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

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  

• Physics 5054 Doubts | Help Page 2

Question 2: [Turning effect of forces]

The diagram shows a plank of mass 5.0 kg and length 3.00 m resting horizontally on two trestles, P and Q, which are a distance of 2.50 m apart. When a student of mass 60.0 kg walks along the plank from one trestle to the other, the plank sags.

(a) Explain why the sag increases as the student walks towards the middle of the plank.

(b) Calculate the downward force exerted on each trestle when the student is at a distance of

(i) 0.50 m from trestle P,

(ii) 1.25 m from trestle P.

(g = 10.0 N/kg)

Reference: Past Exam Paper – N91 / P2 / QA2

Solution 2:

(a)

Assuming the plank is uniform, the weight of the plank acts at the midpoint of the plank causing the sag. As the bog walks from P towards the middle of the plank, the combined centre of gravity of the system of plank and boy acts in between the boy and the midpoint of the plank causing a larger sag. As the boy moves even nearer to the midpoint, the sag increases as the combined c.g. moves towards the centre of the plank.

(b)

(i) When x = 0.50 m from P,

By the Principle of Moments, moment about P

R_Q ×2.50 = (600×0.50) + (50×1.25)

R_Q = 145 N

For vertical equilibrium,

R_P + R_Q = 600 + 50 = 650 N

R_P = 650 – R_Q = 650 – 145 = 505 N

(ii) When x = 1.25 m from P, both the forces 600 N and 50 N act at the centre of mass of the plank.

By symmetry, R_P = R_Q            where  R_P + R_Q = 600 + 50 = 650 N

R_P = R_Q = 650/2 = 325 N

• A-Level Physics Notes Syllabus 2015

Available here is a set of A-Level notes for Physics (9702) syllabus. It contains both AS and A2 chapters.

Even students of the later syllabus may use these notes as most of them is still relevant to the new syllabus.


Link: A-Level Physics Notes Syllabus 2015
Credit: Neha of India


The chapters included are:

1. Units and Measurements
2. Deformation of Solids
3. Electric fields
4. Electric current, potential difference and resistance
5. Resistance and resistivity
6. Practical circuits
7. Waves
8. Superposition of waves
9. Stationary waves
10. Forces, vectors and moments
11. Momentum
12. Waves, energy and power
13. Projectiles
14. Radioactivity
15. Accelerated motion
17. Circuit motion
18. Gravitational fields
19. Oscillations
20. Communication systems
21 Thermal physics
22. Ideal gases
23. Coulomb's law
24. Capacitance
25. Electronics
26. Magnetic fields
27. Charged particles
28. Electromagnetic induction
29. Alternating current
30. Quantum physics
31. Nuclear physics
32. Medical imaging


The notes were written by Neha of India and were made available here with permission. You are also welcomed to help the community by contributing such materials for any subject.

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