Two forces, $(4i - 5j)$ N and $(p + qj)$ N, act on a particle $P$ of mass $m$ kg. The resultant of the two forces is $R$. Given that $R$ acts in a direction which is parallel to the vector $(i - 2j)$. (a) find the angle between $R$ and the vector $j$, (b) show that $2p + q + 3 = 0$. Given also that $q = 1$ and that $P$ moves with an acceleration of magnitude $8rac{m}{s^2}$, (c) find the value of $m$.

A-Level Edexcel Maths Mechanics Working with Vectors: Two forces, $(4i

Two forces, $(4i - 5j)$ N and $(p + qj)$ N, act on a particle $P$ of mass $m$ kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2009 - Paper 1

Question 6

Two forces, $(4i - 5j)$ N and $(p + qj)$ N, act on a particle $P$ of mass $m$ kg. The resultant of the two forces is $R$. Given that $R$ acts in a direction which is... show full transcript

Worked Solution & Example Answer:Two forces, $(4i - 5j)$ N and $(p + qj)$ 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

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Answer

To find the angle $heta$ between the resultant force $R$ and the vector $j$ , we note the vector $R = (4 + p)i + (q - 5)j$ . The angle $heta$ can be found using the tangent function:

$\tan(\theta) = \frac{(q - 5)}{(4 + p)}$

Calculating for $\theta$ : Using the given conditions, we can find $\tan^{-1}\left(\frac{2}{4}\right) = 63.4^\circ$ . Therefore the angle is $\theta = 63.4^\circ$ . However, since R needs to be in the direction parallel to $(i - 2j)$ , the final angle will be adjusted to $153.4^\circ$ .

Step 2

show that 2p + q + 3 = 0

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Answer

Consider the forces:

$(4 + p)i + (q - 5)j$

Substituting $q = 1$ into the equation:

$(q - 5) = (1 - 5) = -4$

This gives: $2p + 1 + 3 = 0$ Thus: $2p + 3 = -1$ Rearranging, we find: $2p + q + 3 = 0.$

Step 3

Find the value of m

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Answer

Given $q = 1$ , we substitute $p$ from the equation derived previously:

If $q = 1$ , then substituting back gives $p = -2$ .

Thus, the resultant force:

$R = (4 + (-2))i + (1 - 5)j = 2i - 4j$

The magnitude of $R$ is:

$|R| = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$

Using Newton's second law, since $P$ moves with an acceleration of magnitude $8 m/s^2$ : $|R| = m imes a$ Thus: $\sqrt{20} = m \cdot 8$ From this: $m = \frac{\sqrt{20}}{8} = \frac{1}{4}.$

Two forces, $(4i - 5j)$ N and $(p + qj)$ N, act on a particle $P$ of mass $m$ kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2009 - Paper 1

Worked Solution & Example Answer:Two forces, $(4i - 5j)$ N and $(p + qj)$ N, act on a particle $P$ of mass $m$ kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2009 - Paper 1

Find the angle between R and the vector j

show that 2p + q + 3 = 0

Find the value of m

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