Sunday, November 23, 2014

Physics 9702 Doubts | Help Page 21

  • Physics 9702 Doubts | Help Page 21



Question 109: [Energy > Kinetic energy]
Car driver adjusts the pressure on car’s brakes so that car travels at constant speed down a hill from P to Q.

Magnitude of change in the car’s kinetic energy is ΔEk. Magnitude of change in its gravitational potential energy is ΔEp.
Which statement is correct?
A ΔEk > ΔEp               B ΔEk = ΔEp                C ΔEp > ΔEk> 0                      D ΔEk = 0

Reference: Past Exam Paper – November 2002 Paper 1 Q17



Solution 109:
Answer: D.
The question clearly states that the pressure on the car’s brakes is adjusted so that the car travels at constant velocity.
Kinetic energy = ½mv2

If velocity is constant (not changing), the kinetic energy will also be constant. Therefore, the change in kinetic energy ΔEk is zero.  

Additionally, application of the car’s brakes uses the frictional force between the hill and the wheels (work is being done). So, the change in kinetic energy is not equal to the change in potential energy.









Question 110: [Kinematics > Linear motion]
Graph shows how the acceleration of an object moving in straight line varies with time.

Object starts from rest.
Which graph shows variation with time of velocity of the object over the same time interval?

Reference: Past Exam Paper – June 2014 Paper 11 Q8



Solution 110:
Answer: A.
Area under the (acceleration – time) graph represents the velocity. The gradient of the (acceleration – time) graph represents the rate of change [can be either an increase or a decrease] of acceleration.
As the choices deal with velocity – time graphs, we may need to consider the area under the acceleration – time graph, not its gradient {we actually won’t be considering any of these as the question is straight-forward}.

Both acceleration and velocity are vectors. Acceleration is the rate of change of velocity. For a positive acceleration (in a straight line motion as in this case), the velocity increases in the same direction.
Since the acceleration is always positive, it means that the velocity keeps on increasing. [B and C are wrong]

Also, at the end of the time interval being considered, the acceleration is zero. So, the gradient of the velocity – time graph {which gives the acceleration} should become zero at the end of the time interval being considered. [D is incorrect since the gradient is not zero at the end]









Question 111: [Waves > Stationary Wave]
Horizontal glass tube, closed at one end, has layer of dust laid inside it on its lower side. Sound is emitted from loudspeaker that is placed near the open end of tube.
Frequency of sound is varied and, at one frequency, a stationary wave is formed inside tube so that the dust forms small heaps.
Distance between four heaps of dust is 30 cm.

What is the frequency of sound emitted by loudspeaker?
A 1650 Hz                  B 2200 Hz                   C 3300 Hz                   D 6600 Hz

Reference: Past Exam Paper – November 2012 Paper 12 Q29



Solution 111:
Answer: A.
A heap formed from the dust corresponds to a node (a point with zero displacement).
A wavelength of the sound wave would form 3 heaps (a wavelength of the sound wave contains 3 nodes – so, the distance formed by 3 successive heaps would be equal to 1 wavelength).
Note that the heaps are only POINTS and so, the distance between 2 successive heaps actually gives a distance equal to half the wavelength. Thus, the distance formed between 3 successive heaps is equal to 1 wavelength.

The 30cm distance shown contains 4 heaps. Label these heaps from 1 to 4.
Distance between 1st and 3rd heap = 1 wavelength (1λ)
Distance between 3rd and 4th heap = 0.5 wavelength (0.5λ)
Therefore, 30cm (= 0.3m) corresponds to (1 + 0.5 =) 1.5λ.

Wavelength of sound wave, λ = 0.3 / 1.5 = 0.2m

Speed of wave, v = fλ
Frequency, f = v / λ = 330 / 0.2 = 1650Hz







Question 112: [Kinematics > Linear motion]
In order that train can stop safely, it will always pass signal showing a yellow light before it reaches a signal showing red light. Drivers apply brake at the yellow light and this results in uniform deceleration to stop exactly at the red light.
Distance between red and yellow lights is x.
What must be the minimum distance between lights if train speed is increased by 20 %, without changing the deceleration of the trains?
A 1.20 x                      B 1.25 x                      C 1.44 x                      D 1.56 x

Reference: Past Exam Paper – November 2010 Paper 11 Q9



Solution 112:
Answer: C.
For 1st case:
Let initial speed = u. Final speed = 0. Acceleration (deceleration) = a. Distance between the lights = x.
Consider the equation for uniformly accelerated motion: v2 = u2 + 2as
This becomes:  0 = u2 + 2ax

