| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, horizontal |
| Difficulty | Moderate -0.3 This is a straightforward connected particles question with clearly structured parts that guide students through standard applications of F=ma. Parts (i)-(iv) involve routine calculations with given values, while part (v) requires recognizing the tow-bar changes from tension to thrust—a small conceptual step but still within standard M1 scope. Slightly easier than average due to the scaffolded structure and explicit 'show that' parts. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| B1 | \(2000 = 1000a\) so \(a = 2\) so \(2 \text{ m s}^{-2}\) |
| M1 | Use of appropriate UVAST for \(t\) |
| A1 | \(12.5 = 5 + 2t\) so \(t = 3.75\) so \(3.75 \text{ s}\) (cao) |
| Answer | Marks |
|---|---|
| M1 | \(2000 - R = 1000 \times 1.4\) |
| E1 | N2L. Accept \(F = mga\). Accept sign errors. Both forces present. Must use \(a = 1.4\) |
| A1 | \(R = 600\) so \(600 \text{ N}\) (AG) |
| Answer | Marks |
|---|---|
| M1 | N2L overall or 2 paired equations. \(F = ma\) and use \(0.7\). Mass must be correct. Allow sign errors and \(600\) omitted. |
| A1 | \(2000 - 600 - S = 1800 \times 0.7\) |
| E1 | All correct. Clearly shown |
| A1 | \(S = 140\) so \(140 \text{ N}\) (AG) |
| Answer | Marks |
|---|---|
| M1 | N2L on trailer (or car). \(F = 800a\) (or \(1000a\)). Condone missing resistance otherwise all forces present. Condone sign errors. |
| B1 | Use of \(140\) (or \(2000 - 600\)) and \(0.7\) |
| A1 | \(T - 140 = 800 \times 0.7\) so \(T = 700\) so \(700 \text{ N}\) |
| Answer | Marks |
|---|---|
| M1 | Use of \(F = 1800a\) to find new accn. Condone \(2000\) included but not \(T\). Allow missing forces. |
| A1 | \(-600 - 140 - 610 = 1800a\) (All forces present; no extra ones. Allow sign errors. Accept \(\pm\)) |
| A1 | \(a = -0.75\) (cao) |
| M1 | N2L with their \(a\) (\(\neq 0.7\)) on trailer or car. Must have correct mass and forces. Accept sign errors |
| A1 | \(T - 140 = -0.75 \times 800\) so \(T = -460\) so \(460\) (cao. Accept \(\pm 460\)) |
| F1 | Dep on M1. Take tension as \(+\)ve unless clear other convention. Thrust |
# Question 1
## 1(i)
B1 | $2000 = 1000a$ so $a = 2$ so $2 \text{ m s}^{-2}$
M1 | Use of appropriate UVAST for $t$
A1 | $12.5 = 5 + 2t$ so $t = 3.75$ so $3.75 \text{ s}$ (cao)
**Total: 3**
## 1(ii)
M1 | $2000 - R = 1000 \times 1.4$
E1 | N2L. Accept $F = mga$. Accept sign errors. Both forces present. Must use $a = 1.4$
A1 | $R = 600$ so $600 \text{ N}$ (AG)
**Total: 2**
## 1(iii)
M1 | N2L overall or 2 paired equations. $F = ma$ and use $0.7$. Mass must be correct. Allow sign errors and $600$ omitted.
A1 | $2000 - 600 - S = 1800 \times 0.7$
E1 | All correct. Clearly shown
A1 | $S = 140$ so $140 \text{ N}$ (AG)
**Total: 3**
## 1(iv)
M1 | N2L on trailer (or car). $F = 800a$ (or $1000a$). Condone missing resistance otherwise all forces present. Condone sign errors.
B1 | Use of $140$ (or $2000 - 600$) and $0.7$
A1 | $T - 140 = 800 \times 0.7$ so $T = 700$ so $700 \text{ N}$
**Total: 3**
## 1(v)
M1 | Use of $F = 1800a$ to find new accn. Condone $2000$ included but not $T$. Allow missing forces.
A1 | $-600 - 140 - 610 = 1800a$ (All forces present; no extra ones. Allow sign errors. Accept $\pm$)
A1 | $a = -0.75$ (cao)
M1 | N2L with their $a$ ($\neq 0.7$) on trailer or car. Must have correct mass and forces. Accept sign errors
A1 | $T - 140 = -0.75 \times 800$ so $T = -460$ so $460$ (cao. Accept $\pm 460$)
F1 | Dep on M1. Take tension as $+$ve unless clear other convention. Thrust
**Total: 6**
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**GRAND TOTAL: 17**1 A car of mass 1000 kg is travelling along a straight, level road.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-1_150_868_316_602}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
(i) Calculate the acceleration of the car when a resultant force of 2000 N acts on it in the direction of its motion.
How long does it take the car to increase its speed from $5 \mathrm {~ms} ^ { - 1 }$ to $12.5 \mathrm {~ms} ^ { - 1 }$ ?
The car has an acceleration of $1.4 \mathrm {~ms} ^ { - 2 }$ when there is a driving force of 2000 N .\\
(ii) Show that the resistance to motion of the car is 600 N .
A trailer is now atached to the car, as shown in Fig. 6.2. The car still has a driving force of 2000 N and resistance to motion of 600 N . The trailer has a mass of 800 kg . The tow-bar connecting the car and the trailer is light and horizontal. The car and trailer are accelerating at $0.7 \mathrm {~ms} ^ { 2 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-1_165_883_1279_554}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}
(iii) Show that the resistance to the motion of the trailer is 140 N .\\
(iv) Calculate the force in the tow bar.
The driving force is now removed and a braking force of 610 N is applied to the car. All the resistances to motion remain as before. The trailer has no brakes.\\
(v) Calculate the new acceleration. Calculate also the force in the tow-bar, stating whether it is a tension or a thrust (compression).
\hfill \mbox{\textit{OCR MEI M1 Q1 [17]}}