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  

Saturday, November 19, 2016

Physics 5054 Doubts | Help Page 2

  • 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
RQ ×2.50 = (600×0.50) + (50×1.25)
RQ = 145 N

For vertical equilibrium,
RP + RQ = 600 + 50 = 650 N
RP = 650 – RQ = 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, RP = RQ            where  RP + RQ = 600 + 50 = 650 N
RP = RQ = 650/2 = 325 N





Tuesday, November 15, 2016

A-Level Physics Notes Syllabus 2015

  • 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.

You may also try to complete this collection by providing notes for the new topics added in the new syllabus.
Share your ideas on how the notes may be improved.

Hope this helps you in your studies.




Thursday, November 3, 2016

Physics 9702 Doubts | Help Page 244

  • Physics 9702 Doubts | Help Page 244



Question 1127: [Deformation > Hooke’s law]
For a wire, Hooke’s law is obeyed for a tension F and extension x. The Young modulus for the material of the wire is E.
Which expression represents the elastic strain energy stored in the wire?
A ½ Ex                        B Ex                C ½ Fx                                    D Fx

Reference: Past Exam Paper – November 2008 Paper 1 Q23



Solution 1127:
Answer: C.
As Hooke’s law is obeyed, a graph of tension against extension would be a straight line passing through the origin.

The elastic strain energy is given by the area under the graph. This would correspond to that of a triangle.
Elastic strain energy = ½ Fx











Question 1128: [Work, Energy and Power]
Which statement is correct?
A A ball lands on the ground and bounces. The kinetic energy changes sign, because the ball changes direction.
B A car drives up a slope at a steady speed. The power generated by the engine equals the potential energy gained per unit time.
C An electric heater can be 100% efficient.
D It is impossible for momentum to be conserved in a collision.

Reference: Past Exam Paper – June 2015 Paper 11 Q17



Solution 1128:
Answer: C.
Choice A: Kinetic energy is a scalar and does not have a direction.

Choice B: The power from the engine does not just increase the potential energy of the car. It is also providing for the kinetic energy – even if it is constant, it is not created in its own. The KE comes from the fuel. Additionally, there would be a frictional drag on the car and energy is required to overcome this. This energy comes from the engine.

Choice C: This is correct. Usually, energy is dissipated as heat in systems but for an electric heater this heat is useful. So, all the output power is useful in this situation.

Choice D. This is obviously wrong.








Question 1129: [Kinematics > Terminal velocity]
A stone of mass m is dropped from a tall building. There is significant air resistance. The acceleration of free fall is g.
When the stone reaches its terminal velocity, which information is correct?


Reference: Past Exam Paper – November 2011 Paper 12 Q7



Solution 1129:
Answer: B.
At terminal velocity, the speed is constant. So, there is no acceleration and the resultant force on the stone is zero.

The force of gravity always acts on the stone and is equal to mg.
For the resultant force to be zero, the air resistance should be of the same magnitude (= mg).

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