Wednesday, December 31, 2014

Physics 9702 Doubts | Help Page 42

  • Physics 9702 Doubts | Help Page 42


Question 255: [Physical Quantities & Units > Units]
An alternative form of the unit of resistance, the ohm (Ω), is VA-1.
Which of the following examples shows a similar correct alternative form of unit?
            Unit                             Alternative form       
A         coulomb (C)                As-1
B         farad (F)                      VC-1
C         pascal (Pa)                   Nm-2
D         volt (V)                       JC
E          watt (W)                      Js

Reference: Past Exam Paper – J91 / I / 1



Solution 255:
Answer: C.
Coulomb is the unit of charge. Charge = Current (I) x Time (t). Units of charge = As

Farad is the unit of capacitance. Capacitance = Charge (Q) / Voltage (V). Units of capacitance = CV-1

Pascal is the unit of pressure. Pressure = Force (F) / Area (A). Units of pressure = Nm-2 









Question 256: [Dynamics > Laws of motion]
When an object moves up in the sky, the resultant force acting on the object is upwards right?



Solution 256:
This depends on how the object is moving. Consider the following cases [for all the cases, the velocity being considered is that after the different forces acting on the object have already been considered (e.g. force due to engine, air resistance, gravity, … )]:

Case 1:
The object is moving always upwards at constant velocity. The acceleration is thus zero. Since Force F = ma, the resultant force on the object is zero.

Case 2:
As the object moves upwards, its velocity is increasing. Therefore, there is an upward acceleration. The resultant force is upwards.

Case 3:
As the object moves upwards, its velocity is decreasing. Therefore, there is a downward acceleration, causing the resultant force to be downward.








Question 257: [Energy > Conservation of energy]
When bungee jumping, student starts with maximum gravitational potential energy (position 1), then falls freely until the rope fully unwinds (position 2), after which the rope starts to stretch until the lowest point of the jump is reached (position 3).

What are kinetic and elastic potential energies at position 3?


Reference: Past Exam Paper – November 2007 Paper 1 Q16



Solution 257:
Answer: C.
From the conservation of energy, the sum of energies at position 2 or 3 should be equal to the initial amount of gravitational potential energy the student has at position 1.

Stretching of the rope causes it to have elastic potential energy. Kinetic energy is associated with motion (speed).
At position 2, the rope fully unwinds but it does not stretch. So, it does not have elastic potential energy. The kinetic energy is maximum at position 2.

At position 3, the rope is stretched. So, it has elastic potential energy. Also, at position 3, the lowest point of the jump is reached. So, the student will not move further below (no further motion below). Thus, the kinetic energy is minimum. (As stated before, the total energy should be equal to the initial gravitational potential energy, so the kinetic energy cannot be maximum too. Energy would not be conserved.)









Question 258: [Measurements > Graphs]
Calibration graph is produced for faulty ammeter.

Which ammeter reading will be nearest to correct value?
A 0.2 A                       B 0.4 A                       C 0.6 A                       D 0.8 A

Reference: Past Exam Paper – November 2008 Paper 1 Q5



Solution 258:
Answer: D.
The calibration graph is a graph of ‘ammeter reading / A’ on the y-axis against ‘true current / A’ on the x-axis.
For the ammeter reading to be nearest to the correct value, the y-coordinate of a point (its ammeter reading) should be closest to the x-coordinate of the same point (the true current).
Of the 4 choices, only the point at 0.8A has the x- and y-coordinates to be almost the same.










Question 259: [Measurements]
A metre rule is used to measure length of a piece of wire. It is found to be 70 cm long to nearest millimetre.
How should this result be recorded in table of results?
A 0.7 m                       B 0.70 m                     C 0.700 m                   D 0.7000 m

Reference: Past Exam Paper – November 2010 Paper 12 Q4



Solution 259:
Answer: C.
1 mm = 0.1 cm = 0.001m
Measurement = 70.0cm = 0.700m
(Since the metre rule can read to the nearest millimeter, here 3 significant figures are necessary)










Question 260: [Kinematics > Linear motion]
A ball was dropped from the window of an apartment located at the tenth storey of a high rise building. 1 second after release, the ball was observed to have fallen by exactly 2 storeys. At which storey will the ball be 2 seconds after it was released?

Reference: ???



Solution 260:
Let the distance between 2 consecutive storeys be x. (Assume this is constant for every storey)

There is an acceleration due to gravity downwards. Let acceleration due to gravity, a = 9.81ms-2. Assume that there gravity is the only force acting on the ball (resistive forces are negligible).

Distance travelled from window, s = ut + ½ at2
Initial velocity, u = 0. 

When time t = 1s, distance s = 2x.
2x = 0 + 0.5 (9.81) (1)2 = 4.905m

Distance x (= 4.905 / 2 = 2.4525m) corresponds to the distance between 2 consecutive storeys.

When time t = 2s,
Distance s = 0 + 0.5 (9.81) (2)2 = 19.62m

Number of storeys = 19.62 / 2.4525 = 8








Question 261: [Waves > Damping > Graphs]
Refer to November 2007 paper 4 Q3(b)(ii)2 at http://physics-ref.blogspot.com/2014/09/9702-november-2007-paper-4-worked.html

Reference: Past Exam Paper – November 2007 Paper 4 Q3 (b) (ii) 2



Solution 261:
EITHER
The total energy is now 1.0mJ, but regardless of the total energy, the graph would have the same basic shape. The amplitude is read at the value of x (for the new graph) where the kinetic energy is zero – that is energy is potential.
[I recommend you draw a rough diagram on the graph as you are reading the explanation below to better understand.]
So, just imagine a graph of the same shape, with the total energy = 1.00mJ (that is, at x = 0, kinetic energy is 1.0mJ). This curve would have a lower value of x when the kinetic energy is zero. 

Now, imagine the point where the kinetic energy is zero on the new curve. Draw a dotted vertical line from that point until it reaches another point on the curve ALREADY given. At what value of kinetic energy would this correspond on the previous curve? It would correspond to a value of (2.56 – 1.00)mJ on the previous curve because the new curve has the same shape but has its peak value of kinetic energy at 1.00mJ.

So, read the value of x at (2.56 – 1.00)mJ on the previous curve. This would give the amplitude for a total energy of 1.00mJ. The value of x is about 0.5cm.

OR
Maximum kinetic energy = ½ mω2a2 
At maximum amplitude a, the energy is entirely potential.
From the formula, it can be seen that the energy is directly proportional to (amplitude, a)2. Refer to the case given (before damping) as case 1 and refer to the case with reduced energy as case 2.

For case 1: amplitude, a12 is proportional to energy E1. Amplitude a1 = 0.8cm and energy E1 = 2.56mJ.
For case 2: amplitude, a22 is proportional to energy E2. Amplitude a2 = ??? and energy E2 = 1.00mJ.

Now, consider the ratio of case 2 to case 1,
(a2 / a1)2 = E2 / E1
Amplitude a2 = [√( E2 / E1)] x a1 = [√(1.00 / 2.56)] x 0.8 = 0.5cm





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