| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Lift with passenger or load |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's second law to vertical motion problems. Parts (a)-(c) are standard equilibrium and F=ma calculations with no air resistance. Part (d) adds air resistance but remains a direct application of F=ma=T-mg-R with given values. The conceptual step about direction of motion is minor. Slightly easier than average due to being entirely procedural with no problem-solving insight required. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No acceleration so we require the weight \(15g\) (147 N) | B1 | 3.3 — Accept \(15g\), \(15g\text{ N}\), \(147\text{ N}\) etc |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tension \(T\) N, N2L \(\uparrow\) \(T-147=15\times 2\) | M1 | 3.4 — Application of N2L |
| So \(T=177\) and tension is \(177\text{ N}\) | A1 | 1.1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(177\text{ N}\) | B1 | 3.4 — FT from (ii) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let the air resistance be \(R\uparrow\) | M1 | 3.3 |
| N2L \(\uparrow\) gives \(R+140-15g=15\times(-1.5)\) | M1 | 1.1 — Finding \(R\) using \(a\) and then \(T\) |
| \(R-7=-22.5\) | ||
| \(R=-15.5\) | A1 | 1.1 |
| Hence magnitude is \(15.5\text{ N}\) | A1 | 3.4 |
| \(R\) is downwards so motion of the box is upwards | E1 | 3.4 |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let total upward force be \(F\) N and air resistance \(R\uparrow\); N2L \(\uparrow\) gives \(F-147=15\times(-1.5)\) | M1 | 3.3 — Finding \(F\) and hence \(R\) using \(a\) and then \(T\) |
| So \(F=124.5\) | A1 | 1.1 |
| Also \(F=R+140\) so \(R=124.5-140=-15.5\) | A1 | 1.1 |
| Hence magnitude is \(15.5\text{ N}\) | A1 | 3.4 |
| \(R\) is downwards so motion of the box is upwards | A1 | 3.4 |
| [5] |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No acceleration so we require the weight $15g$ (147 N) | B1 | 3.3 — Accept $15g$, $15g\text{ N}$, $147\text{ N}$ etc |
| **[1]** | | |
---
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tension $T$ N, N2L $\uparrow$ $T-147=15\times 2$ | M1 | 3.4 — Application of N2L |
| So $T=177$ and tension is $177\text{ N}$ | A1 | 1.1 |
| **[2]** | | |
---
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $177\text{ N}$ | B1 | 3.4 — FT from (ii) |
| **[1]** | | |
---
## Question 12(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let the air resistance be $R\uparrow$ | M1 | 3.3 |
| N2L $\uparrow$ gives $R+140-15g=15\times(-1.5)$ | M1 | 1.1 — Finding $R$ using $a$ and then $T$ |
| $R-7=-22.5$ | | |
| $R=-15.5$ | A1 | 1.1 |
| Hence magnitude is $15.5\text{ N}$ | A1 | 3.4 |
| $R$ is downwards so motion of the box is upwards | E1 | 3.4 |
| **[5]** | | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let total upward force be $F$ N and air resistance $R\uparrow$; N2L $\uparrow$ gives $F-147=15\times(-1.5)$ | M1 | 3.3 — Finding $F$ and hence $R$ using $a$ and then $T$ |
| So $F=124.5$ | A1 | 1.1 |
| Also $F=R+140$ so $R=124.5-140=-15.5$ | A1 | 1.1 |
| Hence magnitude is $15.5\text{ N}$ | A1 | 3.4 |
| $R$ is downwards so motion of the box is upwards | A1 | 3.4 |
| **[5]** | | |12 A box hangs from a balloon by means of a light inelastic string. The string is always vertical. The mass of the box is 15 kg .
Catherine initially models the situation by assuming that there is no air resistance to the motion of the box. Use Catherine's model to calculate the tension in the string if:
\begin{enumerate}[label=(\alph*)]
\item the box is held at rest by the tension in the string,
\item the box is instantaneously at rest and accelerating upwards at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$,
\item the box is moving downwards at $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and accelerating upwards at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Catherine now carries out an experiment to find the magnitude of the air resistance on the box when it is moving.\\
At a time when the box is accelerating downwards at $1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, she finds that the tension in the string is 140 N .
\item Calculate the magnitude of the air resistance at that time.
Give, with a reason, the direction of motion of the box.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 Q12 [9]}}