Since the deceleration is constant in both cases, make it the subject of formula.
Deceleration, a = - (u2 / 2x)

For 2nd case:
Initial speed = 1.2u {increased by 20%}. Final speed = 0. Acceleration (deceleration) = a = - (u2 / x). Let minimum distance = s.
Considering the same equation for uniformly accelerated motion: v2 = u2 + 2as
0 = (1.2u)2 + 2as
Minimum distance, s = - [1.44u2 / 2a] = [1.44u2 (2x) / 2u2] = 1.44x  









Question 113: [Current of Electricity > Variable resistor]
Diagram shows potential divider circuit which, by adjustment of contact X, can be used to provide a variable potential difference between the terminals P and Q.

What are the limits of this potential difference?
A 0 and 5 V                B 0 and 20 V              C 0 and 25 V              D 5 V and 25 V
                                                             
Reference: Past Exam Paper – November 2011 Paper 12 Q37



Solution 113:
Answer: B.
Let the resistance of the variable resistor be R1 and let the resistance of the fixed resistor be R2. R1 varies from 0V – 4kΩ and R2 = 1kΩ. The e.m.f. in the circuit is V = 25V.
Let the potential difference between the terminals P and Q be V2.

From the potential divider equation,
V2 = [R1 / (R1 + R2)] x V

The p.d. across a component is proportional to its resistance. The higher the resistance, the greater is the p.d.
For V2 to be maximum, the resistance of the variable resistor, R1 should be 4kΩ.
Maximum V­2 = [4 / (1+4)] x 25 = 20V           [A, C and D are incorrect]

For V2 to be minimum, the smallest resistance is chosen. So, here R1 is zero {the connections of a variable resistor can be used such as no resistance is taken}.
Minimum V­2 = [0 / (1+0)] x 25 = 0V              [B is confirmed]









Question 114: [Matter > Density]
A mass of liquid of density ρ is thoroughly mixed with an equal mass of another liquid of density 2ρ. No change of total volume occurs.
What is the density of the liquid mixture?
A 4/3ρ                         B 3/2ρ                         C 5/3ρ                         D 3ρ

Reference: Past Exam Paper – June 2002 Paper 1 Q21



Solution 114:
Answer: A.
Density = mass / volume
Volume = mass / density

For 1st liquid: Let the mass of the 1st liquid = m.
Volume of 1st liquid = m / ρ

For 2nd liquid: mass = m and density = 2ρ
Volume of 2nd liquid = m / 2ρ

For mixture:
Total volume of mixture = (m / ρ) + (m / 2ρ) = 3m / 2ρ
Total mass of mixture = m + m = 2m
Density of mixture = mass / volume = 2m / [3m / 2ρ] = 4ρ / 3










Question 115: [Energy > Potential energy]
Bow of mass 400 g shoots arrow of mass 120 g vertically upwards. Potential energy stored in the bow just before release is 80 J. System has an efficiency of 28%.
What is the height reached by arrow when air resistance is neglected?
A 4 m              B 19 m                        C 187 m                      D 243 m

Reference: Past Exam Paper – June 2013 Paper 12 Q16



Solution 115:
Answer: B.
Initial (elastic) potential energy stored = 80J
Efficiency = 28% = 0.28
Energy available for arrow = 0.28x80 = 22.4J.

Only the arrow will be moving upwards, so we consider the mass of arrow only. Since air resistance is negligible, as the arrow moves upwards, its energy is converted to gravitational potential energy.
At maximum height h,
Gravitational potential energy = mgh = 22.4
Height h = 22.4 / (0.12x9.81) = 19.0m




3 comments:

  1. Salame admin,
    In as 133 how do I know and for other exam qs...how do I know that R1 is a variable resistor..? Can you explain how moving the contact can make R1 have 0 resistance leaving total resistance to 1 ohms.
    Thank you

    ReplyDelete
    Replies
    1. Wslm.
      The resistor with an arrow (the moving jack) at the middle is itself the symbol for a variable resistor.

      When the jack is at one end, all the resistance is being considered and at the other end, no resistance is being considered. It's basically how a variable resistor works.

      I remember explaining it in greater details but have forgotten for which solution it was.

      Delete
  2. Thank you so much for all the help. :)

    ReplyDelete

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