OCR MEI M1 — Question 1 18 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeTrain with coupled trucks/carriages
DifficultyStandard +0.3 This is a standard connected particles question with multiple parts requiring Newton's second law applications. While it involves several steps (horizontal motion, inclined plane, and equilibrium on a slope), each part uses routine mechanics techniques without requiring novel insight. The calculations are straightforward once the correct free-body diagrams are drawn, making it slightly easier than average for M1.
Spec3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes
1 A train consists of a locomotive pulling 17 identical trucks. The mass of the locomotive is 120 tonnes and the mass of each truck is 40 tonnes. The locomotive gives a driving force of 121000 N . The resistance to motion on each truck is \(R \mathrm {~N}\) and the resistance on the locomotive is \(5 R \mathrm {~N}\).
Initially the train is travelling on a straight horizontal track and its acceleration is \(0.11 \mathrm {~ms} ^ { - 2 }\).
  1. Show that \(R = 1500\).
  2. Find the tensions in the couplings between
    (A) the last two trucks,
    (B) the locomotive and the first truck. The train now comes to a place where the track goes up a straight, uniform slope at an angle \(\alpha\) with the horizontal, where \(\sin \alpha = \frac { 1 } { 80 }\). The driving force and the resistance forces remain the same as before.
  3. Find the magnitude and direction of the acceleration of the train. The train then comes to a straight uniform downward slope at an angle \(\beta\) to the horizontal.
    The driver of the train reduces the driving force to zero and the resistance forces remain the same as before. The train then travels at a constant speed down the slope.
  4. Find the value of \(\beta\).
(i)
B1 Total mass of train 800 000 kg (Allow 800 tonnes)
B1 Total resistance \(5R + 17R = 22R\)
M1 Newton's 2nd Law in the direction of motion: \(121000 - 22R = 800000 \times 0.11\)
(The right elements must be present, consistent with the candidate's answers above for total resistance and mass. No extra forces.)
E1 \(22R = 121000 - 88000\), \(R = 1500\)
(Perfect answer required)
(ii)(A)
Either (Last truck)
B1 Resultant force on last truck: \(40000 \times 0.11 = 4400\) N
(Award this mark for \(40000 \times 0.11\) or \(400 \times 0.11\) seen)
M1 Use of Newton's 2nd Law: \(T - 1500 = 40000 \times 0.11\)
(The right elements must be present and consistent with the answer above; no extra forces.)
A1 \(T = 5900\)
(Fully correct equation, or equivalent working)
A1 The tension is 5900 N
(Cao)
Or (Rest of the train)
B1 Resultant force on rest of train: \(760000 \times 0.11 = 83600\) N
(Award this mark for \(760000 \times 0.11\) or \(7600 \times 0.11\) seen)
M1 Use of Newton's 2nd Law: \(121000 - 31500 - T = 760000 \times 0.11\)
(The right elements must be present consistent with the answer above; no extra forces.)
A1 \(T = 5900\)
(Fully correct equation, or equivalent working)
A1 The tension is 5900 N
(Cao)
Special case: Award SC2 to a candidate who instead provides a perfect argument that the tension in the penultimate coupling is 11800 N.
(ii)(B)
Either (Rest of the train)
M1 Newton's 2nd Law is applied to the trucks: \(S - 25500 = 680000 \times 0.11\)
(The right elements must be present; no extra forces)
A1 \(S = 100300\)
(Cao)
A1 The tension is 100300 N
(Cao)
Or (Locomotive)
M1 Newton's 2nd Law is applied to the locomotive: \(121000 - S - 51500 = 120000 \times 0.11\)
(The right elements must be present; no extra forces)
A1 \(S = 100300\)
(Cao)
A1 The tension is 100300 N
(Cao)
Or (By argument)
M1 Each of the 17 trucks has the same mass, resistance and acceleration. So the tension in the first coupling is 17 times that in the last coupling
(Cao. For this statement on its own with no supporting argument allow SC2)
A1 \(T = 17 \times 5900 = 100300\)
(Cao)
A1 The tension is 100300 N
(Cao)
(iii)
B1 Resolved component of weight down slope: \(\frac{1}{80} \times 800000 \times 9.8 = 98000\) N
(\(\frac{1}{m} \times 9.8\) where \(m\) is the mass of the object the candidate is considering. Do not award if \(g\) is missing. Evaluation need not be seen)
M1 Let the acceleration be \(a\) m s\(^{-2}\) up the slope. Newton's 2nd Law to the whole train: \(121000 - 33000 - 98000 = 800000a\)
(The right elements must be present consistent with the candidate's component of the weight down the slope. No extra forces allowed)
A1 \(a = 0.0125\)
(Cao but allow an answer rounding to \(-0.012\) or \(-0.013\) following earlier premature rounding.)
A1 Magnitude 0.0125 m s\(^{-2}\), down the slope
(The negative sign must be interpreted so "Down the slope" or "decelerating" must be seen)
(iv)
M1 Taking the train as a whole, Force down the slope = Resistance force: \(800000 \times 9.8 \times \sin\theta = 33000\)
(Equilibrium of whole train required. The evidence for this mark may be obtained from a correct force diagram)
A1 \(\sin\theta = 0.042...\)
(Allow missing \(g\) for this mark only)
A1 \(\theta = 0.24\) (radians or approximately 14°)
(Cao)
## (i)
B1 Total mass of train 800 000 kg (Allow 800 tonnes)

B1 Total resistance $5R + 17R = 22R$

M1 Newton's 2nd Law in the direction of motion: $121000 - 22R = 800000 \times 0.11$
(The right elements must be present, consistent with the candidate's answers above for total resistance and mass. No extra forces.)

