| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Train with coupled trucks/carriages |
| Difficulty | Moderate -0.3 This is a straightforward connected particles problem requiring standard application of Newton's second law and SUVAT equations. Part (a) uses basic kinematics with given values, parts (b) and (c) involve setting up F=ma equations for the system and individual components. All steps are routine with no conceptual challenges beyond standard A-level mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(12950 = 15t + 0.5(0.01)t^2\) leading to \(t = \ldots\) | M1, A1 | Use of \(s = ut + \frac{1}{2}at^2\). Correct equation and attempt to solve. Consistent values used |
| \(11.7\) (minutes) | A1 | BC cao (exact value is \(11\frac{2}{3}\)) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(D - 1250 - 9000 = 70000(0.01)\) | M1 | N2L applied to the whole system, correct number of terms (allow sign errors) |
| \(10950\ \text{(N)}\) | A1 | cao |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \('10950' - 9000 - T_C = 50000(0.01)\) or \(T_C - 1250 = 20000(0.01)\) | M1 | N2L applied correctly to either \(A\) or \(B\), correct number of terms (allow sign errors) |
| \(1450\ \text{(N)}\) | A1 | cao |
| [2] |
## Question 11:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $12950 = 15t + 0.5(0.01)t^2$ leading to $t = \ldots$ | M1, A1 | Use of $s = ut + \frac{1}{2}at^2$. Correct equation and attempt to solve. Consistent values used |
| $11.7$ (minutes) | A1 | BC cao (exact value is $11\frac{2}{3}$) |
| **[3]** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $D - 1250 - 9000 = 70000(0.01)$ | M1 | N2L applied to the whole system, correct number of terms (allow sign errors) |
| $10950\ \text{(N)}$ | A1 | cao |
| **[2]** | | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $'10950' - 9000 - T_C = 50000(0.01)$ or $T_C - 1250 = 20000(0.01)$ | M1 | N2L applied correctly to either $A$ or $B$, correct number of terms (allow sign errors) |
| $1450\ \text{(N)}$ | A1 | cao |
| **[2]** | | |
---11\\
\includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-08_451_1340_251_244}
A train consists of an engine $A$ of mass 50000 kg and a carriage $B$ of mass 20000 kg . The engine and carriage are connected by a rigid coupling. The coupling is modelled as light and horizontal.
The resistances to motion acting on $A$ and $B$ are 9000 N and 1250 N respectively (see diagram).\\
The train passes through station $P$ with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and moves along a straight horizontal track with constant acceleration $0.01 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ towards station $Q$. The distance between $P$ and $Q$ is 12.95 km .
\begin{enumerate}[label=(\alph*)]
\item Determine the time, in minutes, to travel between $P$ and $Q$.
For the train's motion between $P$ and $Q$, determine the following.
\item The driving force of the engine.
\item The tension in the coupling between $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q11 [7]}}