CAIE M1 2013 June — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force up incline, find speed
DifficultyModerate -0.3 This is a straightforward work-energy theorem application with clearly stated values. Students must account for three energy changes (kinetic, gravitational potential, work against resistance) and solve for final speed using standard formulas. The calculation is multi-step but follows a routine template with no conceptual subtleties or problem-solving insight required.
Spec6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle
2 A car of mass 1250 kg travels from the bottom to the top of a straight hill of length 600 m , which is inclined at an angle of \(2.5 ^ { \circ }\) to the horizontal. The resistance to motion of the car is constant and equal to 400 N . The work done by the driving force is 450 kJ . The speed of the car at the bottom of the hill is \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the speed of the car at the top of the hill.
Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Increase in \(PE = 1250 \times 10 \times 600\sin 2.5°\)B1
Decrease in \(KE = \frac{1}{2}1250(30^2 - v_{top}^2)\)B1
WD against resistance \(= 400 \times 600\)B1
\([562500 - 625v_{top}^2 = 327145 + 240000 - 450000]\)M1 For using \(WD\) by \(DF =\) Increase in \(PE -\) decrease in \(KE +\) WD against resistance
Speed is \(26.7\ ms^{-1}\)A1 [5]
Special Ruling (assuming constant DF, maximum mark 4):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([DF - \text{Weight component} - \text{Resistance} = \text{Mass} \times \text{Accel'n}]\)M1 For applying Newton's second law
\(750 - 545 - 400 = 1250a\)A1
\(v^2 = 30^2 + 2 \times (-0.156) \times 600\)B1ft ft value of \(a\)
Speed is \(26.7\ ms^{-1}\)B1 [4] 
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Increase in $PE = 1250 \times 10 \times 600\sin 2.5°$ | B1 | |
| Decrease in $KE = \frac{1}{2}1250(30^2 - v_{top}^2)$ | B1 | |
| WD against resistance $= 400 \times 600$ | B1 | |
| $[562500 - 625v_{top}^2 = 327145 + 240000 - 450000]$ | M1 | For using $WD$ by $DF =$ Increase in $PE -$ decrease in $KE +$ WD against resistance |
| Speed is $26.7\ ms^{-1}$ | A1 | [5] |

**Special Ruling** (assuming constant DF, maximum mark 4):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[DF - \text{Weight component} - \text{Resistance} = \text{Mass} \times \text{Accel'n}]$ | M1 | For applying Newton's second law |
| $750 - 545 - 400 = 1250a$ | A1 | |
| $v^2 = 30^2 + 2 \times (-0.156) \times 600$ | B1ft | ft value of $a$ |
| Speed is $26.7\ ms^{-1}$ | B1 | [4] |

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2 A car of mass 1250 kg travels from the bottom to the top of a straight hill of length 600 m , which is inclined at an angle of $2.5 ^ { \circ }$ to the horizontal. The resistance to motion of the car is constant and equal to 400 N . The work done by the driving force is 450 kJ . The speed of the car at the bottom of the hill is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the speed of the car at the top of the hill.

\hfill \mbox{\textit{CAIE M1 2013 Q2 [5]}}
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