CAIE M1 2013 June — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeMotion down rough slope
DifficultyModerate -0.3 This is a straightforward mechanics problem requiring resolution of forces on a slope, application of F=ma with friction, and use of constant acceleration equations. While it involves multiple steps (resolving perpendicular to find normal reaction, calculating friction force, finding net force down slope, then using kinematics), these are all standard procedures that follow a well-established method for this topic type. The small coefficient of friction and given initial velocity add minor computational complexity but no conceptual difficulty.
Spec3.03v Motion on rough surface: including inclined planes

1 A straight ice track of length 50 m is inclined at \(14 ^ { \circ }\) to the horizontal. A man starts at the top of the track, on a sledge, with speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). He travels on the sledge to the bottom of the track. The coefficient of friction between the sledge and the track is 0.02 . Find the speed of the sledge and the man when they reach the bottom of the track.

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([(W/g)a = W\sin\alpha - 0.02W\cos\alpha]\)M1 For using Newton's second law
\(a = (\sin 14° - 0.02\cos 14°)g = 2.225...\)A1
\([v^2 = 8^2 + 2 \times 2.225... \times 50]\)M1 For using \(v^2 = u^2 + 2as\)
Speed is \(16.9\ \text{m s}^{-1}\)A1 [4]
Alternative Scheme:
AnswerMarks Guidance
Answer/WorkingMark Guidance
WD against friction \(= 0.02W\cos\alpha \times 50\)B1
PE loss \(= W \times 50\sin\alpha\)B1
M1For using Gain in KE = Loss in PE − WD against friction
Speed is \(16.9\ \text{m s}^{-1}\)A1 [4]
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[(W/g)a = W\sin\alpha - 0.02W\cos\alpha]$ | M1 | For using Newton's second law |
| $a = (\sin 14° - 0.02\cos 14°)g = 2.225...$ | A1 | |
| $[v^2 = 8^2 + 2 \times 2.225... \times 50]$ | M1 | For using $v^2 = u^2 + 2as$ |
| Speed is $16.9\ \text{m s}^{-1}$ | A1 [4] | |

**Alternative Scheme:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| WD against friction $= 0.02W\cos\alpha \times 50$ | B1 | |
| PE loss $= W \times 50\sin\alpha$ | B1 | |
| | M1 | For using Gain in KE = Loss in PE − WD against friction |
| Speed is $16.9\ \text{m s}^{-1}$ | A1 [4] | |

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1 A straight ice track of length 50 m is inclined at $14 ^ { \circ }$ to the horizontal. A man starts at the top of the track, on a sledge, with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. He travels on the sledge to the bottom of the track. The coefficient of friction between the sledge and the track is 0.02 . Find the speed of the sledge and the man when they reach the bottom of the track.

\hfill \mbox{\textit{CAIE M1 2013 Q1 [4]}}