Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg. The resultant of these two forces is parallel to the vector $(2i + j)$. (a) Find the value of $q$. At time $t = 0$, $P$ is moving with velocity $(-2i + 4j) \text{ m s}^{-1}$. (b) Find the speed of $P$ at time $t = 2$ seconds.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: Two forces $(4i

Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2014 - Paper 1

Question 2

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Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg. The resultant of these two forces is parallel to the vector $(2i +... show full transcript

Worked Solution & Example Answer:Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2014 - Paper 1

Step 1

Find the value of q.

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Answer

To determine the value of $q$ , we can find the resultant force of the two given forces. The forces acting on the particle are:

$F_1 = (4i - 2j) \text{ N}$
$F_2 = (2i + qj) \text{ N}$

The resultant force $F_R$ can be calculated as follows: $F_R = F_1 + F_2 = (4i - 2j) + (2i + qj) = (6i + (q - 2)j) \text{ N}$

For the resultant to be parallel to the vector $(2i + j)$ , we can set up the ratio: $\frac{6}{2} = \frac{q - 2}{1}$

This simplifies to: $6 = 2(q - 2)$ Expanding this gives: $6 = 2q - 4$ So, $2q = 10 \\ q = 5$

Step 2

Find the speed of P at time t = 2 seconds.

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Answer

To find the speed of particle $P$ at time $t = 2$ seconds, we first need to determine the acceleration, which we'll find from the resultant force calculated earlier.

The resultant force $F_R$ is: $F_R = (6i + (5 - 2)j) = (6i + 3j) \text{ N}$

Using Newton's second law, we find the acceleration $a$ : $F = ma \implies a = \frac{F_R}{m} = \frac{(6i + 3j)}{1.5} = (4i + 2j) \text{ m s}^{-2}$

The initial velocity $u$ of the particle is: $u = -2i + 4j \text{ m s}^{-1}$

Using the equation of motion: $v = u + at$

Substituting the values for $t = 2$ seconds: $v = (-2i + 4j) + 2(4i + 2j)$ This gives: $v = (-2i + 4j) + (8i + 4j) = (6i + 8j) \text{ m s}^{-1}$

To find the speed, we calculate the magnitude of the velocity vector: $\text{speed} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m s}^{-1}$

Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2014 - Paper 1

Worked Solution & Example Answer:Two forces $(4i - 2j) \text{ N}$ and $(2i + qj) \text{ N}$ act on a particle $P$ of mass $1.5$ kg - Edexcel - A-Level Maths Mechanics - Question 2 - 2014 - Paper 1

Find the value of q.

Find the speed of P at time t = 2 seconds.

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