CAIE M1 2014 November — Question 5

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2014
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyModerate -0.5 This appears to be a corrupted/unreadable question text, but based on the topic (SUVAT & Travel Graphs) and exam context (CAIE M1), this would typically involve standard kinematics equations with constant acceleration. M1 questions on SUVAT are generally straightforward applications of formulae, placing this slightly below average difficulty for A-level maths overall.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.02a Kinematics language: position, displacement, velocity, acceleration
5 A box of mass 8 kg is on a rough plane inclined at \(5 ^ { \circ }\) to the horizontal. A force of magnitude \(P \mathrm {~N}\) acts on the box in a direction upwards and parallel to a line of greatest slope of the plane. When \(P = 7 X\) the box moves up the line of greatest slope with acceleration \(0.15 \mathrm {~ms} ^ { - 2 }\) and when \(P = 8 X\) the box moves up the line of greatest slope with acceleration \(1.15 \mathrm {~ms} ^ { - 2 }\). Find the value of \(X\) and the coefficient of friction between the box and the plane.
Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([P - 8g\sin 5° - F = 8a]\)M1 For using Newton's 2nd law (either case)
\(7X - 8g\sin 5° - F = 8 \times 0.15\) and \(8X - 8g\sin 5° - F = 8 \times 1.15\)A1
\(X = 8\)A1
M1For obtaining a numerical expression for \(F\)
\(F = 56 - 8g\sin 5° - 8 \times 0.15\) or \(F = 64 - 8g\sin 5° - 8 \times 1.15\) or \(F = 56 \times 1.15 - 64 \times 0.15 - 8g\sin 5°\) or \(F = 47.8(275...)\)A1\(\checkmark\) ft \(X\) either from error for one term in \(X/F\) equation or from error in solution of correct \(X/F\) equations
\(R = 8g\cos 5°\ (= 79.695...)\)B1
\([\mu = 47.8 \div 79.7]\)M1 For using \(\mu = \dfrac{F}{R}\)
Coefficient is \(0.600\) (accept \(0.6\))A1 [8 marks] 
## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P - 8g\sin 5° - F = 8a]$ | M1 | For using Newton's 2nd law (either case) |
| $7X - 8g\sin 5° - F = 8 \times 0.15$ **and** $8X - 8g\sin 5° - F = 8 \times 1.15$ | A1 | |
| $X = 8$ | A1 | |
| | M1 | For obtaining a numerical expression for $F$ |
| $F = 56 - 8g\sin 5° - 8 \times 0.15$ **or** $F = 64 - 8g\sin 5° - 8 \times 1.15$ **or** $F = 56 \times 1.15 - 64 \times 0.15 - 8g\sin 5°$ **or** $F = 47.8(275...)$ | A1$\checkmark$ | ft $X$ either from error for one term in $X/F$ equation or from error in solution of correct $X/F$ equations |
| $R = 8g\cos 5°\ (= 79.695...)$ | B1 | |
| $[\mu = 47.8 \div 79.7]$ | M1 | For using $\mu = \dfrac{F}{R}$ |
| Coefficient is $0.600$ (accept $0.6$) | A1 | **[8 marks]** |

---
5 A box of mass 8 kg is on a rough plane inclined at $5 ^ { \circ }$ to the horizontal. A force of magnitude $P \mathrm {~N}$ acts on the box in a direction upwards and parallel to a line of greatest slope of the plane. When $P = 7 X$ the box moves up the line of greatest slope with acceleration $0.15 \mathrm {~ms} ^ { - 2 }$ and when $P = 8 X$ the box moves up the line of greatest slope with acceleration $1.15 \mathrm {~ms} ^ { - 2 }$. Find the value of $X$ and the coefficient of friction between the box and the plane.

\hfill \mbox{\textit{CAIE M1 2014 Q5}}
This paper (6 questions)
View full paper
Q2 Q3 Q4 Q5 Q6 Q7