CAIE M1 2015 June — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeMotion up rough slope
DifficultyStandard +0.3 This is a standard mechanics problem requiring resolution of forces on a slope, application of friction laws (F = μR), and use of equations of motion. While it involves multiple steps (resolving perpendicular and parallel to slope, finding net force, then using kinematics), these are routine procedures for M1 students with no novel insight required. The unusual angle and coefficient values are straightforward to work with algebraically.
Spec3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle
3 A block of weight 6.1 N slides down a slope inclined at \(\tan ^ { - 1 } \left( \frac { 11 } { 60 } \right)\) to the horizontal. The coefficient of friction between the block and the slope is \(\frac { 1 } { 4 }\). The block passes through a point \(A\) with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find how far the block moves from \(A\) before it comes to rest.
Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = 0.25\left(6.1 \times \frac{60}{61}\right)\ [= 1.5]\)B1 Allow \(F = 0.25(6.1\cos 10.4)\)
\([W\sin\alpha - F = ma]\)M1 For using Newton's 2nd law
\(6.1 \times \left(\frac{11}{61}\right) - 0.25\left(6.1 \times \frac{60}{61}\right) = 0.61a\)A1 \(\left[a = -\frac{40}{61} = -0.656\right]\) — value of \(a\) may be seen but is not a required answer
or \(6.1\sin 10.4 - 0.25 \times 6.1\cos 10.4 = 0.61a\)
M1For using \(0 = v_A^2 + 2as\)
Distance is \(4 \div \left(2 \times \frac{40}{61}\right) = 3.05\) mA1
Total: 5 marks
Alternative Method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = 0.25\left(6.1 \times \frac{60}{61}\right)\ [= 1.5]\)B1 Allow \(F = 0.25(6.1\cos 10.4)\)
\(\text{KE loss} = \frac{1}{2} \times 0.61 \times 2^2\)B1 Finding loss of KE
\(\text{PE loss} = 0.61 \times 10 \times x\left(\frac{11}{61}\right)\)B1 Finding loss of PE
\([1.5x = 1.22 + 1.1x]\)M1 Using \(WD\) against \(F =\) KE loss + PE loss
\(0.4x = 1.22 \rightarrow \text{distance} = 3.05\) mA1
Total: 5 marks 
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.25\left(6.1 \times \frac{60}{61}\right)\ [= 1.5]$ | B1 | Allow $F = 0.25(6.1\cos 10.4)$ |
| $[W\sin\alpha - F = ma]$ | M1 | For using Newton's 2nd law |
| $6.1 \times \left(\frac{11}{61}\right) - 0.25\left(6.1 \times \frac{60}{61}\right) = 0.61a$ | A1 | $\left[a = -\frac{40}{61} = -0.656\right]$ — value of $a$ may be seen but is not a required answer |
| or $6.1\sin 10.4 - 0.25 \times 6.1\cos 10.4 = 0.61a$ | | |
| | M1 | For using $0 = v_A^2 + 2as$ |
| Distance is $4 \div \left(2 \times \frac{40}{61}\right) = 3.05$ m | A1 | |
| **Total: 5 marks** | | |

### Alternative Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.25\left(6.1 \times \frac{60}{61}\right)\ [= 1.5]$ | B1 | Allow $F = 0.25(6.1\cos 10.4)$ |
| $\text{KE loss} = \frac{1}{2} \times 0.61 \times 2^2$ | B1 | Finding loss of KE |
| $\text{PE loss} = 0.61 \times 10 \times x\left(\frac{11}{61}\right)$ | B1 | Finding loss of PE |
| $[1.5x = 1.22 + 1.1x]$ | M1 | Using $WD$ against $F =$ KE loss + PE loss |
| $0.4x = 1.22 \rightarrow \text{distance} = 3.05$ m | A1 | |
| **Total: 5 marks** | | |

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3 A block of weight 6.1 N slides down a slope inclined at $\tan ^ { - 1 } \left( \frac { 11 } { 60 } \right)$ to the horizontal. The coefficient of friction between the block and the slope is $\frac { 1 } { 4 }$. The block passes through a point $A$ with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find how far the block moves from $A$ before it comes to rest.

\hfill \mbox{\textit{CAIE M1 2015 Q3 [5]}}
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