| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Two connected particles, horizontal surface |
| Difficulty | Moderate -0.8 This is a straightforward connected particles problem requiring standard application of Newton's second law. Parts (a)-(c) involve routine force diagrams and solving F=ma with given values. Part (d) adds one extra step (setting tension equal to weight) but requires no novel insight—it's a typical textbook exercise testing basic mechanics concepts with clear numerical setup. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| 400N, 1200N, 2400N with skier and boat diagram showing tension T | B1, B1 | Driving force and common tension correct and labelled (boat shown on right). Both resistances correct and labelled |
| Answer | Marks | Guidance |
|---|---|---|
| N2L for skier \(T - 400 = 65a\) | B1 | Allow \(-T - 400 = 65a\) if consistent with their diagram |
| Answer | Marks | Guidance |
|---|---|---|
| N2L for the boat \(2400 - 1200 - T = 985a\); \(1200 - T = 985a\) | M1, A1 | Attempt to use N2L with \(m = 985\) for the boat. Allow missing or incorrect resistance. Allow \(-T\) for \(T\) if consistent with their diagram. Allow SC1 for \(2400 - 1200 - 400 = 1050a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Add equations to give acceleration \(\frac{800}{1050} = 0.762\ \text{m s}^{-2}\) | B1 | cao Allow if resistances interchanged |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_{\max} = 65g\); \(65g - 400 = 65a\) gives \(a = 3.65\ \text{m s}^{-2}\) | M1, A1 | Substitute \(65g\) for \(T\) in their equation from (b)(i). Accept awrt 3.6 or 3.7 |
## Question 10:
### Part (a):
400N, 1200N, 2400N with skier and boat diagram showing tension T | B1, B1 | Driving force and common tension correct and labelled (boat shown on right). Both resistances correct and labelled
**[2]**
### Part (b)(i):
N2L for skier $T - 400 = 65a$ | B1 | Allow $-T - 400 = 65a$ if consistent with their diagram
**[1]**
### Part (b)(ii):
N2L for the boat $2400 - 1200 - T = 985a$; $1200 - T = 985a$ | M1, A1 | Attempt to use N2L with $m = 985$ for the boat. Allow missing or incorrect resistance. Allow $-T$ for $T$ if consistent with their diagram. Allow SC1 for $2400 - 1200 - 400 = 1050a$
**[2]**
### Part (c):
Add equations to give acceleration $\frac{800}{1050} = 0.762\ \text{m s}^{-2}$ | B1 | cao Allow if resistances interchanged
**[1]**
### Part (d):
$T_{\max} = 65g$; $65g - 400 = 65a$ gives $a = 3.65\ \text{m s}^{-2}$ | M1, A1 | Substitute $65g$ for $T$ in their equation from (b)(i). Accept awrt 3.6 or 3.7
**[2]**
---10 A boat pulls a water skier of mass 65 kg with a light inextensible horizontal towrope. The mass of the boat is 985 kg . There is a driving force of 2400 N acting on the boat. There are horizontal resistances to motion of 400 N and 1200 N acting on the skier and the boat respectively.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram showing all the horizontal forces acting on the skier and the boat.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the equation of motion of the skier.
\item Find the equation of motion of the boat.
\end{enumerate}\item Find the acceleration of the skier and the boat.
The driving force of the boat is increased. The skier can only hold on to the towrope when the tension is no greater than her weight.
\item Determine her greatest acceleration, assuming that the resistances to motion stay the same.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q10 [8]}}