Question 3 - A-Level Maths

CAIE M1 2015 November — Question 3 7 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2015
Session	November
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Resultant of coplanar forces
Difficulty	Standard +0.3 This is a standard resultant force problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. While it involves multiple steps (resolving horizontally and vertically, then solving for two unknowns), the method is straightforward and commonly practiced in M1. It's slightly above average difficulty due to the algebraic manipulation required, but doesn't require novel insight.
Spec	3.03p Resultant forces: using vectors

3 \includegraphics[max width=\textwidth, alt={}, center]{f23ea8e7-9b81-4192-8c20-8c46aabfecca-2_296_735_1685_705} Three horizontal forces of magnitudes \(150 \mathrm {~N} , 100 \mathrm {~N}\) and \(P \mathrm {~N}\) have directions as shown in the diagram. The resultant of the three forces is shown by the broken line in the diagram. This resultant has magnitude 120 N and makes an angle \(75 ^ { \circ }\) with the 150 N force. Find the values of \(P\) and \(\theta\).

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Question 3:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
M1	For resolving forces in the \(x\) or \(-x\) direction
\(120\cos75° = 150 - 100 - P\cos\theta°\)	A1
M1	For resolving forces in the \(y\) direction
\(120\sin75° = P\sin\theta°\)	A1
\(P^2 = 14400 - 12000\cos75° + 2500\) or \(\tan\theta = \frac{120\sin75°}{(50 - 120\cos75°)}\)	M1	For using \(P^2 = (P\cos\theta)^2 + (P\sin\theta)^2\) or for using \(P\sin\theta / P\cos\theta = \tan\theta\)
\(P = 117\) or \(\theta = 80.7\)	A1
\(\theta = 80.7\) or \(P = 117\)	B1	Total: 7 marks

## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces in the $x$ or $-x$ direction |
| $120\cos75° = 150 - 100 - P\cos\theta°$ | A1 | |
| | M1 | For resolving forces in the $y$ direction |
| $120\sin75° = P\sin\theta°$ | A1 | |
| $P^2 = 14400 - 12000\cos75° + 2500$ or $\tan\theta = \frac{120\sin75°}{(50 - 120\cos75°)}$ | M1 | For using $P^2 = (P\cos\theta)^2 + (P\sin\theta)^2$ or for using $P\sin\theta / P\cos\theta = \tan\theta$ |
| $P = 117$ or $\theta = 80.7$ | A1 | |
| $\theta = 80.7$ or $P = 117$ | B1 | Total: 7 marks |

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\includegraphics[max width=\textwidth, alt={}, center]{f23ea8e7-9b81-4192-8c20-8c46aabfecca-2_296_735_1685_705}

Three horizontal forces of magnitudes $150 \mathrm {~N} , 100 \mathrm {~N}$ and $P \mathrm {~N}$ have directions as shown in the diagram. The resultant of the three forces is shown by the broken line in the diagram. This resultant has magnitude 120 N and makes an angle $75 ^ { \circ }$ with the 150 N force. Find the values of $P$ and $\theta$.

\hfill \mbox{\textit{CAIE M1 2015 Q3 [7]}}

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