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

First, we need to find the acceleration of particle $P$. The resultant force is:

$F = (6i + 3j) \, \text{N}$ (using $q = 5$).

Using Newton's second law:

$F = ma,$

where $m = 1.5 \, \text{kg}$, we have:

$6i + 3j = 1.5a.$

Thus, the acceleration $a$ is:

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

Next, we apply the equation of motion to find the velocity at $t = 2$ seconds:

$v = u + at,$

where:

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

Substituting the values:

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

Finally, we calculate the speed $|v|$:

$|v| = \sqrt{(6^2 + 8^2)} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m s}^{-1}.$