CAIE M1 2024 June — Question 3 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force up incline, find work done by engine/force
DifficultyStandard +0.3 This is a straightforward energy conservation problem requiring students to apply the work-energy principle with three standard components: kinetic energy change, gravitational potential energy change, and work against resistance. All values are given directly, requiring only substitution into W_engine = ΔKE + ΔPE + W_resistance with no conceptual difficulty or problem-solving insight needed.
Spec6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae
3 A train of mass 180000 kg ascends a straight hill of length 1.5 km , inclined at an angle of \(1.5 ^ { \circ }\) to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is 12000 kJ and the speed of the train decreases from \(45 \mathrm {~ms} ^ { - 1 }\) to \(40 \mathrm {~ms} ^ { - 1 }\). Find the work done by the engine of the train as it ascends the hill, giving your answer in kJ .
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
PE gained \(= 180000g \times 1500\sin 1.5\) \([= 70677760.4...]\)B1 \(180000g \times 39.2654...\)
\(\text{KE}_{\text{Initial}} = \frac{1}{2} \times 180000 \times 45^2\) \([= 182\,250\,000]\)B1 For initial KE or final KE (for reference: difference in KE is \(38\,250\,000\))
\(\text{KE}_{\text{Final}} = \frac{1}{2} \times 180000 \times 40^2\) \([= 144\,000\,000]\)
\(WD = 180000g \times 1500\sin 1.5 + 12000000 - \left(\frac{1}{2}\times 180000\times 45^2 - \frac{1}{2}\times 180000\times 40^2\right)\)M1 Correct number of terms; dimensionally correct; allow sign errors and minor slip(s) in values; allow sin/cos mix on PE term. Work done \(= (70677760.4...+12000000-38250000)\), \(J=(70677.7...+12000-38250)\text{ kJ}\)
\(= 44400\text{ kJ}\) \([44427.7604...]\)A1 Must be in kJ
Alternative Method (Newton's second law):
AnswerMarks Guidance
AnswerMark Guidance
\(a = -0.142\) \([=-0.141666...]\)(B1) Correct acceleration from \(40^2 = 45^2 + 2a(1500)\). Allow AWRT \(-0.14\) or exact \(-\frac{17}{120}\)
\(1500\!\left(D - \frac{12\,000\,000}{1500} - 180000g\sin 1.5\right) = 180000 \times -\frac{17}{120} \times 1500\)(M1) M1 for applying Newton's second law parallel to the hill. Correct number of terms, allow sin/cos mix on weight component, dimensionally correct and multiplying both sides by 1500. Allow *their* \(a\) or \(a\) for acceleration (and minor slip(s) in values)
(B1)Correct weight component multiplied by 1500
\(WD\ (=1500D) = 44400\text{ kJ}\) \([44427.7604...]\)(A1) Must be in kJ
4 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| PE gained $= 180000g \times 1500\sin 1.5$ $[= 70677760.4...]$ | B1 | $180000g \times 39.2654...$ |
| $\text{KE}_{\text{Initial}} = \frac{1}{2} \times 180000 \times 45^2$ $[= 182\,250\,000]$ | B1 | For initial KE or final KE (for reference: difference in KE is $38\,250\,000$) |
| $\text{KE}_{\text{Final}} = \frac{1}{2} \times 180000 \times 40^2$ $[= 144\,000\,000]$ | | |
| $WD = 180000g \times 1500\sin 1.5 + 12000000 - \left(\frac{1}{2}\times 180000\times 45^2 - \frac{1}{2}\times 180000\times 40^2\right)$ | M1 | Correct number of terms; dimensionally correct; allow sign errors and minor slip(s) in values; allow sin/cos mix on PE term. Work done $= (70677760.4...+12000000-38250000)$, $J=(70677.7...+12000-38250)\text{ kJ}$ |
| $= 44400\text{ kJ}$ $[44427.7604...]$ | A1 | Must be in kJ |

**Alternative Method (Newton's second law):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $a = -0.142$ $[=-0.141666...]$ | (B1) | Correct acceleration from $40^2 = 45^2 + 2a(1500)$. Allow AWRT $-0.14$ or exact $-\frac{17}{120}$ |
| $1500\!\left(D - \frac{12\,000\,000}{1500} - 180000g\sin 1.5\right) = 180000 \times -\frac{17}{120} \times 1500$ | (M1) | M1 for applying Newton's second law parallel to the hill. Correct number of terms, allow sin/cos mix on weight component, dimensionally correct and multiplying both sides by 1500. Allow *their* $a$ or $a$ for acceleration (and minor slip(s) in values) |
| | (B1) | Correct weight component multiplied by 1500 |
| $WD\ (=1500D) = 44400\text{ kJ}$ $[44427.7604...]$ | (A1) | Must be in kJ |
| | **4** | |

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3 A train of mass 180000 kg ascends a straight hill of length 1.5 km , inclined at an angle of $1.5 ^ { \circ }$ to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is 12000 kJ and the speed of the train decreases from $45 \mathrm {~ms} ^ { - 1 }$ to $40 \mathrm {~ms} ^ { - 1 }$.

Find the work done by the engine of the train as it ascends the hill, giving your answer in kJ .\\

\hfill \mbox{\textit{CAIE M1 2024 Q3 [4]}}
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