• 9702 November 2012 Paper 22 Worked Solutions | A-Level Physics

Question 1

{Detailed explanations for this question is available as Solution 687 at Physics 9702 Doubts | Help Page 139 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-139.html}

Question 2

{Detailed explanations for this question is available as Solution 990 at Physics 9702 Doubts | Help Page 205 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-205.html}

Question 3

{Detailed explanations for this question is available as Solution 643 at Physics 9702 Doubts | Help Page 128 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-128.html}

Question 4

(a)

Arrangement for producing stationary waves in tube that is closed at one end.

How waves from loudspeaker produce stationary waves in tube:

The waves travel along the tube and are reflected at the closed end (end of tube). The incident and reflected waves are in opposite directions and so, interfere and stationary wave is formed if the tube length is equivalent to λ/4, 3λ/4, etc.

(b)

(i)

Motion of air particles in tube at 1. Point P    2. Point S

1.

No motion of air particle as P is a node (zero amplitude).

2.

Vibration (motion) of the air particles is backwards and forwards / maximum amplitude along length.

(ii)

Speed of sound in tube = 330ms^-1 and frequency of waves from loudspeaker = 880Hz. Calculate length of tube:

L = 3λ/4

v = f λ.                        So, λ = 330 / 880 = 0.375m

L = ¾ x (0.375) = 0.28 (0.281)m

Question 5

{Detailed explanations for this question is available as Solution 685 at Physics 9702 Doubts | Help Page 138 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-138.html}




Question 6

(a)

Hooke’s law states that the extension of a spring is proportional to the force/load applied to it.

(b)

Object M of mass 0.41kg attached to lower end of spring. Spring extends until M is at rest at R.

Spring constant, k = 25Nm^-1. Show that extension is about 0.16m:

F = mg = kx

x = mg / k = 0.41x9.81 / 25 = 0.16m

(c)

Object pulled down a further 0.060m to S and then release. For M, just as it is released:

(i)

Forces acting on M:

Weight and (reaction) force from spring (which is equal to tension in spring)

(ii)

Acceleration of M:

F – weight = ma                                  or (25x0.06) = ma

F = (0.1609 + 0.06) x 25 = 5.52N       or (0.16+0.06) x 25 = 5.5N

a = (5.52 – (0.41x9.81)) / 0.41            or 1.5 / 0.41 and (5.5 – 4.02)

a = 3.7 (3.66)ms^-2                                gives 3.6ms^-2

(d)

Describe energy changes from time object is released to time it first returns to R:

The elastic potential energy / strain energy is converted to kinetic and gravitational potential energy. The stretching / extension reduces while the velocity of object M, along with its height from the ground increases.

Question 7
{Detailed explanations for this question is available as Solution 716 at Physics 9702 Doubts | Help Page 145 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-145.html}