Plk Vicwood K.T. Chong Sixth Form College

99’ AL Physics/Structural Questions/Marking/P.4

PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE

99’ AL Physics : Structural Questions

Marking Scheme

1. (a) 4.6 ± 0.1 cm 1 1

(b) Avoid the formation of bubbles adhering the ball bearings, 1

which will affect the volume measured. 1

Or Tilting the measuring cylinder and let the bearings run along the wall of the cylinder.

Or Avoid the splashing of water when putting the bearings into the cylinder. 2

(ii) density = 1

= 1

= 7830 kg×m-3 1 3

(iii) % uncertainty in r = % uncertainty in mass + % uncertainty in volume 1

= 1

= 0.04% + 1% + 2%

= 3% 1 3

(d) Use micrometer screw gauge to measure the diameter of the ball bearings. 1

Average diameter of ball bearings = 1

= 6.73 ´ 10-3 m or 6.73 mm

% uncertainty in volume = = 0.4 % 1 3

2. (a) (i) 3, 11, 19 1 1

(ii) l = 5 cm 1

f =

= 50 kHz 1

v = fl

= 50 ´ 103 ´ 5 ´ 10-2 1

= 2500 ms-1 1 4

(b) (i) A compression is reflected as a compression at the bench surface. 1 1

(ii)

For a certain point along PQ, the sound waves coming directly from the

loudspeaker interfere with those reflected from the bench. Due to the path 1

difference between the waves, constructive interference and destructive

interference occur alternately as the path difference changes. 1 3

(iii) path difference D = 1 + 1

= 0.658 m

D = 4l = 0.658

l = 0.165 m 1

v = fl

= 2 ´ 103 ´ 0.165

= 330 ms-1 1 4

3. (a)

(b) (i) qV0 = 1

As qvB = 1

r =

d = 2r = 1

For a proton, d = 0.18 m 1 4

(ii) Time taken T = 1

= (by r = in (i)) 1

= 0.66 ms (or 6.6 ´ 10-7 s) 1 3

mx = 4 mp 1

Nucleus X is a helium nucleus () or an a-particle. (Deduction should be consistent 1 3

with the result calculated.)

4. (a) (i) Gain =

= 1

Rf = 30 kW 1 2

(ii) The circuit amplifies the input signal linearly by 3 times and 1 + 1

with a change of sign, i.e. a negative feedback inverting amplifier. 1 3

(b) = 1

R = 25 kW 1 2

For Vin < VQ = 4.5 V, Op amp 2 gives a high output that makes LED Y light up 1

For Vin > VP = 5.5 V, Op amp 1 gives a high output that makes LED X light up. 1 3

(ii) 15 kW : R1 : R2 = 4.5 : 1 : 6.5 1

R1 = 3.3 kW 1

R2 = 21.7 kW 1 3

5. (a) a-source 1

- normally the source can hardly enter the body and it is harmful only when it has entered

the body 1

- short range with low activity and the source is sealed, therefore it is safe in normal use 1 3

(b) (i) I0 = 2000 ´ 5 ´ 104 ´ 1.6 ´ 10-19 1

= 1.6 ´ 10-11 A 1

Dust or air particles may carry away some of the ions. 1 3

(ii) I = I0e-kt 1

5 ´ 10-12 = 1.6 ´ 10-11 e-k(10 yr) 1

k = 0.116 yr-1

t1/2 =

= 5.96 yr (or 1.88 ´ 108 s) 1 3

6. (a) f ’ = 1

= (as ® 0) 1 2

(b) (i) When the satellite is close to the tracking station, its direction of motion (or

velocity) makes an angle with the line joining the satellite and the station. The

Doppler effect is thus due to the component of its velocity, which varies. 2 2

(ii) Df1 = f ’ – f0 = ……….. (1)

Df2 = f ” – f0 = ……….. (2) 1

Df1 - Df2 = 1

7000 – 2000 =

v = 7500 ms-1 1 3

(iii) By (1), 7000 = 1

f0 = 99.9955 MHz 1 2

(iv) No, it is only the velocity ‘observed’ from the tracking station, the rotation of the

earth together with the tracking station is not taken into account. 2 2

7. (a) (i) A : frictional force, B : normal reaction, C : weight 1

Force C (weight) 1 2

(ii) The ball bearing is initially at rest and all the forces acting on it are on a vertical

plane. 2 2

(iii) w2 = = 1

w = 6.9 rads-1

T = 1

= 0.91 s 1 3

(iv) The hollow sphere has a greater moment of inertia, so for the same loss in 1

potential energy, its speed of rotation and thus the linear speed when passing the 1

centre of the lens are smaller than those of the bearing. 1 3

(b) (i)

(ii) When the downward acceleration of the platform is equal to (or greater than) the

gravitational acceleration, the slotted weight will lose contact with the platform

(normal reaction = zero) 1

Maximum acceleration = w2A = g 1

(2pf0)2A = g

f0 = 1

= 4.1 Hz 1 4

(iii) f0 remains unchanged as the acceleration is independent of mass. 1 + 1 2

8. (a) (i) At low frequency, the reactance of C and hence the impedance of the circuit is 1

high so the ammeter reading is small.

As frequency increases, both the reactance of C and the impedance of the circuit 1

decrease so the ammeter reading increases. 1 3

(ii) (I) T = = = 0.5 ms 1

time base = 0.125 ms per cm 1

VR leads VO by = 72° (or or ) 1 + 1

(or using cos q = ) 4

(II) r.m.s. of VR =

= 2.8 V

r.m.s. of VO =

= 9.2 V

Either peak value correct 1

Able to find r.m.s. values 1

Correct r.m.s. values 1 3

(III) tan 72° = 1

C = 1

= 4.7 mF 1 3

(b) (i) At low frequency, bulb Y lights up while bulb X is dim as the reactance of L is 1

low but that of C is high. 1

As frequency increases, bulb X glows gradually and bulb Y decreases in 1 3

brightness.

(ii)

At resonance (1 kHz), large and nearly equal currents flow through C and L

approximately at antiphase, therefore the total current through bulb Z is very

small. 1 3

9. (a) (i) VO =

= 1

= - 10 ´ 6400 ´ 103

= - 6.4 ´ 107 Jkg-1 1 2

(ii) V¥ = 0 Jkg-1 1

Minimum energy per kg = 0 – (- 6.4 ´ 107)

= 6.4 ´ 107 J 1 2

(b) 1989 N1 is too close to Neptune that it is lost in the glare of the reflected sunlight from

Neptune. 1 1

or 1989 N1 and Neptune cannot be resolved when observed from earth.

Or Assume Neptune is stationary.

Or Neglect the gravitational pull of nearby planets/satellites. 1

(ii)

T2/hr2 / 177 / 64 / 90 / 56 / 50
r3/109 km3 / 3.99 ´ 105 / 1.45 ´ 105 / 2.16 ´ 105 / 1.25 ´ 105 / 1.12 ´ 105

Appropriate quantities are chosen for plotting the graph 1

Points correctly plotted 1

Correct graph (best straight line) 1

slope = 1

= 1.79 ´ 1014 (m3s-2) (or 2.32 ´ 1012 km3 hr-2)

\ slope = = 1.79 ´ 1014 1

M =

= 1.05 ´ 1026 kg 1 6