CAIE M1 2014 November — Question 7 11 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2014
SessionNovember
Marks11
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 work-energy problem requiring students to apply the work-energy principle with multiple forces (applied force at an angle, weight, resistance). While it involves several components and careful bookkeeping of energy terms, it follows a standard template with no novel insight required—slightly easier than average due to its methodical nature.
Spec6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle

7 \includegraphics[max width=\textwidth, alt={}, center]{ffefbc81-402f-4048-8741-23c8bae30d5a-3_246_1006_1781_571} A block of mass 60 kg is pulled up a hill in the line of greatest slope by a force of magnitude 50 N acting at an angle \(\alpha ^ { \circ }\) above the hill. The block passes through points \(A\) and \(B\) with speeds \(8.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) respectively (see diagram). The distance \(A B\) is 250 m and \(B\) is 17.5 m above the level of \(A\). The resistance to motion of the block is 6 N . Find the value of \(\alpha\).
[0pt] [11]

Question 7:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(Obtain PE change or KE change)M1 To obtain PE change or KE change
PE change \(= 60g \times 17.5\) or KE change \(= \frac{1}{2} \times 60(8.5^2 - 3.5^2)\)A1 \([PE = 10500]\)
KE change \(= \frac{1}{2} \times 60(8.5^2 - 3.5^2)\) or PE change \(= 60g \times 17.5\)B1 \([KE = 1800]\)
WD against resistance \(= 6 \times 250\)B1 \([= 1500]\)
WD by pulling force \(= 50\cos\alpha \times 250\)B1
(WD by pulling force is linear combination of PE change, KE change and WD against resistance)M1 For using 'WD by the pulling force is a linear combination of PE change, KE change and WD against resistance'
\(WD = 10500 - 1800 + 1500\)A1\(\checkmark\)
WD by the pulling force is \(10200\) J or \(10.2\) kJA1
For using \(WD = Fd\cos\alpha\)M1
\(10200 = 50 \times 250\cos\alpha\)A1
\(\alpha = 35.3\)A1 [Total: 11]
Alternative Solution:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Using \(v^2 = u^2 + 2as\)M1 Using \(v^2 = u^2 + 2as\)
\((3.5)^2 = (8.5)^2 + 2a(250)\)A1
\(a = -3/25 = -0.12\)A1
M2Applying Newton's \(2^{nd}\) law with 4 relevant terms [Allow M1 with 3 relevant terms]
\(50\cos\alpha - 6 - 60g(17.5/250) = 60(-0.12)\)A4 One mark for each correct term
\([\cos\alpha = 102/125]\)M1 Solve for \(\cos\alpha\)
\(\alpha = 35.3\)A1 11
## Question 7:

| Answer/Working | Mark | Guidance |
|---|---|---|
| (Obtain PE change or KE change) | M1 | To obtain PE change or KE change |
| PE change $= 60g \times 17.5$ **or** KE change $= \frac{1}{2} \times 60(8.5^2 - 3.5^2)$ | A1 | $[PE = 10500]$ |
| KE change $= \frac{1}{2} \times 60(8.5^2 - 3.5^2)$ **or** PE change $= 60g \times 17.5$ | B1 | $[KE = 1800]$ |
| WD against resistance $= 6 \times 250$ | B1 | $[= 1500]$ |
| WD by pulling force $= 50\cos\alpha \times 250$ | B1 | |
| (WD by pulling force is linear combination of PE change, KE change and WD against resistance) | M1 | For using 'WD by the pulling force is a linear combination of PE change, KE change and WD against resistance' |
| $WD = 10500 - 1800 + 1500$ | A1$\checkmark$ | |
| WD by the pulling force is $10200$ J or $10.2$ kJ | A1 | |
| For using $WD = Fd\cos\alpha$ | M1 | |
| $10200 = 50 \times 250\cos\alpha$ | A1 | |
| $\alpha = 35.3$ | A1 | **[Total: 11]** |

## Alternative Solution:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $v^2 = u^2 + 2as$ | M1 | Using $v^2 = u^2 + 2as$ |
| $(3.5)^2 = (8.5)^2 + 2a(250)$ | A1 | |
| $a = -3/25 = -0.12$ | A1 | |
| | M2 | Applying Newton's $2^{nd}$ law with 4 relevant terms [Allow M1 with 3 relevant terms] |
| $50\cos\alpha - 6 - 60g(17.5/250) = 60(-0.12)$ | A4 | One mark for each correct term |
| $[\cos\alpha = 102/125]$ | M1 | Solve for $\cos\alpha$ |
| $\alpha = 35.3$ | A1 | **11** |
7\\
\includegraphics[max width=\textwidth, alt={}, center]{ffefbc81-402f-4048-8741-23c8bae30d5a-3_246_1006_1781_571}

A block of mass 60 kg is pulled up a hill in the line of greatest slope by a force of magnitude 50 N acting at an angle $\alpha ^ { \circ }$ above the hill. The block passes through points $A$ and $B$ with speeds $8.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively (see diagram). The distance $A B$ is 250 m and $B$ is 17.5 m above the level of $A$. The resistance to motion of the block is 6 N . Find the value of $\alpha$.\\[0pt]
[11]

\hfill \mbox{\textit{CAIE M1 2014 Q7 [11]}}