Question 1 - A-Level Maths

CAIE M1 2013 November — Question 1 3 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2013
Session	November
Marks	3
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Motion on a slope
Type	String at angle to slope
Difficulty	Standard +0.3 This is a standard equilibrium problem on an inclined plane requiring resolution of forces in two perpendicular directions (parallel and perpendicular to the slope). Students must use given sine values to find components and solve simultaneous equations. While it involves multiple steps, it follows a routine procedure taught in M1 with no novel insight required, making it slightly easier than average.
Spec	3.03b Newton's first law: equilibrium 3.03e Resolve forces: two dimensions 3.03m Equilibrium: sum of resolved forces = 0

1 \includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-2_346_583_255_781} A small block of weight 5.1 N rests on a smooth plane inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac { 8 } { 17 }\). The block is held in equilibrium by means of a light inextensible string. The string makes an angle \(\beta\) above the line of greatest slope on which the block rests, where \(\sin \beta = \frac { 7 } { 25 }\) (see diagram). Find the tension in the string.

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Question 1:

Main Scheme:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
(resolving parallel to line of greatest slope)	M1	For resolving forces parallel to the line of greatest slope
\(T\cos\beta = W\sin\alpha\)	A1	\(T(24/25) = 5.1(8/17)\) or \(T\cos 16.26 = 5.1\sin 28.07\)
Tension is \(2.5\) N	A1	3 marks total

First Alternative Scheme:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
(resolving forces vertically or horizontally)	M1	For resolving forces vertically or horizontally
\(R\cos\alpha + T\sin(\alpha+\beta) = W\) and \(R\sin\alpha = T\cos(\alpha+\beta)\)	A1	\(R\cos 28.07 + T\sin 44.33 = 5.1\) and \(R\sin 28.07 = T\cos 44.33\)
Tension is \(2.5\) N	A1	3 marks total

Second Alternative Scheme:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
(Triangle of forces)	M1	Using Triangle of forces
\(T/\sin\alpha = 5.1/\sin(90+\beta)\)	A1	\(T/\sin 28.07 = 5.1/\sin 106.26\)
Tension is \(2.5\) N	A1	3 marks total

## Question 1:

**Main Scheme:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| (resolving parallel to line of greatest slope) | M1 | For resolving forces parallel to the line of greatest slope |
| $T\cos\beta = W\sin\alpha$ | A1 | $T(24/25) = 5.1(8/17)$ or $T\cos 16.26 = 5.1\sin 28.07$ |
| Tension is $2.5$ N | A1 | **3 marks total** |

**First Alternative Scheme:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| (resolving forces vertically or horizontally) | M1 | For resolving forces vertically or horizontally |
| $R\cos\alpha + T\sin(\alpha+\beta) = W$ and $R\sin\alpha = T\cos(\alpha+\beta)$ | A1 | $R\cos 28.07 + T\sin 44.33 = 5.1$ and $R\sin 28.07 = T\cos 44.33$ |
| Tension is $2.5$ N | A1 | **3 marks total** |

**Second Alternative Scheme:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| (Triangle of forces) | M1 | Using Triangle of forces |
| $T/\sin\alpha = 5.1/\sin(90+\beta)$ | A1 | $T/\sin 28.07 = 5.1/\sin 106.26$ |
| Tension is $2.5$ N | A1 | **3 marks total** |

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Show LaTeX source

1\\
\includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-2_346_583_255_781}

A small block of weight 5.1 N rests on a smooth plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 8 } { 17 }$. The block is held in equilibrium by means of a light inextensible string. The string makes an angle $\beta$ above the line of greatest slope on which the block rests, where $\sin \beta = \frac { 7 } { 25 }$ (see diagram). Find the tension in the string.

\hfill \mbox{\textit{CAIE M1 2013 Q1 [3]}}

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