Wednesday, August 12, 2015

Physics 9702 Doubts | Help Page 186

  • Physics 9702 Doubts | Help Page 186



Question 908: [Simple harmonic motion]
The vibrations of a mass of 150 g are simple harmonic. Fig.1 shows the variation with displacement x of the kinetic energy Ek of the mass.

(a) On Fig.1, draw lines to represent the variation with displacement x of
(i) potential energy of the vibrating mass (label this line P),
(ii) total energy of the vibrations (label this line T).
                                                             
(b) Calculate angular frequency of the vibrations of the mass.

(c) The oscillations are now subject to damping.
(i) Explain what is meant by damping.
(ii) The mass loses 20% of its vibrational energy. Use Fig.1 to determine the new amplitude of oscillation. Explain your working.

Reference: Past Exam Paper – November 2004 Paper 4 Q3



Solution 908:
(a)
(i) The line should have a reasonable shape as the ‘inverse’ of the k.e. line

(ii) This is a straight line, parallel to x-axis at 15 mJ

                                                             
(b)
EITHER
(Maximum) kinetic energy (= ½ mv2) = ½ mω2a02
15 × 10-3 = ½ × 0.15 × ω2 × (5.0×10-2)2
Angular frequency, ω = 8.9(4) rad s-1

OR
(k.e. = ½ mv2), Speed v = 0.44(7) m s-1
Angular frequency, ω = v / a = (0.447) / (5.0 × 10-2)
Angular frequency, ω = 8.9(4) rad s-1

(c)
(i)
Damping is EITHER loss of energy (from the system) or amplitude decreases OR the additional force acting (on the mass).
EITHER there is a continuous/gradual loss OR the force is always opposing motion

(ii)
{The total energy of the system is now 80% of the previous maximum value since 20% has been lost. Previous maximum energy = 15mJ. New maximum energy = (80/100) × 15 = 12mJ. OR equivalently, this corresponds to a loss of 15 – 12 = 3mJ of energy.}
EITHER (now has 80% of its) p.e. / k.e. = 12 mJ      OR loss in k.e. = 3 mJ
{For the EITHER case (the EITHER case is more time-consuming)
(Maximum) kinetic energy (= ½ mv2) = ½ mω2a02
12 × 10-3 = ½ × 0.15 × (8.94)2 × a2
New amplitude, a = 0.045m = 4.5cm

For the OR case (The OR case requires a proper understanding of the situation)
Amplitude is the maximum displacement. For amplitude, energy is totally potential, not kinetic. The 3mJ corresponds to 7.5 squares vertically on the graph.
For the following explanations, you are HIGHLY RECOMMENDED to make a rough diagram of what is being explained as you proceed.
This is important: the graph given in the question is that of KINETIC energy. IF we want the amplitude, the energy should be zero. This is one important point.
Second point: The graph shown in the question is that of the previous case (with no damping). Assume a new graph is drawn accounting for the damping. This NEW (kinetic) energy curve would have all the values of energy at 3mJ LESS than the energy of the previous curve.
Negative values of energies in the new curve are rejected. In the previous curve, all values of energy that were less than 3mJ would correspond to negative energies in the new curve.
For amplitude, energy is totally potential, not kinetic. So, when the energy of the new curve is zero, this would indicate the value of the amplitude.
But for the value of energy of the new curve to be zero, it should be 3mJ in the previous curve. So, we look for the corresponding value of displacement in the curve given in the question for energy = 3mJ.}
new amplitude = 4.5 cm (allow ± 0.1 cm)










Question 909: [Dynamics]
What is the definition of the force on a body?
A the mass of the body multiplied by its acceleration
B the power input to the body divided by its velocity
C the rate of change of momentum of the body
D the work done on the body divided by its displacement

Reference: Past Exam Paper – November 2011 Paper 11 Q10 & Paper 13 Q11



Solution 909:
Answer: C.
The force on a body is DEFINED as the rate of change of momentum of the body (from Newton’s 2nd law of motion).

This does equal to the product of mass and acceleration but this is NOT accepted as a DEFINITION.











Question 910 [Kinematics > Projectile motion]
A bomber flies with a constant velocity of 50ms-1 at a constant height of 1000m such that it will fly above a target on the ground. What is the horizontal distance of the bomber from the target so that a bomb released from it will hit the target?

Reference: ???



Solution 910:
This problem is on projectile motion. So, we can consider the vertical and horizontal motion separately.

Initially the bomb will have the same initial velocity as the bomber. The bomb is also at the same initial height. Since the bomber is moving at constant height, it only has a horizontal component of velocity. The same goes for the bomb.

The quantity which is common for both the vertical and horizontal components of velocity of the bomb is the time at which the bomb hits the target, t.

