Question 1 - A-Level Maths

OCR MEI Further Mechanics A AS 2024 June — Question 1 4 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2024
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Resultant of coplanar forces
Difficulty	Moderate -0.8 This is a straightforward two-force resolution problem requiring basic trigonometry and component addition. Students resolve both forces into x and y components, set the y-component to zero (since resultant is horizontal), solve for θ, then find the resultant magnitude. It's more routine than average A-level mechanics questions, requiring only standard force resolution techniques with no problem-solving insight or multi-step reasoning.
Spec	3.03a Force: vector nature and diagrams 3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces

1 Two horizontal forces of magnitudes 7 N and 15 N act at a point O .
The 15 N force acts an angle of \(\theta ^ { \circ }\) above the positive \(x\)-axis.
The 7 N force acts at an angle of \(70 ^ { \circ }\) below the negative \(x\)-axis (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-2_606_773_402_239} The resultant of the two forces acts only in the positive \(x\)-direction.

Calculate the value of \(\theta\).
Calculate the magnitude of the resultant of the two forces.

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1	(a)	15sin𝜃 = 7sin70°
2 6 .0   =	M1

A1

Answer	Marks
[2]	3.1b
1.1	Resolving both forces in 𝑦-direction and forming an

equation.

Allow sign errors

sin𝜃 sin110°

OR Vector triangle =

7 15

Accept 0.454 (rad)

Answer	Marks
(b)	15cos𝜃−7cos70°
Magnitude of resultant force is 11.1 (N)	M1

A1

Answer	Marks
[2]	1.1
1.1	Resolving both forces in 𝑥-direction and combining

Allow sign errors, and sin/cos interchange if consistent

with (a)

𝑅 15 7

OR Vector triangle = ( = )

sin(70−𝜃) sin110° sin𝜃

𝑅2 = 152+72−2(15)(7)cos(70−𝜃)

152 = 𝑅2+72−2𝑅(7)cos110° etc

11.086656

cao and www e.g. A0 if obtained from 𝜃 = −26

Question 1:
1 | (a) | 15sin𝜃 = 7sin70°
2 6 .0   = | M1
A1
[2] | 3.1b
1.1 | Resolving both forces in 𝑦-direction and forming an
equation.
Allow sign errors
sin𝜃 sin110°
OR Vector triangle =
7 15
Accept 0.454 (rad)
(b) | 15cos𝜃−7cos70°
Magnitude of resultant force is 11.1 (N) | M1
A1
[2] | 1.1
1.1 | Resolving both forces in 𝑥-direction and combining
Allow sign errors, and sin/cos interchange if consistent
with (a)
𝑅 15 7
OR Vector triangle = ( = )
sin(70−𝜃) sin110° sin𝜃
𝑅2 = 152+72−2(15)(7)cos(70−𝜃)
152 = 𝑅2+72−2𝑅(7)cos110° etc
11.086656
cao and www e.g. A0 if obtained from 𝜃 = −26

Show LaTeX source

1 Two horizontal forces of magnitudes 7 N and 15 N act at a point O .\\
The 15 N force acts an angle of $\theta ^ { \circ }$ above the positive $x$-axis.\\
The 7 N force acts at an angle of $70 ^ { \circ }$ below the negative $x$-axis (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-2_606_773_402_239}

The resultant of the two forces acts only in the positive $x$-direction.
\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $\theta$.
\item Calculate the magnitude of the resultant of the two forces.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2024 Q1 [4]}}

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Q1 4 Q2 11 Q3 13 Q4 13 Q5 9 Q6 10