OCR MEI M1 2015 June — Question 6 18 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2015
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeModelling assumptions and refinements
DifficultyStandard +0.3 This is a straightforward M1 mechanics question involving forces on slopes with clear step-by-step parts. Part (i) is simple F=ma application, part (ii) requires comparing forces to weight components on a slope (standard technique), and part (iii) appears to continue similarly. The angles are small, calculations are routine, and no novel problem-solving insight is required—slightly easier than average A-level maths.
Spec3.03c Newton's second law: F=ma one dimension3.03v Motion on rough surface: including inclined planes

6 The battery on Carol and Martin's car is flat so the car will not start. They hope to be able to "bump start" the car by letting it run down a hill and engaging the engine when the car is going fast enough. Fig. 6.1 shows the road leading away from their house, which is at A . The road is straight, and at all times the car is steered directly along it.
  • From A to B the road is horizontal.
  • Between B and C, it goes up a hill with a uniform slope of \(1.5 ^ { \circ }\) to the horizontal.
  • Between C and D the road goes down a hill with a uniform slope of \(3 ^ { \circ }\) to the horizontal. CD is 100 m . (This is the part of the road where they hope to get the car started.)
  • From D to E the road is again horizontal.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-4_241_1134_808_450} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure} The mass of the car is 750 kg , Carol's mass is 50 kg and Martin's mass is 80 kg .
Throughout the rest of this question, whenever Martin pushes the car, he exerts a force of 300 N along the line of the car.
  1. Between A and B , Martin pushes the car and Carol sits inside to steer it. The car has an acceleration of \(0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Show that the resistance to the car's motion is 100 N . Throughout the rest of this question you should assume that the resistance to motion is constant at 100 N .
  2. They stop at B and then Martin tries to push the car up the hill BC. Show that Martin cannot push the car up the hill with Carol inside it but can if she gets out.
    Find the acceleration of the car when Martin is pushing it and Carol is standing outside.
  3. While between B and C , Carol opens the window of the car and pushes it from outside while steering with one hand. Carol is able to exert a force of 150 N parallel to the surface of the road but at an angle of \(30 ^ { \circ }\) to the line of the car. This is illustrated in Fig. 6.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-4_218_426_2133_831} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{figure} Find the acceleration of the car.
  4. At C, both Martin and Carol get in the car and, starting from rest, let it run down the hill under gravity. If the car reaches a speed of \(8 \mathrm {~ms} ^ { - 1 }\) they can get the engine to start. Does the car reach this speed before it reaches D ?

Question 6:
Part (i):
AnswerMarks Guidance
\(F - R = ma\)M1 Use of Newton's 2nd Law
\(300 - R = (750 + 50) \times 0.25\)A1 Correct elements present
\(R = 100\)A1 This is a given result
Part (ii):
AnswerMarks Guidance
Component of weight down slope \(= 800g\sin 1.5° = 205.2\text{ N}\)M1, A1 Resolving down the slope. Accept use of 750 instead of 800. Condone no \(g\) and allow sin-cos interchange. Give M1 A1 for \(800g\sin 15°\) seen
Martin has to overcome 305.2 N; \(300 < 305.2\) Martin cannot manageA1 May be awarded for argument based on Newton's 2nd law leading to \(a = -0.006\)
Carol out: Martin has to overcome \(750g\sin 1.5° + 100 = 292.4\text{ N}\); \(300 > 292.4\) so Martin managesB1 Explanation based on correct working that Martin can manage. Can be given retrospectively with comment on positive value of \(a\)
\(300 - 292.4 = 7.6 = 750a\)M1 Use of Newton's 2nd Law
The acceleration is \(0.010\text{ ms}^{-2}\)A1 Cao. Accept 0.01 or answer rounding to 0.01
Part (iii):
AnswerMarks Guidance
Component of Carol's force parallel to line of car \(= 150\cos 30° = 129.9\)M1, A1 For attempt at resolution in correct direction. Condone sin-cos interchange. Give M1 A1 for \(150\cos 30°\) seen
Resultant forward force \(= 7.6 + 129.9 = 137.5\); \(750a = 137.5\)M1 All forces parallel to slope present and correct. Sign errors condoned
The acceleration is \(0.183\text{ ms}^{-2}\)A1 FT their force parallel to slope from part (ii) (correct value 7.6 N)
Part (iv):
AnswerMarks Guidance
Component of weight down slope \(= (750+50+80) \times 9.8 \times \sin 3°\)
\(880a = 451.3 - 100\)M1 Newton's 2nd law with correct elements. No sin-cos interchange. Same mass used in both places
\(a = 0.399\)A1
\(v^2 - u^2 = 2as\); When \(v = 8\), \(s = 8^2 \div (2 \times 0.399)\)M1 Selection and use of appropriate formula (unless \(a = g\))
\(s = 80.1\)A1 FT their value of \(a\)
\(80.1 < 100\) so Yes they get the car startedA1 FT their value of \(a\)
Part (iv) Alternative: Finding speed after 100 m:
AnswerMarks Guidance
\((750+50+80) \times 9.8 \times \sin 3°\); \(880a = 451.3 - 100\); \(a = 0.399\)M1, A1 Newton's 2nd law with correct elements. No sin-cos interchange
\(v^2 = (0^2) + 2 \times 0.399 \times 100\)M1 Selection and use of appropriate formula (unless \(a = g\))
\(v = \sqrt{79.8} = 8.93...\)A1 FT their value of \(a\)
\((v > 8)\) so they get the car startedA1 FT their value of \(a\)
# Question 6:

