A positively charged oil droplet falls in air in a uniform electric field that is vertically upwards. The droplet has a constant terminal speed v0 and the electric field strength is E.

Question 41

A positively charged oil droplet falls in air in a uniform electric field that is vertically upwards. The droplet has a constant terminal speed v0 and the electric field strength is E.

The magnitude of the force due to air resistance acting on the droplet is proportional to the speed of the droplet.

Which graph shows the variation with E of v0?

Reference: Past Exam Paper – November 2019 Paper 12 Q10

Solution:

Answer: A.

Since the oil droplet is falling with a constant speed, it has no acceleration.

The resultant force on the oil droplet is zero.

Sum of downward forces = Sum of upward forces

The weight of the droplet acts downwards.

Weight = mg

Since the droplet is falling (moving downwards), the air resistance is upwards. This force is proportional to the speed of the droplet.

F_R = kv₀                       where k is a constant

The electric field is vertically upwards. Since the direction of an electric field gives the direction of the force on a positive charge, the force on the positively-charged oil droplet is upwards.

Electric field strength E = F / q

Electric force F = Eq

Sum of downward forces = Sum of upward forces

Weight = Electric force + Force due to air resistance

mg = Eq + kv₀

A graph of speed against field strength is plotted. So, v₀ the subject of formula.

Eq + kv₀ = mg

v₀ = mg/k – Eq/k

Comparing with the equation of a straight line (y = mx + c) with v₀ on the y-axis and E on the x-axis,

Gradient = - q/k

Thus, the graph is a straight line with negative constant gradient.

The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube.

Question 16

The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube.

The tube is under tension, caused by a force T, parallel to the axis of the tube. To reduce the stress in the material of the tube, it is proposed to thicken the wall.

The tube diameter and the tension being constant, which wall thickness gives half the stress?

A w / 2                        B √2 w                        C 2w                           D 4w

Reference: Past Exam Paper – June 2015 Paper 12 Q23

Solution:

Answer: C.

Stress = Force / Area

The steel tube is hollow in the centre with its wall having thickness w. When we calculate stress, we need to consider the area of the steel ‘material’, not the hollow space.

Suppose the length of the tube is L and the diameter is d. Imagine that we have cut the steel tube (along the length) so that it is now in the form of a cuboid (a rectangular sheet with thickness w).

Area of annulus = (circumference of tube) × w

Area of annulus = 2π(d/2) × w = πdw

Area = πdw

Since π and d are constant, the area of the annulus is proportion to w.

Because the wall is thin in comparison with the diameter of the tube, the area of the annulus (representing the wall thickness) is proportional to w.

Stress = Force / Area

The tension (force) is kept constant.

So, the stress is inversely proportional to the area.

To obtain half the stress, the area should be doubled. But we have seen above that the area is directly proportional to w.

Thus, if the area is doubled, w also needs to be twice greater.

The maximum useful output power of a car travelling on a horizontal road is P. The total resistive force acting on the car is kv2, where v is the speed of the car and k is a constant.

Question 42

The maximum useful output power of a car travelling on a horizontal road is P. The total resistive force acting on the car is kv², where v is the speed of the car and k is a constant.

 

Which equation is correct when the car is travelling at maximum speed?

A v³ = P / k                 B v² = P / k                 C v = (P/k)²                 D v = (P/k)³

 

Reference: Past Exam Paper – November 2019 Paper 13 Q17

 

Solution:

Answer: A.

Power = P

Resistive force F = kv²

For a body moving with constant speed,

Power P = Force × velocity

P = F × v

P = (kv²) × v

P = kv³

Make v the subject of formula,

v³ = P / k