Wednesday, November 4, 2015

Physics 9702 Doubts | Help Page 223

  • Physics 9702 Doubts | Help Page 223



Question 1060: [Dynamics > Resultant force]
A ball is falling at terminal speed in still air. The forces acting on the ball are upthrust, viscous drag and weight.
What is the order of increasing magnitude of these three forces?
A upthrust → viscous drag → weight
B viscous drag → upthrust → weight
C viscous drag → weight → upthrust
D weight → upthrust → viscous drag

Reference: Past Exam Paper – June 2008 Paper 1 Q12 & November 2014 Paper 11 & 12 Q11



Solution 1060:
Answer: A.
Sketching a force diagram would have greatly helped in answering this question. The weight is downward, while the viscous drag and the upthrust would be upwards, opposing the weight.

At terminal speed, the resultant force on the ball is zero. As weight is the only downward force, it must be the largest of the three.

The upthrust is a result of the force due to the difference in pressure at the top and bottom of the ball.
Pressure P = hρg         and Force = Pressure / Area
Air is not very dense (ρ is small) and the difference in height, h, from the top of the ball to its bottom is also very small. Thus, the upthrust force is small.

Many students would think that the Archimedean upthrust on a body in air is quite large. This is a mistake. The reverse is the case. The upthrust will only be approximately a thousandth or less than the weight of a solid.

Viscous drag (air resistance), on the other hand, will be almost as large as the weight when a body is at terminal velocity.










Question 1061: [Kinematics > Air resistance]
A golf ball is hit with the same force and direction on the Earth and on the Moon.
Which diagram best represents the shapes of the paths taken by the golf ball?


Reference: Past Exam Paper – November 2012 Paper 11 Q10



Solution 1061:
Answer: C.
On both the Earth and the Moon, the force of gravity would pull the ball towards the surface. [D is incorrect] However this force is greater on Earth, so the ball is pulled back to the surface in a shorter range.

Consider the path of the ball on Earth. Gravity pulls the ball to the surface while the air resistance opposes the motion. Thus, the path of the ball is not symmetric.

As for the path on Moon, we need to take into account the lack of air resistance (since it is in vacuum). Hence, the path of the golf ball is symmetric on Moon.










Question 1062: [Matter > Pressure of gas]
A mass of gas enclosed in a cylinder by a piston is heated gently. At the same time, the piston is moved so that the pressure remains constant.
As a result of this, what will not occur?
A The average velocity of the molecules will increase.
B The mean separation of the molecules will increase.
C The molecules will travel greater distances between collisions.
D The number of collisions per second of the molecules on the piston will increase.

Reference: Past Exam Paper – June 2012 Paper 12 Q20



Solution 1062:
Answer: D.
The average velocity of the molecules will increase since the (kinetic) energy of the molecules is increased from the heating (heat energy).

As the gas is heated, the molecules would tend to hot with the walls of the cylinder and piston, causing the pressure to increase. For the pressure to remain constant, the volume available for the gas should be increased.
Since volume is increased, both B (the mean separation of the molecules will increase) and C (the molecules will travel greater distances between collisions) occurs.

The gas molecules has now a larger volume to move about, so the probability of collisions with the piston will decrease.











Question 1063: [Waves > Phase difference]
Two light waves of the same frequency are represented by the diagram.

What could be the phase difference between the two waves?
A 150°                         B 220°                         C 260°                         D 330°

Reference: Past Exam Paper – June 2015 Paper 12 Q24



Solution 1063:
Answer: C.
The 2 light waves are said to have the same frequency. The waves have a sinusoidal form.

Let the wave with the larger amplitude be wave A and the one with smaller amplitude be wave B.

Since both waves have a sinusoidal form, we can assume that any of the 2 waves will start at displacement = 0 (we take this point as the reference point), and then move up – just like the wave A at (0, 0).
{In fact, any point (at any displacement) could be taken as the reference point, but in this question, it is easier to consider that point.}



On the graph, the x-axis gives the phase angle.
At a phase of 0°, wave B has not yet reaching the starting (reference) point while wave A is already at that point. It is only at a phase of 100° that wave B reached the reference point (or we could say that wave B reaches this point AGAIN at 100°).
The solution of 100° is not available in the 4 choices, so we can say that wave B had actually already reached this point before – we need to find at which phase this was, according to the x-axis in the diagram.

The phase angle is actually between 0° and 359°. A phase difference of 360° is the same as a phase difference of 0° and a phase difference of 1° is the same as a phase difference of 361°, ….

The wavelength of a wave corresponds to a phase difference of 360°. Since the 2 light waves have the same frequency, it means that they are the same wavelength.

So, if wave B reached the reference point again at a phase of 100°, according to the x-axis, going back by a wavelength (by a phase difference of 360°), we can say that previously, wave B had reached the reference point at the phase of
Phase = 100 – 360 = – 260°

As for wave A, it reaches the reference point at a phase of 0° (according to the x-axis).

Thus, phase difference between wave A and B = 0 – (–260) = 260°




3 comments:

  1. this is really a big help for physics students .
    thanks

    ReplyDelete
  2. props to whoever runs this website

    ReplyDelete
  3. this is amazing.
    I finally understand physics.

    ReplyDelete

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