Friday, December 12, 2014

Physics 9702 Doubts | Help Page 36

  • Physics 9702 Doubts | Help Page 36

Question 221: [Kinematics > Linear motion]
On a particular railway, train driver applies brake of the train at yellow signal, a distance of 1.0 km from a red signal, where the train stops.
Maximum deceleration of train is 0.20 m s–2.
Assuming uniform deceleration, what is maximum safe speed of train at the yellow signal?
For November 2009:
A 20 m s–1                   B 40 m s–1                   C 200 m s–1                 D 400 m s–1
For November 2013:
A 14 m s–1                   B 20 m s–1                   C 40 m s–1                   D 400 m s–1

Reference: Past Exam Paper – November 2009 Paper 11 Q5 & November 2013 Paper 13 Q8



Solution 221:
For Nov 2009 Paper 11 Q5 – Answer: A.
For Nov 2013 Paper 13 Q8 – Answer: B.
Consider the equation for uniformly accelerated motion: v2 = u2 + 2as
Final velocity, v = 0
Acceleration, a = -0.2ms-2
Distance, s = 1000
Initial velocity, u = ???

So, 0 = u2 + 2(-0.2)(1000) giving u = 20ms-1









Question 222: [Forces > Electric field]
Tiny oil droplet with mass 6.9 × 10–13kg is at rest in electric field of electric field strength 2.1 × 107N C–1, as shown.

Weight of the droplet is exactly balanced by the electrical force on droplet.
What is the charge on droplet?
A 3.3 × 10–20C
B –3.3 × 10–20C
C 3.2 × 10–19C
D –3.2 × 10–19C

Reference: Past Exam Paper – June 2014 Paper 12 Q12



Solution 222:
Answer: D.
The weight of the droplet acts downwards. For the droplet to be at rest, the electric force on the droplet should act upwards. Electric field is from positive to negative. So, the droplet is attracted towards the positive upper plate. The drop should be negatively-charged so that it moves upwards, against the force of gravity.

The electric force F should be equal in magnitude to the weight of the droplet (F = mg).
Electric field strength, E = F / Q
Charge, Q = F / E = mg / E = (6.9x10-13 x 9.81) / 2.1x107 = 3.2x10-19C










Question 223: [Forces > Moment]
Uniform beam is pivoted at P as shown. Weights of 10 N and 20 N are attached to its ends.
Length of beam is marked at 0.1 m intervals. Weight of the beam is 100 N.
At which point should further weight of 20 N be attached to achieve equilibrium?


Reference: Past Exam Paper – November 2005 Paper 1 Q12 & June 2014 Paper 13 Q15



Solution 223:
Answer: D.
Since the beam is uniform, its weight acts at the centre, that is, 0.1m to the left of P (where B is).

For equilibrium, the sum of clockwise moment should be equal to the sum of anticlockwise moment.
Clockwise moment = 20(0.4) Nm
Anticlockwise moment = 10(0.6) + 100(0.1) Nm

Since the anticlockwise moment is greater, the further weight of 20N should contribute to the sum of clockwise moments.  That is, it should be to the right of P. Let the position of the further weight of 20N be x to the right of P.

Sum of clockwise moment = Sum of anticlockwise moment
20(0.4) + 20(x) = 100(0.1) + 10(0.6)
Position x = 0.4m










Question 224: [Waves > Phase difference]
Frequency of a certain wave is 500 Hz and its speed is 340 m s–1.
What is the phase difference between motions of two points on the wave 0.17 m apart?
A π / 4 rad                   B π / 2 rad                   C 3π / 4 rad                 D π rad

Reference: Past Exam Paper – June 2006 Paper 1 Q25



Solution 224:
Answer: B.
Speed, v = fλ
Wavelength, λ = v / f = 340 / 500 = 0.68m
The wavelength λ corresponds to a phase difference of 2π rad (= 360 deg) – that is, 2 points separated by a distance equal to λ will have a phase difference of 2π rad – they are in phase.

Therefore,
(Separation of λ =) 0.68m corresponds to phase difference of 2π rad
Separation of 0.17m will correspond to (0.17 / 0.68) x 2π = π / 2 rad










Question 225: [Measurements > CRO]
Y-input terminals of cathode-ray oscilloscope (c.r.o.) are connected to supply of peak value 5.0 V and of frequency 50 Hz. Time-base is set at 10 ms per division and Y-gain at 5.0 V per division.
Which trace is obtained?

Reference: Past Exam Paper – November 2006 Paper 1 Q4



Solution 225:
Answer: D.
Y-gain is set at 5.0V per division. Since the peak value of the supply is 5.0V, the amplitude should occupy only 1 division. [A and B are incorrect]

Time-base setting is 10ms per division.
Frequency = 1 / Period
Period = 1 / 50 = 0.02s = 20ms
So, the period should occupy 2 divisions. [C is incorrect]










Question 226: [Dynamics > Conservation of Linear momentum]
A stationary thoron nucleus (A = 220, Z = 90) emits an alpha particle with kinetic energy Eα. Which is the kinetic energy of the recoiling nucleus?
A Eα / 108       B Eα / 110        C Eα / 54          D Eα / 55         E Eα

{Note that the relative atomic mass of thoron, A is 220, not 200 as given in some books for this question}

Reference: Past Exam Paper – J84 / II / 37



Solution 226:
Answer: C.
A is the relative atomic mass and Z is the proton number.

Since the thoron nucleus is initially stationary (its speed is zero), it has zero momentum. From the conservation of momentum, the sum of momentum after the emission of the alpha particle should also be zero.

An alpha particle has A = 4 and Z = 2.
Mass of alpha particle = 4u where u is the unified atomic mass.
Let the velocity of the alpha particle = v

Momentum = mass x velocity
Kinetic energy = ½ mv2

Momentum of alpha particle = 4u(v) = 4uv
Kinetic energy of alpha particle = 0.5 (4u) v2 = 2uv2 = Eα

The recoiling nucleus has A = 220 – 4 = 216 and Z = 90 – 2 = 88
Mass of recoiling nucleus = 216u
Let the velocity of the recoiling nucleus = y
Momentum of recoiling nucleus = 216u (y) = 216uy

From conservation of momentum,
Momentum of recoiling nucleus + Momentum of alpha particle = Momentum of thoron nucleus
216uy + 4uv = 0
Velocity, y = - 4v / 216 = - v/54
(The negative velocity means that the recoiling nucleus moves in a direction opposite to the particle particle)

Kinetic energy of recoiling nucleus = 0.5 (216u) (- v/54)2 = uv2 / 27 = 2uv2 / 54
Since Eα = 2uv2,
Kinetic energy of recoiling nucleus = Eα / 54








Question 227: [Measurements > Estimates]
Which estimate is realistic?
A Kinetic energy of a bus travelling on an expressway is 30 000 J.
B Power of a domestic light is 300 W.
C Temperature of a hot oven is 300 K.
D Volume of air in a car tyre is 0.03 m3.

Reference: Past Exam Paper – November 2013 Paper 13 Q1



Solution 227:
Answer: D.
The kinetic energy of the bus might well be about 106 J (KE = ½ mv2. Consider the mass of the bus and the square of its velocity on an expressway). The powers of domestic lights are usually around 60 W (often less) and an oven at 300 K is not hot (this temperature is 300 – 273 = 27 deg).

The correct answer, D, indicates just how large a cubic metre is. By making a rough estimate of the size of a tyre and the volume could be determined from it.




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