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 2

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 2

Step 1

Find the value of q.

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Answer

To determine the value of $q$ , we first find the resultant force from the two given forces:

The first force is ${\mathbf{F_1}} = (4i - 2j) \text{ N}$ .

The second force is ${\mathbf{F_2}} = (2i + qj) \text{ N}$ .

The resultant force ${\mathbf{F}}$ is given by:

$\mathbf{F} = \mathbf{F_1} + \mathbf{F_2} = (4i - 2j) + (2i + qj) = (6i + (q-2)j) \text{ N}$

Since the resultant is parallel to the vector $(2i + j)$ , we can write the relationship:

$\frac{6}{2} = \frac{q-2}{1}$

This simplifies to:

$6 = 2(q - 2)$

Expanding this yields:

$6 = 2q - 4$

Rearranging gives:

$2q = 10$

Thus, we find:

$q = 5.$

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 start with the acceleration. The resultant force ${\mathbf{F}}$ we calculated earlier is:

$\mathbf{F} = (6i + (5-2)j) = (6i + 3j) \text{ N}.$

Using Newton's second law, $\mathbf{F} = ma$ , where $m = 1.5 \, \text{kg}$ , we find the acceleration:

$\mathbf{a} = \frac{\mathbf{F}}{m} = \frac{(6i + 3j)}{1.5} = (4i + 2j) \text{ m s}^{-2}.$

Next, we calculate the velocity after $2$ seconds:

The initial velocity $\mathbf{u} = (-2i + 4j) \text{ m s}^{-1}$ , and the acceleration is $\mathbf{a} = (4i + 2j) \text{ m s}^{-2}$ .

Using the equation of motion:

$\mathbf{v} = \mathbf{u} + \mathbf{a}t,$

Substituting in the values gives:

$\mathbf{v} = (-2i + 4j) + (4i + 2j)(2) = (-2i + 4j) + (8i + 4j) = (6i + 8j) \text{ m s}^{-1}.$

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 2

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 2

Find the value of q.

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

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