| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a standard kinematics question testing routine application of SUVAT equations for vertical motion under gravity. All four parts follow textbook procedures: (i) uses v²=u²+2as with v=0, (ii) uses the same equation to find v at given height, (iii) uses s=ut+½at² for downward motion, and (iv) combines symmetry with previous results. The question requires no problem-solving insight—just systematic application of memorized formulas with straightforward arithmetic, making it easier than the average A-level question which typically demands more integration of concepts or multi-step reasoning. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0 = 28^2 - 2 \times 9.8 \times h\) | M1 | For use of const acc formula(s) to find \(h\) |
| Hence maximum height is 40 m | A1 | For correct value 40 |
| Subtotal: 2 marks | ||
| (ii) \(v^2 = 28^2 - 2 \times 9.8 \times 30\) | M1 | For use of const acc formula(s) to find \(v\) |
| Hence speed is 14 m s\(^{-1}\) | A1, A1 | For correct equation in \(v\); For correct value 14 |
| Subtotal: 3 marks | ||
| (iii) \(10 = \frac{1}{4} \times 9.8 t^2\) | M1 | For use of const acc formula(s) to find \(t\) |
| Hence time is \(\frac{10}{7} = 1.43\) s | A1, A1 | For correct equation in \(t\); For correct value \(\frac{10}{7}\) or equivalent |
| Subtotal: 3 marks | ||
| (iv) Length of time is \(2 \times \frac{10}{7} = \frac{20}{7}\) s | M1, A1 | For doubling, or equiv longer method; For correct value, i.e. double their (iii) |
| Subtotal: 2 marks | ||
| Total: 10 marks |
**(i)** $0 = 28^2 - 2 \times 9.8 \times h$ | M1 | For use of const acc formula(s) to find $h$
Hence maximum height is 40 m | A1 | For correct value 40
| | **Subtotal: 2 marks**
**(ii)** $v^2 = 28^2 - 2 \times 9.8 \times 30$ | M1 | For use of const acc formula(s) to find $v$
Hence speed is 14 m s$^{-1}$ | A1, A1 | For correct equation in $v$; For correct value 14
| | **Subtotal: 3 marks**
**(iii)** $10 = \frac{1}{4} \times 9.8 t^2$ | M1 | For use of const acc formula(s) to find $t$
Hence time is $\frac{10}{7} = 1.43$ s | A1, A1 | For correct equation in $t$; For correct value $\frac{10}{7}$ or equivalent
| | **Subtotal: 3 marks**
**(iv)** Length of time is $2 \times \frac{10}{7} = \frac{20}{7}$ s | M1, A1 | For doubling, or equiv longer method; For correct value, i.e. double their (iii)
| | **Subtotal: 2 marks**
| | **Total: 10 marks**A particle is projected vertically upwards, from the ground, with a speed of $28 \text{ m s}^{-1}$. Ignoring air resistance, find
\begin{enumerate}[label=(\roman*)]
\item the maximum height reached by the particle, [2]
\item the speed of the particle when it is 30 m above the ground, [3]
\item the time taken for the particle to fall from its highest point to a height of 30 m, [3]
\item the length of time for which the particle is more than 30 m above the ground. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 Q3 [10]}}