Saturday, December 10, 2016

The diagram shows an oil droplet that has become charged by gaining five electrons.





Question 1
The diagram shows an oil droplet that has become charged by gaining five electrons. The droplet remains stationary between charged plates.



What is the magnitude and direction of the electrostatic force on the oil droplet?
A 5.0 × 10–15 N upwards
B 5.0 × 10–15 N downwards
C 5.0 × 10–13 N upwards
D 5.0 × 10–13 N downwards





Reference: Past Exam Paper – June 2008 Paper 1 Q31





Solution 1:
Answer: C.

Consider the forces acting on the charged oil droplet.

The weight of the oil droplets causes a downward force.
Being negatively charged (due to the electrons), the oil droplet is under an upward electrostatic force attracting it to the positive plate.


Electrostatic force = EQ        
where E is the electric field strength and Q is the total charge of the droplet


Electric field strength = V / d = 5000 / (0.8×10-2)
Total charge = 5 × charge of an electric = 5 × (1.6×10-19)


Electrostatic force = EQ = (V/d) × Q = 5.0 × 10-13 N

Tuesday, December 6, 2016

A small ball rests at point P on a curved track of radius r ...







Question 1
(a) State what is meant by simple harmonic motion. [2]

(b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 4.1.


Fig. 4.1
The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression
a = − gx / r
where g is the acceleration of free fall.

(i) Show that the ball undergoes simple harmonic motion. [2]

(ii) The radius r of curvature of the track is 28 cm.
Determine the time interval τ between the ball passing point P and then returning to point P. [3]

(c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 4.2.


Fig. 4.2
Some moisture now forms on the track, causing the ball to come to rest after approximately 15 oscillations.
On the axes of Fig. 4.2, sketch the variation with time t of the displacement x of the ball for the first two periods after the moisture has formed. Assume the moisture forms at time t = 0. [3]



Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q4

 

Solution 1:
(a) For a simple harmonic motion, the acceleration / force is proportional to the displacement (from a fixed point) and the
EITHER acceleration and displacement are in opposite directions                OR acceleration is always directed towards a fixed point

(b)
(i) Acceleration due to gravity, g and radius r are constant, so the acceleration a is proportional to the displacement x.
The negative sign shows that a and x are in opposite directions.

(ii)
{For simple harmonic motion, acceleration a = – ω2x
a = – ω2x = – gx / r giving}
ω2 = g / r                      and      ω = 2π / T
ω2 = 9.8 / 0.28 = 35                
T = {2π / ω2 =} 2π / √35 = 1.06 s
{The ‘T’ calculated above is the period. The question asks for only the time interval τ between the ball passing point P and then returning to point P. This is half the period T.}
time interval τ = {T/2 =} 0.53 s

(c)
sketch: time period constant (or increases very slightly)
drawn line always ‘inside’ given loops
successive decrease in peak height



Saturday, December 3, 2016

Car X is travelling at half the speed of car Y. Car X has twice the mass of car Y.






Question 1
Car X is travelling at half the speed of car Y. Car X has twice the mass of car Y.
Which statement is correct?
A Car X has half the kinetic energy of car Y.
B Car X has one quarter of the kinetic energy of car Y.
C Car X has twice the kinetic energy of car Y.
D The two cars have the same kinetic energy.



Reference: Past Exam Paper – June 2004 Paper 1 Q17




Solution 1:
Answer: A.

Let the speed of car X = v.                 Speed of car Y = 2v
Let the mass of car X = 2m.               Mass of car Y = m

Kinetic energy = ½ mv2
KE of car X = ½ (2m) v2 = mv2
KE of car Y = ½ (m) (2v)2 = 2 mv2  
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