| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring standard application of kinematic equations for vertical motion under gravity. Parts (i) and (ii) involve direct substitution into familiar formulas (v² = u² + 2as and s = ut + ½at²), while part (iii) tests basic understanding of motion graphs. No problem-solving insight or multi-step reasoning required—purely routine mechanics practice. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = (+ /-5)^2 + 2 \times 9.8 \times 2.5\) \(\text{Speed (or } v) = 8.6(0) \text{ ms}^{-1}\) OR \(0 = s^2 - 2 \times 9.8 \times s\) with \(v^2 = (0) + 2 \times 9.8(s+2.5)\) \(v^2 = 2 \times 9.8 \times (2.5 + 1.28)\) \(\text{Speed} = 8.6(0) \text{ ms}^{-1}\) | M1 A1 A1 [3] | Uses \(v^2 = u^2 \pm 2gs\), u non-zero Accept \(\sqrt{74}\) Do not accept -8.6(0) \(s = 1.2755...\) \(19.8 \times 3.7755...\) Or rounds to 8.6 Uses v(from (i)) = +/- 5 +/- 9.8t CV(8.60 from (i)) |
| \(8.6 = -5 + 9.8t\) \(\text{Time} = 1.39 \text{ s}\) OR \(9.8t^2 - 10t - 5 = 0\) \(\text{Time} = 1.39 \text{ s}\) OR \(2.5 = (8.6-5)t/2\) \(\text{Time} = 1.39 \text{ s}\) OR \(t = 5/9.8 + 8.6/9.8\) \(\text{Time} = 1.39\) | M1 A1 A1 [3] | \(+/-2.5 = 5t +/-\) gt\(^2/2\) Times to top and ground found and added CV(8.60 from (i)) |
| \(+/-2.5 = +/- (5 - \text{Speed from (i)) x t / 2\) CV(8.60 from (i)) | ||
| a) \(v, \text{ms}^{-1}\) | B1 B1 [4] | Straight descending line to t axis Continues straight below t axis |
| b) \(x, m\) | B1 B1 | Inverted "parabolic" curve, starts anywhere on t=0 Ends below t = 0 level, need not be below t axis |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $v^2 = (+ /-5)^2 + 2 \times 9.8 \times 2.5$ $\text{Speed (or } v) = 8.6(0) \text{ ms}^{-1}$ OR $0 = s^2 - 2 \times 9.8 \times s$ with $v^2 = (0) + 2 \times 9.8(s+2.5)$ $v^2 = 2 \times 9.8 \times (2.5 + 1.28)$ $\text{Speed} = 8.6(0) \text{ ms}^{-1}$ | M1 A1 A1 [3] | Uses $v^2 = u^2 \pm 2gs$, u non-zero Accept $\sqrt{74}$ Do not accept -8.6(0) $s = 1.2755...$ $19.8 \times 3.7755...$ Or rounds to 8.6 Uses v(from (i)) = +/- 5 +/- 9.8t CV(8.60 from (i)) | **It is common to see the upwards and downwards motion treated separately. Both parts must be attempted for M1, and both parts must be attempted accurately with cvs for the A1** |
| $8.6 = -5 + 9.8t$ $\text{Time} = 1.39 \text{ s}$ OR $9.8t^2 - 10t - 5 = 0$ $\text{Time} = 1.39 \text{ s}$ OR $2.5 = (8.6-5)t/2$ $\text{Time} = 1.39 \text{ s}$ OR $t = 5/9.8 + 8.6/9.8$ $\text{Time} = 1.39$ | M1 A1 A1 [3] | $+/-2.5 = 5t +/-$ gt$^2/2$ Times to top and ground found and added CV(8.60 from (i)) | **It is common to see the upwards and downwards motion treated separately. Both parts must be attempted for M1, and both parts must be attempted accurately with cvs for the A1** |
| | | $+/-2.5 = +/- (5 - \text{Speed from (i)) x t / 2$ CV(8.60 from (i)) |
| **a)** $v, \text{ms}^{-1}$ | B1 B1 [4] | Straight descending line to t axis Continues straight below t axis | **Ignore values written on diagrams** |
| **b)** $x, m$ | B1 B1 | Inverted "parabolic" curve, starts anywhere on t=0 Ends below t = 0 level, need not be below t axis |3 A particle is projected vertically upwards with velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 2.5 m above the ground.\\
(i) Calculate the speed of the particle when it strikes the ground.\\
(ii) Calculate the time after projection when the particle reaches the ground.\\
(iii) Sketch on separate diagrams
\begin{enumerate}[label=(\alph*)]
\item the $( t , v )$ graph,
\item the $( t , x )$ graph,\\
representing the motion of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2011 Q3 [10]}}