## Part (i):
| $F - R = ma$ | M1 | Use of Newton's 2nd Law |
| $300 - R = (750 + 50) \times 0.25$ | A1 | Correct elements present |
| $R = 100$ | A1 | This is a given result |

## Part (ii):
| Component of weight down slope $= 800g\sin 1.5° = 205.2\text{ N}$ | M1, A1 | Resolving down the slope. Accept use of 750 instead of 800. Condone no $g$ and allow sin-cos interchange. Give M1 A1 for $800g\sin 15°$ seen |
| Martin has to overcome 305.2 N; $300 < 305.2$ Martin cannot manage | A1 | May be awarded for argument based on Newton's 2nd law leading to $a = -0.006$ |
| **Carol out:** Martin has to overcome $750g\sin 1.5° + 100 = 292.4\text{ N}$; $300 > 292.4$ so Martin manages | B1 | Explanation based on correct working that Martin can manage. Can be given retrospectively with comment on positive value of $a$ |
| $300 - 292.4 = 7.6 = 750a$ | M1 | Use of Newton's 2nd Law |
| The acceleration is $0.010\text{ ms}^{-2}$ | A1 | Cao. Accept 0.01 or answer rounding to 0.01 |

## Part (iii):
| Component of Carol's force parallel to line of car $= 150\cos 30° = 129.9$ | M1, A1 | For attempt at resolution in correct direction. Condone sin-cos interchange. Give M1 A1 for $150\cos 30°$ seen |
| Resultant forward force $= 7.6 + 129.9 = 137.5$; $750a = 137.5$ | M1 | All forces parallel to slope present and correct. Sign errors condoned |
| The acceleration is $0.183\text{ ms}^{-2}$ | A1 | FT their force parallel to slope from part (ii) (correct value 7.6 N) |

## Part (iv):
| Component of weight down slope $= (750+50+80) \times 9.8 \times \sin 3°$ | | |
| $880a = 451.3 - 100$ | M1 | Newton's 2nd law with correct elements. No sin-cos interchange. Same mass used in both places |
| $a = 0.399$ | A1 | |
| $v^2 - u^2 = 2as$; When $v = 8$, $s = 8^2 \div (2 \times 0.399)$ | M1 | Selection and use of appropriate formula (unless $a = g$) |
| $s = 80.1$ | A1 | FT their value of $a$ |
| $80.1 < 100$ so Yes they get the car started | A1 | FT their value of $a$ |

### Part (iv) Alternative: Finding speed after 100 m:
| $(750+50+80) \times 9.8 \times \sin 3°$; $880a = 451.3 - 100$; $a = 0.399$ | M1, A1 | Newton's 2nd law with correct elements. No sin-cos interchange |
| $v^2 = (0^2) + 2 \times 0.399 \times 100$ | M1 | Selection and use of appropriate formula (unless $a = g$) |
| $v = \sqrt{79.8} = 8.93...$ | A1 | FT their value of $a$ |
| $(v > 8)$ so they get the car started | A1 | FT their value of $a$ |

---
6 The battery on Carol and Martin's car is flat so the car will not start. They hope to be able to "bump start" the car by letting it run down a hill and engaging the engine when the car is going fast enough. Fig. 6.1 shows the road leading away from their house, which is at A . The road is straight, and at all times the car is steered directly along it.

\begin{itemize}
  \item From A to B the road is horizontal.
  \item Between B and C, it goes up a hill with a uniform slope of $1.5 ^ { \circ }$ to the horizontal.
  \item Between C and D the road goes down a hill with a uniform slope of $3 ^ { \circ }$ to the horizontal. CD is 100 m . (This is the part of the road where they hope to get the car started.)
  \item From D to E the road is again horizontal.
\end{itemize}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-4_241_1134_808_450}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}

The mass of the car is 750 kg , Carol's mass is 50 kg and Martin's mass is 80 kg .\\
Throughout the rest of this question, whenever Martin pushes the car, he exerts a force of 300 N along the line of the car.\\
(i) Between A and B , Martin pushes the car and Carol sits inside to steer it. The car has an acceleration of $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

Show that the resistance to the car's motion is 100 N .

Throughout the rest of this question you should assume that the resistance to motion is constant at 100 N .\\
(ii) They stop at B and then Martin tries to push the car up the hill BC.

Show that Martin cannot push the car up the hill with Carol inside it but can if she gets out.\\
Find the acceleration of the car when Martin is pushing it and Carol is standing outside.\\
(iii) While between B and C , Carol opens the window of the car and pushes it from outside while steering with one hand. Carol is able to exert a force of 150 N parallel to the surface of the road but at an angle of $30 ^ { \circ }$ to the line of the car. This is illustrated in Fig. 6.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-4_218_426_2133_831}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}

Find the acceleration of the car.\\
(iv) At C, both Martin and Carol get in the car and, starting from rest, let it run down the hill under gravity. If the car reaches a speed of $8 \mathrm {~ms} ^ { - 1 }$ they can get the engine to start.

Does the car reach this speed before it reaches D ?

\hfill \mbox{\textit{OCR MEI M1 2015 Q6 [18]}}