Saturday, November 1, 2014

9702 June 2009 Paper 21 Worked Solutions | A-Level Physics

  • 9702 June 2009 Paper 21 Worked Solutions | A-Level Physics


Paper 21


Question 1
{Detailed explanations for this question is available as Solution 1009 at Physics 9702 Doubts | Help Page 210 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-210.html}





Question 2
Ball B of mass 1.2 kg travelling at constant velocity collides head-on with stationary ball S of mass 3.6 kg, as shown. Frictional forces are negligible. Variation with time t of velocity v of ball B before, during and after colliding with ball S shown.
(a)
Significance of positive and negative values for v:
The negative value indicates that the ball is moving in the opposite direction (after collision)

(b)
Use Fig to determine, for ball B during collision with ball S,
(i)
Change in momentum of ball B:
Change in momentum = 1.2(4.0 + 0.8) = 5.76Ns

(ii)
Magnitude of force acting on ball B:
Force = Δp / Δt (or mΔv / Δt) = 5.76 / 0.08 = 72N

(c)
Speed of ball S after collision:
5.76 = 3.6 x v
Speed, v = 1.6ms-1

(d)
Using answer in (c) and information from Fig, deduce quantitatively whether collision is elastic or inelastic:
EITHER
Relative speed of approach = 4.0ms-1 and relative speed of separation = (0.8 + 1.6 =) 2.4ms-1. These are not equal, so the collision is inelastic.
OR
Kinetic energy before collision = 9.6J and Kinetic energy after collision = 4.99J. Kinetic energy is less after the collision / not conserved. So, the collision is inelastic.





Question 3

{Detailed explanations for this question is available as Solution 1091 at Physics 9702 Doubts | Help Page 232 - http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-232.html}





Question 4
Spring having spring constant k hangs vertically from fixed point. Load of weight L, when hung from spring, causes extension e. Elastic limit of spring is not exceeded.
(a)
(i)
An elastic deformation is the change of shape / size / length / dimension when a (deforming) force is removed, returns to original shape / size

(ii)
Relation between k, L and e:
L = ke

(b)
Some identical springs, each with spring constant k, are arranged as shown. Load on each of the arrangements is L. For each arrangement, complete table by determining (i) total extension in terms of e, (ii) spring constant in terms of k.
{Force, L = (spring constant, k) x( extension, e). L is constant in all cases. e can be determined from the diagrams. E.g. 2 springs in parallel causes extension to be halved while 2 springs in series causes extension to be doubled. Spring constant, k is given by k = F / extension. In this question, F is constant – so, k is the reciprocal of the extension}
Arrangement 1:
Extension = 2e
Spring constant = ½ k

Arrangement 2:
Extension = ½ e
Spring constant = 2k

Arrangement 3:
Extension = (3/2) e
Spring constant = (2/3) k



Question 5
2 sources S1 and S2 of sound are situated 80 cm apart in air, as shown. Frequency of vibration can be varied. The 2 sources always vibrate in phase but have different amplitudes of vibration. Microphone M is situated distance 100 cm from S1 along line that is normal to S1S2. As frequency of S1 and S2 is gradually increased, microphone M detects maxima and minima of intensity of sound.
(a)
2 conditions that must be satisfied for intensity of sound at M to be zero:
Condition 1: EITHER phase difference is π rad / 180° OR path difference (between the waves from S1 and S2) is ½λ / (n + ½)λ
Condition 2: EITHER the waves have the same amplitude / intensity at M OR the ratio of amplitudes is 1.28 / ratio of intensities is 1.282

(b)
Speed of sound in air is 330 m s–1. Frequency of sound from S1 and S2 is increased. Number of minima that will be detected at M as frequency is increased from 1.0 kHz to 4.0 kHz:
Path difference between the waves from S1 and S2 = (128 – 100 =) 28cm

(when frequency = 1kHz, wavelength = 330/1000 = 0.33m = 33cm. When frequency = 4kHz, wavelength = 330/4000 = 0.0825m = 8.25cm)
The wavelength changes from 33cm to 8.25cm
((n + ½)λ = 28. So, λ = 28 / (n + ½). 
When n = 0, λ = 56cm. When n = 1, λ = 18.7cm. When n =2, λ = 11.2cm. When n =3, λ = 8.0cm)
Minima will be detected when λ = (56cm,) 18.7cm, 11.2cm, (8.0cm). So there are 2 minima.



Question 6
2 vertical parallel metal plates are situated 2.50 cm apart in vacuum. Potential difference between plates is 350 V, as shown. An electron is initially at rest close to negative plate and in uniform electric field between plates.
(a)
(i)
Magnitude of electric field between plates:
E = V / d = 350 / (2.5x10-2) = 1.4x104NC-1

(ii)
Show that force on electron due to electric field is 2.24x10-15N:
Force = Eq = (1.4x104) (1.6x10-19) = 2.24x10-15N

(b)
Electron accelerates horizontally across space between plates.
(i)
Horizontal acceleration of electron:
Force, F = ma
a = (F/m =) (2.24x10-15) / (9.1x10-31) = 2.46x1015ms-2

(ii)
Time to travel horizontal distance 2.50cm between plates:
s = ½at2
2.5x10-2 = ½ (2.46x1015) t2  
Time, t = 4.5x10-9s

(c)
Why gravitational effects on electron need not be taken into consideration in calculation in (b):
EITHER The gravitational force is normal to the electric force OR The electric force is horizontal while the gravitational force is vertical.



Question 7
{Detailed explanations for this question is available as Solution 595 at Physics 9702 Doubts | Help Page 117 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-117.html}






Question 8
Spontaneous and random decay of radioactive substance involves emission of either α-radiation or β-radiation and/or γ-radiation.
(a)
For a spontaneous decay, the rate of decay / activity / decay (of nucleus) is not affected by external factors / environment / surroundings

(b)
Type of emission that:
(i)
Not affected by electric and magnetic fields:
Gamma / γ

(ii)
Produces greatest density of ionization in medium:
Alpha / α

(iii)
Does not directly result in change in proton number of nucleus:
Gamma / γ

(iv)
Has range of energies, rather than discrete values:
Beta / β




12 comments:

  1. I don't understand the method you use to find the number and minima of question 5)(b)
    I can find the phase difference and the wavelength range but I got stucked here.

    ReplyDelete
  2. what do they mean when they say "number of minima detected"?do they mean the what ORDER of minima has been detected?how can more than one minima be detected at the same point?
    if the wavelength had been 8.0 cm, would the answer be 3?

    ReplyDelete
    Replies
    1. It;s pretty much the same thing. When the order = 1, there's a minima. when order = 2, there's another minima. So, the total number of minima = 1+1 = 2.
      When order = 5, total number of minima = 5.

      Only one minima can be found at a point.

      The question gives a range of frequencies in which minima can be detected. These correspond to a range of wavelengths. Even if there's a minima when the wavelength is 8.0cm, it's outside the range and so, it cannot be DETECTED even if it is present.

      Delete
    2. Hi
      But for order 3 , there is minima at wavelength 8.0 cm which is less than 8.25 cm then how it is outside the range?

      Delete
    3. the corresponding range of wavelength is, as calculated, from 8.25cm to 33cm. A wavelength of 8.0cm is out of the range as it is too small

      Delete
  3. Please explain Q7 b in detail

    ReplyDelete
  4. question 1b how is the percentage uncertainty in density 0.18?

    ReplyDelete
  5. How came phase diffrence, 128-100=28 in 5(b)???

    ReplyDelete

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 | 9702 June 2009 Paper 21 Worked Solutions | A-Level Physics