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Two forces, $(4i - 5j) \text{ N}$ and $(pi + qj) \text{ N}$, act on a particle $P$ of mass $m$ kg - Edexcel - A-Level Maths: Mechanics - Question 6 - 2009 - Paper 1
Question 6
Two forces, $(4i - 5j) \text{ N}$ and $(pi + qj) \text{ N}$, act on a particle $P$ of mass $m$ kg. The resultant of the two forces is $R$. Given that $R$ acts in a d... show full transcript
Worked Solution & Example Answer:Two forces, $(4i - 5j) \text{ N}$ and $(pi + qj) \text{ N}$, act on a particle $P$ of mass $m$ kg - Edexcel - A-Level Maths: Mechanics - Question 6 - 2009 - Paper 1
Step 1
find the angle between R and the vector j
Answer
To find the angle between the resultant vector and the vector , we first need to express in terms of its components:
Given the forces: .
To find the angle between and the -direction, we use the definition of the tangent function:
tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{q - 5}{4 + p}\
Using the relationship yields:
So,
Step 2
Step 3
find the value of m
Answer
Given , substituting into the equation:
\Rightarrow -2p = 4\ \Rightarrow p = -2.$$ Next, we can find $R$: $$R = (4 - 2i) + (-4j) = 2i - 4j,$$ Calculating the magnitude of $R$: $$|R| = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}.$$ Given the acceleration $a = 8\sqrt{5} \text{ m s}^{-2}$, we apply Newton's second law: $$F = ma\ |R| = m \cdot a\ \Rightarrow 2\sqrt{5} = m \cdot 8\sqrt{5} \implies m = \frac{2}{8} = \frac{1}{4}.$$