OCR MEI AS Paper 1 2021 November — Question 3 2 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2021
SessionNovember
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: equilibrium (find unknowns)
DifficultyEasy -1.2 This is a straightforward equilibrium problem requiring only the basic principle that forces sum to zero. Students simply add the two given force vectors and negate the result. It's a direct application of a fundamental concept with no problem-solving complexity or multi-step reasoning required.
Spec1.10b Vectors in 3D: i,j,k notation3.03b Newton's first law: equilibrium
3 Forces \(\mathbf { F } _ { 1 } = ( 2 \mathbf { i } + 9 \mathbf { j } ) \mathbf { N }\) and \(\mathbf { F } _ { 2 } = ( - \mathbf { i } + \mathbf { j } ) \mathbf { N }\) act on a particle. A third force \(\mathbf { F } _ { 3 }\) acts so that the particle is in equilibrium under the action of the three forces. Find the force \(\mathbf { F } _ { 3 }\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} = \mathbf{0}\)M1 soi
\(\mathbf{F_3} = -(2\mathbf{i} + 9\mathbf{j} - \mathbf{i} + \mathbf{j}) = -\mathbf{i} - 10\mathbf{j}\)A1 cao may be written as a column vector
Total: [2]
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} = \mathbf{0}$ | M1 | soi |
| $\mathbf{F_3} = -(2\mathbf{i} + 9\mathbf{j} - \mathbf{i} + \mathbf{j}) = -\mathbf{i} - 10\mathbf{j}$ | A1 | cao may be written as a column vector |

**Total: [2]**

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3 Forces $\mathbf { F } _ { 1 } = ( 2 \mathbf { i } + 9 \mathbf { j } ) \mathbf { N }$ and $\mathbf { F } _ { 2 } = ( - \mathbf { i } + \mathbf { j } ) \mathbf { N }$ act on a particle. A third force $\mathbf { F } _ { 3 }$ acts so that the particle is in equilibrium under the action of the three forces.

Find the force $\mathbf { F } _ { 3 }$.

\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q3 [2]}}
This paper (11 questions)
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Q1 2 Q2 2 Q3 2 Q4 4 Q5 5 Q6 8 Q7 6 Q8 12 Q9 9 Q10 10 Q11 10