(a) (i) Define uniform acceleration.
(ii) Write an equation that relates linear velocity, angular velocity and radius
of path in circular motion.
(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.
(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.
(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:
(i) co-efficient of kinetic friction;
(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\(^{-2}\)]
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]
Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]
By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]
Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]
By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
