Friday, November 14, 2014

Physics 9702 Doubts | Help Page 12

  • Physics 9702 Doubts | Help Page 12




Question 64: [Energy > Potential energy]
Diagram shows two identical vessels X and Y connected by short pipe with a tap.

Initially, X is filled with water of mass m to depth h, and Y is empty. When tap is opened, water flows from X to Y until depths of water in both vessels are equal.
How much potential energy is lost by water during this process? (g = acceleration of free fall)

Reference: Past Exam Paper – June 2009 Paper 1 Q15



Solution 64:
Answer: B.

Consider the water in both vessels as one body with one mass.
Since the mass of the water is spread all over the volume it occupies, the centre of mass should be considered when calculating the (gravitational) potential energy.
Before the tap is opened, the centre of mass of the water in vessel X may be considered to be at its centre (that is, at half the water level [h/2]).
Potential energy of water before tap is opened = m g (h/2) = mgh / 2

When the tap is opened, the total mass of water in the whole body is the same, but the water level is now at a height of h/2 in both vessels.
The centre of mass may be considered to be at the centre of the volume of water, that is, at a height of half the water level [water level = h/2, so centre of gravity is now at a height of h/4]
Potential energy of water after tap has been opened = m g (h/4) = mgh / 4

Lost in potential energy = (mgh / 2) – (mgh / 4) = mgh / 4








Question 65: [Energy > Potential energy]
Calculate how much gravitational potential energy is lost by an aircraft of mass 80000 kg if it descends from an altitude of 10000 m to an altitude of 1000m. What happens to this energy if the pilot keeps the aircraft's speed constant?

Reference: ???



Solution 65:
(Gravitational) Potential energy = mgh
Lost in (gravitational) potential energy = mgΔh = 80000 x 9.81 x (10000 – 1000)
Lost in (gravitational) potential energy = 7.1x109J

This energy is lost as work done against drag forces (to keep the aircraft’s speed constant) and against air resistance (when decreasing in altitude).








Question 66: [Dynamics > Resultant forces]
Mass of 2.0 kg rests on a frictionless surface. It is attached to 1.0 kg mass by a light, thin string which passes over frictionless pulley. 1.0 kg mass is released and accelerates downwards.

What is the speed of 2.0 kg mass as the 1.0 kg mass hits floor, having fallen a distance of 0.50 m?
A 1.8 m s–1      B 2.2 m s–1      C 3.1 m s–1      D 9.8 m s–1

Reference: Past Exam Paper – November 2012 Paper 11 Q13



Solution 66:
Answer: A.
The gravitational force acts on the 1.0kg mass, causing an acceleration of 9.81ms-2.
Weight of 1.0kg box = 1 x 9.81 = 9.81N

But as the 1.0kg mass moves, the 2.0kg also undergoes motion. Therefore, considering the whole system, a force of 9.81N is acting on a total mass of (2.0 + 1.0 =) 3.0kg.
The resultant acceleration, a on this total mass can be obtained as follows
3a = 9.81
Acceleration, a = 9.81 / 3 = 3.27ms-2

Hence, each of the 2 masses undergoes a resultant acceleration of 3.27ms-2.
Initial speed of 2.0kg mass, u = 0
Distance travelled, s = 0.5m
v2 = u2 + 2as = 0 + 2 (3.27) (0.5)
Speed, v = √(2 x 3.27 x 0.5) = 1.8ms-1








Question 67: [Kinetic energy > Speed]
Kinetic energy of particle is increased by a factor of 4. By what factor does its speed increase?
A 2      B 4      C 8      D 16

Reference: Past Exam Paper – November 2004 Paper 1 Q15 & November 2012 Paper 11 Q18



Solution 67:
Answer: A.
Kinetic energy, KE is proportional to v2 (KE = ½ mv2). So, speed v is proportional to √(KE).
Kinetic energy is increased by a factor of 4. (KE becomes 4KE)
So, new speed is √(4KE) = (√4) (√(KE)) = 2√(KE)

Since, speed v is proportional to √(KE),
New speed = 2v
Therefore, speed increases by a factor of (2v / v =) 2









Question 68: [Dynamics > Resultant forces]
A light spring has a mass of 0.20kg suspended from its lower end. A second mass of 0.10kg is suspended from the first by a thread. The arrangement is allowed to come into static equilibrium and then the thread is burned through. At this instant, what is the upward acceleration of the 0.20kg mass? (Take g as 10ms-2.)
A zero             B 3.3ms-2         C 5.0ms-2         D 6.7ms-2         E 10ms-2

Reference: Past Exam Paper – November 1987 / I / 5



Solution 68:
Answer: C.
For static equilibrium, the resultant force is zero. Thus, the (upward) force in the spring is equal to the total (downward) weight of the masses.
Spring force, T = 0.20 (10) + 0.1 (0.10) = 3N

When the thread is burnt, the weight of the 0.10kg mass no longer contributes any forces to the system, but at that instant, the spring force is still 3N.

So, the weight of the 0.20kg acts downwards while the spring force acts upwards. The resultant (upward) force on the 0.20kg mass is given by
Resultant force = ma = T – mg = 3 – 0.20(10) = 1N
Acceleration, a = 1 / 0.20 = 5ms-2 







Question 69: [Dynamics > Resultant forces]
A raindrop of mass m is falling vertically through the air with a steady speed v. The raindrop experiences a retarding force kv due to the air, where k is a constant. The acceleration of free fall is g.
Which expression gives the kinetic energy of the raindrop?
A mg / k                      B mg2 / 2k2                  C m3g2 / k2                   D m3g2 / 2k

Reference: Past Exam Paper – June 1992 / I / 6 & November 1996 / I / 7 & November 2010 Paper 12 Q15



Solution 69:
Answer: D.
For raindrop to fall with a steady speed v, terminal speed must be reached (net force and thus, net acceleration is zero).
This occurs when the retarding force kv equals to the weight mg.
So, mg = kv giving v = mg / k.
Kinetic energy of raindrop, KE = ½ mv2 = ½ m(mg / k)2 = m3g2 / 2k2.



12 comments:

  1. salam! in q65, wont energy be converted to kinetic energy?

    ReplyDelete
    Replies
    1. Wslm.
      The question states that the aircraft's speed is constant. So, kinetic energy is also constant [since the aircraft was initially moving, it has some initial kinetic energy].

      Kinetic energy = (1/2) mv^2. If v is not changing, the kinetic energy is not changing also.

      Delete
    2. so in q65 we assume that the constant speed with which it descends is the same as that with which it was travelling at the altitude of 10000 m?

      Delete
    3. Yes, but we are only assuming this because it is said in the question.

      Delete
  2. Could you please explain question 6 of October/November 2012 p11?

    ReplyDelete
    Replies
    1. See solution 433 at
      http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-82.html

      Delete
  3. Thank you for your post =D

    ReplyDelete
  4. I don't understand why we need to consider centre of gravity in solution 65

    ReplyDelete
    Replies
    1. this has always been the case, but people do not usually pay attention.

      the centre of mass is where all of the mass of the object is assumed to be and it is where the weight would act.

      Delete
    2. Potential energy is due to height so shouldn't it be mgh at the start?

      Delete
    3. yest and 'height' is actually the height of the centre of mass.

      Delete

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