• Physics 9702 Doubts | Help Page 68

Question 374: [Electric field]
A metal sphere of radius 0.1 m was insulated from its surroundings and given a large positive charge. A small charge was brought from a distance point to a point 0.5 m from the sphere’s centre. The work done against the electric field was W and the force on the small charge in its final position is F. If the small charge had been moved to only 1 m from the centre of the sphere, what would have been the values for the work done and the force?
            Work done                  Force
A         W/4                             F/4
B         W/4                             F/2
C         W/2                             F/4
D         W/2                             F/2
E          W/2                             F/√2

Reference: Past Exam Paper – J82 / II / 21



Solution 374:
Answer: C.
Work done = Force x distance
Since the distance through which the small charge is moved against the electric field is halved, the work done is also halved.

The electric force acting on the small charge is due to the electric field produced by the charged metal sphere.

The electric force is inversely proportional to the square of the separation of the charges. Since it is assumed that in both cases, the small charge is initially at the same position, the electric force at that initial position is the same. That is the force used in the equation for work done above (this is only a simple explanation but the force is actually also changing with distance, so a better equation would involve integration. But the final answer would still be the same).

But the force F mentioned in the question is the force at the final position. Since the separation between the charges is now twice than before, the force would be F/4 since the electric force depends on the square of the separation.










Question 375: [Waves > Interference]

(a) Explain the term interference.

(b) Ripple tank is used to demonstrate interference between water waves.

Describe

(i) apparatus used to produce two sources of coherent waves that have circular wavefronts,

(ii) how pattern of interfering waves may be observed.

(c) Wave pattern produced in (b) is shown in Fig.

Solid lines on Fig represent crests.

On Fig,

(i) draw two lines to show where maxima would be seen (label each of these lines with letter X),

(ii) draw one line to show where minima would be seen (label this line with letter N).

Reference: Past Exam Paper – June 2011 Paper 21 Q7



Solution 375:

(a) When waves overlap, the (resultant) displacement is the sum of the individual displacements

(b)

(i)

EITHER

2 (ball-type / spherical) dippers connected to the same vibrating source / motor

OR

One wave source described with 2 slits

(ii)

A lamp with a viewing screen on the opposite side of the tank are used. A means of freezing the picture e.g. strobe

(c)

(i) 2 correct lines are drawn labeled X

(ii) A correct line labeled N

Question 376: [Electric field]
Which one of the following statements about the electric potential at a point is correct?
A The potential is given by the rate of change of electric field strength with distance.
B The potential is defined as the work done per unit positive charge in moving one electron from the point to infinity.
C Alternative units for electrical potential are the joule and the volt.
D The potential due to a system of point of charges is given by the sum of the potentials due to the individual charges.
E Two points in an electric field are at the same potential when a unit positive charge placed anywhere on the line joining them must remain stationary.

Reference: Past Exam Paper – N85 / I / 16



Solution 376:
Answer: D.
The electric field strength is given by the rate of change of potential with distance. [A is incorrect]

The potential is defined as the work done per unit positive charge in moving one electron from infinity to that point. [B is incorrect]

Volt is the unit for electrical potential. Joule is not an alternative unit for electrical potential. [C is incorrect]

D is the only correct choice. The potential due to a system of point of charges is given by the sum of the potentials due to the individual charges. Electrical potential is a scalar and so, can be added directly.

E is only true of the unit positive charge is midway between the 2 points, not anywhere. [E is incorrect]











Question 377: [Dynamics > Linear motion]
(a) Define acceleration.                       [1]

(b) State the two factors that affect the acceleration of an object.                [1]

(c) Fig shows the variation of velocity v with time t for a small rocket.

The rocket is initially at rest and is fired vertically upwards from the ground. All the rocket fuel is burnt after a time of 5.0 s when the rocket has a vertical velocity of 200 ms^-1. Assume that air resistance has a negligible effect on the motion of the rocket.
(i) Without doing any calculations, describe the motion of the rocket
1. from t = 0 to t = 5.0s
2. from t = 5.0s to t = 25s                                           [3]
(ii) Calculate the maximum height reached by the rocket.                 [3]
(iii) Explain why the rocket has a speed greater than 200 ms^-1 as it hits the ground. [1]

Reference: OCR AS GCE PHYSICS A G481/01 Mechanics Friday 11 January 2013 – Question 4



Solution 377:
(a) Acceleration is defined as the rate of change of velocity (or acceleration = change in velocity / time)

(b) Mass and (net) force

(c)
(i)
1. The rocket undergoes (constant) acceleration.

2. The rocket undergoes constant deceleration / negative acceleration
Detail mark: constant used in either 1 or 2 OR reaches maximum height at 25s OR stops at 25s

(ii)
Maximum height = area under velocity-time graph from t = 0 to 25s
{Area of triangle = ½ x base x height}
Maximum height = ½ (25) (200) = 2500 m

(iii)
A sensible suggestion, for example:
EITHER
{The maximum height reached is calculated to be 2500m. At the maximum height, the velocity is zero (u = 0). Afterwards, the rocket falls under gravity. Here, we calculate the speed reached by the rocket at the ground, at a distance of 2500m (s = 2500m).}
Equation for uniformly accelerated motion:   v² = u² + 2as
Speed at ground, v = √[2 (9.81) (2500)] = 220 ms^-1


OR
{Here, we calculate the maximum height that would give a speed of 200ms^-1 at the ground. The rocket is falling under gravity, causing it to accelerate. The longer the distance travelled, the more the rocket would accelerate and hence, the greater the speed. So, if the maximum height that would give a speed of 200ms^-1 at the ground is less than the height calculated, then the speed at the ground would actually be greater than 200ms^-1. }
Equation for uniformly accelerated motion:   v² = u² + 2as
200² = 0 + 2 (9.81) s
Maximum height, s = 2040m


OR
The (burning) rocket fuel does work on the rocket {causing an additional acceleration together with the acceleration due to gravity}