OCR PURE — Question 9 2 marks

Exam BoardOCR
ModulePURE
Marks2
PaperDownload PDF ↗
TopicVectors Introduction & 2D
TypeNewton's second law with vector forces (find acceleration or force)
DifficultyModerate -0.8 This is a straightforward application of Newton's second law (F=ma) with vectors. Students need to find the resultant force using the given mass and acceleration, then subtract the known force to find F. It requires only basic vector arithmetic and direct recall of F=ma, making it easier than average with no problem-solving insight needed.
Spec1.10b Vectors in 3D: i,j,k notation3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors
9 Two forces \(( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F N }\) act on a particle \(P\) of mass 4 kg .
Given that the acceleration of \(P\) is \(( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 2 }\), calculate \(\mathbf { F }\).
Question 9:
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{F}+(3\mathbf{i}+2\mathbf{j})=4(-2\mathbf{i}+3\mathbf{j})\)M1 Using \(\mathbf{F}=m\mathbf{a}\) with correct number of terms; Column vectors allowed
\(\mathbf{F}=4(-2\mathbf{i}+3\mathbf{j})-(3\mathbf{i}+2\mathbf{j})\) \(\Rightarrow \mathbf{F}=-11\mathbf{i}+10\mathbf{j}\)A1 cao – isw if magnitude found 
## Question 9:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{F}+(3\mathbf{i}+2\mathbf{j})=4(-2\mathbf{i}+3\mathbf{j})$ | M1 | Using $\mathbf{F}=m\mathbf{a}$ with correct number of terms; Column vectors allowed |
| $\mathbf{F}=4(-2\mathbf{i}+3\mathbf{j})-(3\mathbf{i}+2\mathbf{j})$ $\Rightarrow \mathbf{F}=-11\mathbf{i}+10\mathbf{j}$ | A1 | cao – isw if magnitude found |

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9 Two forces $( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { N }$ and $\mathbf { F N }$ act on a particle $P$ of mass 4 kg .\\
Given that the acceleration of $P$ is $( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$, calculate $\mathbf { F }$.

\hfill \mbox{\textit{OCR PURE  Q9 [2]}}
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