Consider the vertical motion.
Distance, s = 1000m
Vertical component of initial velocity, u = 0
Acceleration due to gravity, g = 9.81ms-2
Equation of uniformly accelerated motion:    s = ut + ½ at2
1000 = 0 + 0.5(9.81) t2
Time, t = √[1000 / (0.5x9.81)] = 14.278s

Now, consider the horizontal motion.
Distance to travel horizontally = s
Time taken for travel, t = 14.278s
Horizontal component of initial velocity = 50ms-1
Horizontal component of acceleration = 0
s = ut + ½ at2
s = 50(14.278) + 0 = 714m

So, horizontal distance of bomber from target = 714m











Question 911: [Current of Electricity > Resistance]
A battery of e.m.f. 4.50 V and negligible internal resistance is connected in series with a fixed resistor of resistance 1200 Ω and a thermistor, as shown in Fig.1.

(a) At room temperature, thermistor has a resistance of 1800 Ω. Deduce that the potential difference across the thermistor (across AB) is 2.70 V.

(b) A uniform resistance wire PQ of length 1.00 m is now connected in parallel with the resistor and the thermistor, as shown in Fig.2. 

A sensitive voltmeter is connected between point B and a moveable contact M on the wire.
(i) Explain why, for constant current in the wire, the potential difference between any two points on the wire is proportional to the distance between the points.

(ii) Contact M is moved along PQ until the voltmeter shows zero reading.
1. State potential difference between the contact at M and the point Q
2. Calculate the length of wire between M and Q

(iii) Thermistor is warmed slightly. State and explain the effect on the length of wire between M and Q for the voltmeter to remain at zero deflection.

Reference: Past Exam Paper – November 2005 Paper 2 Q7



Solution 911:
(a)
Either
V = ER1 / (R1+R2) = (1800 / 3000) × 4.50 = 2.70V

Or
I = E / (R1+R2)
V = IR1 = (1800 / 3000) × 4.50 = 2.70V

(b)
(i)
For a wire, V = IR = I(ρL / A)
Since I, ρ and A are constant, V is proportional to L.

(ii)
1. 2.70V

2.
L / 100 = 2.70 / 4.50
So, L = 60.0cm

(iii) The thermistor resistance decreases as the temperature rises. So, the length of wire between Q and M is shorter.
{The resistance of the thermistor decreases as the temperature increases. So, the p.d. across the thermistor decreases [V = IR]. This causes the p.d. across the 1200Ω resistor to be greater (since it is in series with the thermistor).

For points B and M to be at the same potential (for zero deflection), the resistance across PM should be increased so that the p.d. across PM is also increased. To increase the resistance of PM, the length of wire between points P and M should be increased.


Current flows from the positive terminal of the battery to the negative one. The negative terminal is normally taken to be at a potential of 0V while the positive terminal is taken to have the value of the e.m.f. of the battery (here, 4.50V). Since points A and Q are connected to the negative terminal of the supply, they are both at zero potentials. Points C and Q are both at a potential of 4.50V.

So, current flows from C to A and from P to Q. We should consider the section of the wire where current starts flowing and increase the resistance of that section.}

6 comments:

  1. Please consider answering ALL of the following questions before October:
    4/O/N/02 Q.3(b)(i),Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.4(c),Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.9(b)(iii),Q.11(b)
    04/O/N/04 Q.3(c)(ii)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.7(b)(i),(c),Q.10(c)
    04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)
    04/O/N/08 Q.7(c)
    04/M/J/09 Q.10(c)(ii),Q.11(b)(iii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    42/O/N/09 Q.5(a),Q.7(b)(ii)
    41/M/J/10 Q.6(a),Q.7(a)
    42/M/J/10 Q.6(a)(ii)
    51/M/J/10 Q.2(d)
    42/O/N/10 Q.2(b)(iii),Q.3(c)

    ReplyDelete
    Replies
    1. For 4/O/N/02 Q.3(b)(i), see solution 912 at
      http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-187.html

      For 42/O/N/10 Q.2(b)(iii), see solution 915 at
      http://physics-ref.blogspot.com/2014/09/9702-november-2010-paper-41-42-worked.html

      Delete
  2. Hi, can you post the solutions of q20,23,24 and 32 of physics paper 12, year 2015

    ReplyDelete
    Replies
    1. Solutions of the 2015 questions may be posted a bit later.

      Delete
  3. Why is the p.d between the contacts 2.7? Shouldn't it be zero because then the reflection would be zero? The question ask for the difference right?

    ReplyDelete
    Replies
    1. the contacts are on a wire that have resistance, so there is a pd between the contacts. Since the voltmeter reading is zero, it means that the p.d. is the same as across AB.

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 186