200 m interval so inconsistent | B1 | A numerical comparison is required for this mark but may refer to values for it stated in part (i). Dependent on previous mark. Special Case Allow up to M1 A0 A1 B1 for a response in which the time at which $v$ becomes constant is near but not equal to 5 (eg 4 or 4.5). |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| The same: initial constant acceleration (of 10 m s$^{-2}$) | B1 | Do not allow statements about the initial speed or the time taken |
| Different: two part motion with constant speed at end | B1 | |
**(iv)**
| Answer | Mark | Guidance |
|--------|------|----------|
| For $0 \leq t \leq 5$, the distance travelled is $\int_0^{t} (10t - t^2) dt$ | M1 | Or equivalent using indefinite integration |
| $= \left[5t^2 - \frac{t^3}{3}\right]_0^t$ | A1 | Limits not required for this mark |
| $5 \times 5^2 - \frac{5^3}{3} = (83 \frac{1}{3})$ | A1 | $A \Rightarrow M$ |
| For $5 < t \leq 8$, the distance travelled is $25 \times 3$ (= 75) | B1 | Seen or implied |
| $d = 83\frac{1}{3} + 75 = 158\frac{1}{3}$ | A1 | CAO |
| This is within the given interval. | B1 | Dependent on previous mark |
**(v)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Similar: constant speed for $5 < t \leq 8$ | B1 | |
| Different: acceleration is not constant for $0 \leq t \leq 5$. | B1 | |
---6 In this question you should take $\boldsymbol { g $ to be $\mathbf { 1 0 } \mathrm { ms } ^ { \boldsymbol { - } \mathbf { 2 } }$.}
Piran finds a disused mineshaft on his land and wants to know its depth, $d$ metres.\\
Local records state that the mineshaft is between 150 and 200 metres deep.\\
He drops a small stone down the mineshaft and records the time, $T$ seconds, until he hears it hit the bottom. It takes 8.0 seconds.
Piran tries three models, $\mathrm { A } , \mathrm { B }$ and C .\\
In model A, Piran uses the formula $d = 5 T ^ { 2 }$ to estimate the depth.\\
(i) Find the depth that model A gives and comment on whether it is consistent with the local records.
Explain how the formula in model A is obtained.
In model B, Piran uses the speed-time graph in Fig. 6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-5_762_1176_1087_424}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(ii) Calculate the depth of the mineshaft according to model B.
Comment on whether this depth is consistent with the local records.\\
(iii) Describe briefly one respect in which model B is the same as model A and one respect in which it is different.
Piran then tries model C in which the speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by
$$\begin{aligned}
& v = 10 t - t ^ { 2 } \text { for } 0 \leqslant t \leqslant 5 \\
& v = 25 \text { for } 5 < t \leqslant 8
\end{aligned}$$
(iv) Calculate the depth of the mineshaft according to model C.
Comment on whether this depth is consistent with the local records.\\
(v) Describe briefly one respect in which model C is similar to model B and one respect in which it is different.
\hfill \mbox{\textit{OCR MEI M1 2016 Q6 [18]}}