Q4. Three forces act on a particle in equilibrium. Two of them are 5 N east and 12 N north. Find the third.
The three must close into a triangle, so the third is −(5 N east + 12 N north): magnitude √(25 + 144) = 13 N, pointing south-west (about 67° south of west, since tan α = 12/5).
Q5. Two vectors of equal magnitude A add to give a resultant also of magnitude A. What is the angle between them?
The two vectors and the resultant form a triangle whose sides are all A, so it is equilateral and each inside angle is 60°. That 60° is between the first vector and the second one drawn from its head. Drawn tail to tail, the vectors are 180° − 60° = 120° apart.
Q6. A particle moves on a circle at constant speed 4 m s. Find the change in its velocity over half a revolution.
After half a turn the velocity is reversed, so Δv = vfinal − vinitial = (−v) − v = −2v, of magnitude 8 m s, opposite to the initial velocity.