Question 5 - A-Level Maths

CAIE M1 2016 June — Question 5 7 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2016
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Motion on a slope
Type	Equilibrium on slope with force at angle to slope
Difficulty	Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces parallel and perpendicular to the slope, with friction at limiting equilibrium. The string at an angle adds mild complexity, but the method is routine: resolve in two directions, use F=μR, solve simultaneous equations. Slightly above average due to the angled string and algebraic manipulation required.
Spec	3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces 3.03r Friction: concept and vector form

5 \includegraphics[max width=\textwidth, alt={}, center]{fd2fbf13-912c-46c5-a470-306b2269aa0b-3_394_531_260_806} A block of mass 2.5 kg is placed on a plane which is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. The block is kept in equilibrium by a light string making an angle of \(20 ^ { \circ }\) above a line of greatest slope. The tension in the string is \(T \mathrm {~N}\), as shown in the diagram. The coefficient of friction between the block and plane is \(\frac { 1 } { 4 }\). The block is in limiting equilibrium and is about to move up the plane. Find the value of \(T\).

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Question 5:

Answer	Marks	Guidance
Answer	Mark	Guidance
M1	For resolving forces perpendicular to the plane (3 term equation)
\(R + T\sin 20 = 2.5g\cos 30\)	A1
\(F = 0.25 \times R\)	B1	May be implied
M1	For resolving forces parallel to the plane (3 term equation)
\(T\cos 20 = F + 2.5g\sin 30\)	A1
M1	For solving and obtaining \(T\)
\(T = 17.5\)	A1	Total: 7

Alternative scheme for Question 5:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(F = 0.25 \times R\)	B1	May be implied
M1	For resolving forces horizontally (3 term equation)
\(T\cos 50 = F\cos 30 + R\sin 30\)	A1
M1	For resolving forces vertically (4 term equation)
\(R\cos 30 + T\sin 50 = F\sin 30 + 2.5g\)	A1
M1	For solving and obtaining \(T\)
\(T = 17.5\)	A1	Total: 7

## Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| | M1 | For resolving forces perpendicular to the plane (3 term equation) |
| $R + T\sin 20 = 2.5g\cos 30$ | A1 | |
| $F = 0.25 \times R$ | B1 | May be implied |
| | M1 | For resolving forces parallel to the plane (3 term equation) |
| $T\cos 20 = F + 2.5g\sin 30$ | A1 | |
| | M1 | For solving and obtaining $T$ |
| $T = 17.5$ | A1 | **Total: 7** |

**Alternative scheme for Question 5:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $F = 0.25 \times R$ | B1 | May be implied |
| | M1 | For resolving forces horizontally (3 term equation) |
| $T\cos 50 = F\cos 30 + R\sin 30$ | A1 | |
| | M1 | For resolving forces vertically (4 term equation) |
| $R\cos 30 + T\sin 50 = F\sin 30 + 2.5g$ | A1 | |
| | M1 | For solving and obtaining $T$ |
| $T = 17.5$ | A1 | **Total: 7** |

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Show LaTeX source

5\\
\includegraphics[max width=\textwidth, alt={}, center]{fd2fbf13-912c-46c5-a470-306b2269aa0b-3_394_531_260_806}

A block of mass 2.5 kg is placed on a plane which is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The block is kept in equilibrium by a light string making an angle of $20 ^ { \circ }$ above a line of greatest slope. The tension in the string is $T \mathrm {~N}$, as shown in the diagram. The coefficient of friction between the block and plane is $\frac { 1 } { 4 }$. The block is in limiting equilibrium and is about to move up the plane. Find the value of $T$.

\hfill \mbox{\textit{CAIE M1 2016 Q5 [7]}}

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