E1 $22R = 121000 - 88000$, $R = 1500$
(Perfect answer required)

## (ii)(A)

**Either (Last truck)**

B1 Resultant force on last truck: $40000 \times 0.11 = 4400$ N
(Award this mark for $40000 \times 0.11$ or $400 \times 0.11$ seen)

M1 Use of Newton's 2nd Law: $T - 1500 = 40000 \times 0.11$
(The right elements must be present and consistent with the answer above; no extra forces.)

A1 $T = 5900$
(Fully correct equation, or equivalent working)

A1 The tension is 5900 N
(Cao)

**Or (Rest of the train)**

B1 Resultant force on rest of train: $760000 \times 0.11 = 83600$ N
(Award this mark for $760000 \times 0.11$ or $7600 \times 0.11$ seen)

M1 Use of Newton's 2nd Law: $121000 - 31500 - T = 760000 \times 0.11$
(The right elements must be present consistent with the answer above; no extra forces.)

A1 $T = 5900$
(Fully correct equation, or equivalent working)

A1 The tension is 5900 N
(Cao)

**Special case:** Award SC2 to a candidate who instead provides a perfect argument that the tension in the penultimate coupling is 11800 N.

## (ii)(B)

**Either (Rest of the train)**

M1 Newton's 2nd Law is applied to the trucks: $S - 25500 = 680000 \times 0.11$
(The right elements must be present; no extra forces)

A1 $S = 100300$
(Cao)

A1 The tension is 100300 N
(Cao)

**Or (Locomotive)**

M1 Newton's 2nd Law is applied to the locomotive: $121000 - S - 51500 = 120000 \times 0.11$
(The right elements must be present; no extra forces)

A1 $S = 100300$
(Cao)

A1 The tension is 100300 N
(Cao)

**Or (By argument)**

M1 Each of the 17 trucks has the same mass, resistance and acceleration. So the tension in the first coupling is 17 times that in the last coupling
(Cao. For this statement on its own with no supporting argument allow SC2)

A1 $T = 17 \times 5900 = 100300$
(Cao)

A1 The tension is 100300 N
(Cao)

## (iii)

B1 Resolved component of weight down slope: $\frac{1}{80} \times 800000 \times 9.8 = 98000$ N
($\frac{1}{m} \times 9.8$ where $m$ is the mass of the object the candidate is considering. Do not award if $g$ is missing. Evaluation need not be seen)

M1 Let the acceleration be $a$ m s$^{-2}$ up the slope. Newton's 2nd Law to the whole train: $121000 - 33000 - 98000 = 800000a$
(The right elements must be present consistent with the candidate's component of the weight down the slope. No extra forces allowed)

A1 $a = 0.0125$
(Cao but allow an answer rounding to $-0.012$ or $-0.013$ following earlier premature rounding.)

A1 Magnitude 0.0125 m s$^{-2}$, down the slope
(The negative sign must be interpreted so "Down the slope" or "decelerating" must be seen)

## (iv)

M1 Taking the train as a whole, Force down the slope = Resistance force: $800000 \times 9.8 \times \sin\theta = 33000$
(Equilibrium of whole train required. The evidence for this mark may be obtained from a correct force diagram)

A1 $\sin\theta = 0.042...$
(Allow missing $g$ for this mark only)

A1 $\theta = 0.24$ (radians or approximately 14°)
(Cao)

---
1 A train consists of a locomotive pulling 17 identical trucks.
The mass of the locomotive is 120 tonnes and the mass of each truck is 40 tonnes. The locomotive gives a driving force of 121000 N .

The resistance to motion on each truck is $R \mathrm {~N}$ and the resistance on the locomotive is $5 R \mathrm {~N}$.\\
Initially the train is travelling on a straight horizontal track and its acceleration is $0.11 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\roman*)]
\item Show that $R = 1500$.
\item Find the tensions in the couplings between\\
(A) the last two trucks,\\
(B) the locomotive and the first truck.

The train now comes to a place where the track goes up a straight, uniform slope at an angle $\alpha$ with the horizontal, where $\sin \alpha = \frac { 1 } { 80 }$.

The driving force and the resistance forces remain the same as before.
\item Find the magnitude and direction of the acceleration of the train.

The train then comes to a straight uniform downward slope at an angle $\beta$ to the horizontal.\\
The driver of the train reduces the driving force to zero and the resistance forces remain the same as before. The train then travels at a constant speed down the slope.
\item Find the value of $\beta$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q1 [18]}}
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