The diagram shows a motorised vehicle for carrying one person. The vehicle has two wheels on one axle. The passenger stands on a platform between the wheels.

Question 7

The diagram shows a motorised vehicle for carrying one person.

The vehicle has two wheels on one axle. The passenger stands on a platform between the wheels.

The weight of the machine is 600 N. Its centre of mass is 200 mm in front of the axle. The wheel radius is 400 mm.

When stationary, a passenger of weight 600 N stands with his centre of mass 200 mm behind the axle to balance the machine.

The motor is now switched on to provide a horizontal force of 90 N at the ground to move the vehicle forwards.

How far and in which direction must the passenger move his centre of mass to maintain balance?

A 60 mm backwards

B 60 mm forwards

C 140 mm backwards

D 140 mm forwards

Reference: Past Exam Paper – November 2017 Paper 11 Q14

Solution:

Answer: B.

To maintain balance,

Anti-clockwise moment = Clockwise moment

The weight of the machine exerts a clockwise moment (= 600×200).

The weight of the person exerts an anti-clockwise moment (= 600×200).

To move forward, the machine exerts a force of 90N in the direction of motion. This provides an anti-clockwise moment (= 90×400).

When in motion,

(600×200) + (90×400) = (600×200)

When the motor is on, the anti-clockwise moment is greater than the clockwise moment.

To maintain balance,

Anti-clockwise moment = Clockwise moment

So, to maintain balance, the anti-clockwise moment should be reduced so that it equals the clockwise moment.

The moment exerted by the weight of the person can be changed by moving his centre of mass. To reduce the anti-clockwise moment, the person should bring his centre of mass towards the pivot (axle). That is, he needs to move forwards. This reduced the distance of his weight from the axle.

Let the distance moved forwards to maintain balance be x.

[600 × (200 – x)] + (90×400) = (600×200)

200 – x = [(600×200) – (90×400)] / 600 = 140

x = 200 – 240 = 60 mm

Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 102 V is applied between the plates. Fig. 5.1 shows a side view of this arrangement.

Question 3

(a) Define the coulomb. [1]

(b) Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 10² V is applied between the plates. Fig. 5.1 shows a side view of this arrangement.

Fig. 5.1

A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10^-15 N and charge –8.0 × 10^-19 C.

(i) Show that the electric force acting on the particle is 4.0 × 10^-15 N. [2]

(ii) On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle. [1]

(iii) The resultant force acting on the particle is F.

Determine

1. the magnitude of F,

2. the angle of F to the horizontal. [3]

(c) The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal

displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T.

On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate.

Fig. 5.2

[2]

[Total: 9]

Reference: Past Exam Paper – November 2017 Paper 22 Q5

Solution:

(a) 1 coulomb is the amount of charges that passes when a current of 1 ampere flows for 1 second.

(b)

(i)

E = V / d          or E = F / Q

F = VQ / d

F = (2.0 × 10² × 8.0 × 10^-19) / 4.0 × 10^-2 = 4.0 × 10^-15 N

(ii)

arrow pointing to the left labelled ‘electric force’ and

arrow pointing downwards labelled ‘weight’   

(iii)

1.

{The resultant force F of the 2 forces mentioned above can be obtained from Pythagoras’ theorem.

F² = (weight)² + (electric force)²}

resultant force = √ [(3.9 × 10^-15)² + (4.0 × 10^-15)²]     

= 5.6 × 10^-15 N                                   

2. angle = tan^-1 (3.9 × 10^-15 / 4.0 × 10^-15) = 44°         

(c)

downward sloping line from (0, 2.0)

magnitude of gradient of line increases with time and line ends at (T, 0)

{The horizontal displacement is affected by the electric force. The particle accelerates towards the left. So, the velocity increases. From the s-t graph, the velocity is obtained by the gradient. Thus, the gradient increases.}