| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| 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 straightforward two-part mechanics question requiring standard SUVAT equations for vertical free fall and then resolving forces on an inclined plane with friction. Both parts use routine methods with clear setups and no conceptual surprises, making it slightly easier than average for M1. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02h Motion under gravity: vector form3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2.5 = 9.8t^2/2\) | M1 | Uses \(s = 0 \pm gt^2/2\) |
| \(t = 0.714\) s or better or \(5/7\) | A1 | Not awarded if \(-\) sign "lost" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = 2 \times 9.8 \times 2.5\) OR \(v = 9.8 \times 0.714\) | M1 | Uses \(v^2 = 0 \pm 2gs\) or \(v = u \pm gt\) |
| \(v = 7 \text{ ms}^{-1}\) or 6.99 or art 7.00 | A1 | Not awarded if \(-\) sign "lost" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 2 \times 9.8\sin60 \, (= 16.97 = 17)\) | B1 | With incorrect angle, e.g. \(R = 2 \times 9.8\cos60 \, (=9.8)\) B0 |
| \(F = 0.2 \times 16.97 \, (=3.395 \text{ or } 3.4)\) | M1 | |
| \(F = 0.2 \times 16.97\) | A1ft | \(F = 0.2 \times 9.8 \, (=1.96)\) M1A1\(\sqrt{}\) |
| Cmpt weight \(= 2 \times 9.8\cos60 \, (=9.8)\) | B1 | Cmpt wt \(= 2 \times 9.8\sin60 \, (=16.97)\) B0 |
| \(2a = 9.8 - 3.395\) | M1 | \(2a = 16.97 - 1.96\) M1 |
| \(a = 3.2 \text{ ms}^{-2}\) | A1ft | \(a = 7.5\) A1\(\sqrt{}\) ft \(cv(R\) and Cmpt weight) |
| Distance down ramp \(= 5\) m | B1 | |
| \(v^2 = 2 \times 3.2 \times 5\) | M1 | \(v^2 = 2 \times 7.5 \times 5\) |
| \(v = 5.66\) or \(5.7\) | A1ft | \(v = 8.66\) or \(8.7\) A1\(\sqrt{}\) ft \(cv(\sqrt{(10a)})\) |
# Question 6:
## Part (i)a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2.5 = 9.8t^2/2$ | M1 | Uses $s = 0 \pm gt^2/2$ |
| $t = 0.714$ s or better or $5/7$ | A1 | Not awarded if $-$ sign "lost" |
## Part (i)b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = 2 \times 9.8 \times 2.5$ OR $v = 9.8 \times 0.714$ | M1 | Uses $v^2 = 0 \pm 2gs$ or $v = u \pm gt$ |
| $v = 7 \text{ ms}^{-1}$ or 6.99 or art 7.00 | A1 | Not awarded if $-$ sign "lost" |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 2 \times 9.8\sin60 \, (= 16.97 = 17)$ | B1 | With incorrect angle, e.g. $R = 2 \times 9.8\cos60 \, (=9.8)$ B0 |
| $F = 0.2 \times 16.97 \, (=3.395 \text{ or } 3.4)$ | M1 | |
| $F = 0.2 \times 16.97$ | A1ft | $F = 0.2 \times 9.8 \, (=1.96)$ M1A1$\sqrt{}$ |
| Cmpt weight $= 2 \times 9.8\cos60 \, (=9.8)$ | B1 | Cmpt wt $= 2 \times 9.8\sin60 \, (=16.97)$ B0 |
| $2a = 9.8 - 3.395$ | M1 | $2a = 16.97 - 1.96$ M1 |
| $a = 3.2 \text{ ms}^{-2}$ | A1ft | $a = 7.5$ A1$\sqrt{}$ ft $cv(R$ and Cmpt weight) |
| Distance down ramp $= 5$ m | B1 | |
| $v^2 = 2 \times 3.2 \times 5$ | M1 | $v^2 = 2 \times 7.5 \times 5$ |
| $v = 5.66$ or $5.7$ | A1ft | $v = 8.66$ or $8.7$ A1$\sqrt{}$ ft $cv(\sqrt{(10a)})$ |
---6 Small parcels are being loaded onto a trolley. Initially the parcels are 2.5 m above the trolley.\\
(i) A parcel is released from rest and falls vertically onto the trolley. Calculate
\begin{enumerate}[label=(\alph*)]
\item the time taken for a parcel to fall onto the trolley,
\item the speed of a parcel when it strikes the trolley.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{470e70de-66ba-4dcc-a205-0c92f29471b1-4_327_723_603_751}
Parcels are often damaged when loaded in the way described, so a ramp is constructed down which parcels can slide onto the trolley. The ramp makes an angle of $60 ^ { \circ }$ to the vertical, and the coefficient of friction between the ramp and a parcel is 0.2 . A parcel of mass 2 kg is released from rest at the top of the ramp (see diagram). Calculate the speed of the parcel after sliding down the ramp.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2009 Q6 [13]}}