| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Train with coupled trucks/carriages |
| Difficulty | Moderate -0.3 This is a standard M1 connected particles question with straightforward application of Newton's second law to systems and individual components. Parts (i) and (ii) are routine calculations with given values, while parts (iii)-(v) require systematic application of F=ma to different parts of the system. The question is slightly easier than average because it's highly structured with clear guidance ('show that' questions provide target values) and uses standard M1 techniques without requiring novel problem-solving insight. |
| Spec | 3.03d Newton's second law: 2D vectors3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10500 + 3000 + 1500\) | M1 | For summing 3 resistances |
| Driving force below 15000 gives retardation | A1 | 2 Accept generalised case or specific instance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(35000 - 15000 = 80000a\) | M1 | Newton's second law for whole train |
| Acceleration is \(0.25\ \text{ms}^{-2}\) | A1 | 2 AG Accept verification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's second law to \(E\) only, at least 2 forces out of the relevant 3 | |
| \(35000 - 10500 - 8500 = 0.25m\) | A1 | |
| Mass is 64000 kg | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's second law with all appropriate forces | |
| \(-15000 - 15000 = 80000a\) OR \(-3000-10500-15000=(80000-m)a\) | A1 | \(a = -0.375\) |
| M1 | For applying Newton's second law to \(B\) only, only 1 force, or cv(\(a\)) | |
| \(-1500 = ma\) | A1 | |
| Mass is 4000 kg | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-15000 - 10500 \pm T = 64000(-0.375)\) | B1ft | Follow through cv(\(m_E\), \(a\)), or accept use of \(m_E\), \(a\) |
| \(T = \pm 1500 \rightarrow\) forward force on \(E\) of 1500 N | B1 | 2 |
| OR working with A and B: \(-1500 - 3000 \pm T = (80000 - 64000)(-0.375)\) | B1ft | Follow through cv(\(m_E\), \(a\)) |
| \(T = \pm 1500 \rightarrow\) forward force on \(E\) of 1500 | B1 |
# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10500 + 3000 + 1500$ | M1 | For summing 3 resistances |
| Driving force below 15000 gives retardation | A1 | **2** Accept generalised case or specific instance |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $35000 - 15000 = 80000a$ | M1 | Newton's second law for whole train |
| Acceleration is $0.25\ \text{ms}^{-2}$ | A1 | **2** AG Accept verification |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's second law to $E$ only, at least 2 forces out of the relevant 3 |
| $35000 - 10500 - 8500 = 0.25m$ | A1 | |
| Mass is 64000 kg | A1 | **3** |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's second law with all appropriate forces |
| $-15000 - 15000 = 80000a$ OR $-3000-10500-15000=(80000-m)a$ | A1 | $a = -0.375$ |
| | M1 | For applying Newton's second law to $B$ only, only 1 force, or cv($a$) |
| $-1500 = ma$ | A1 | |
| Mass is 4000 kg | A1 | **5** |
## Part (v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-15000 - 10500 \pm T = 64000(-0.375)$ | B1ft | Follow through cv($m_E$, $a$), or accept use of $m_E$, $a$ |
| $T = \pm 1500 \rightarrow$ forward force on $E$ of 1500 N | B1 | **2** |
| OR working with A and B: $-1500 - 3000 \pm T = (80000 - 64000)(-0.375)$ | B1ft | Follow through cv($m_E$, $a$) |
| $T = \pm 1500 \rightarrow$ forward force on $E$ of 1500 | B1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{8ee41313-b516-48cb-bc87-cd8e54245d28-4_314_997_267_577}
A train of total mass 80000 kg consists of an engine $E$ and two trucks $A$ and $B$. The engine $E$ and truck $A$ are connected by a rigid coupling $X$, and trucks $A$ and $B$ are connected by another rigid coupling $Y$. The couplings are light and horizontal. The train is moving along a straight horizontal track. The resistances to motion acting on $E , A$ and $B$ are $10500 \mathrm {~N} , 3000 \mathrm {~N}$ and 1500 N respectively (see diagram).\\
(i) By modelling the whole train as a single particle, show that it is decelerating when the driving force of the engine is less than 15000 N .\\
(ii) Show that, when the magnitude of the driving force is 35000 N , the acceleration of the train is $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iii) Hence find the mass of $E$, given that the tension in the coupling $X$ is 8500 N when the magnitude of the driving force is 35000 N .
The driving force is replaced by a braking force of magnitude 15000 N acting on the engine. The force exerted by the coupling $Y$ is zero.\\
(iv) Find the mass of $B$.\\
(v) Show that the coupling $X$ exerts a forward force of magnitude 1500 N on the engine.
\hfill \mbox{\textit{OCR M1 2006 Q6 [14]}}