• 9702 June 2004 Paper 2 Worked Solutions | A-Level Physics

Paper 2

Question 1

(a)

Difference between scalar quantity and vector quantity:

A scalar quantity has a magnitude only while a vector quantity has both a magnitude and a direction

(b)

2 forces of magnitude 6.0N and 8.0N act at point P. both forces act away from point P and angle between them is 40^o. Fig shows 2 lines at angle of 40^o to one another. On Fig, draw vector diagram to determine magnitude of resultant of 2 forces:

The diagram has a correct shape (parallelogram or triangle) with the arrows in the correct directions (parallel to the given ones)

(Note that the dotted lines do not represent the size of the 2 vectors mentioned – you need to choose the appropriate scale)

Resultant = 13.2 ± 0.2 N

Question 2

{Detailed explanations for this question is available as Solution 435 at Physics 9702 Doubts | Help Page 83 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-83.html}

Question 3

Student has been asked to determine linear acceleration of toy car as it moves down a slope. He sets up apparatus as shown.

Time t to move from rest through a distance d found for different values of d. Graph of d (y-axis) plotted against t² (x-axis) as shown.

(a)

Theory suggests that graph is a straight line through origin. Name feature on Fig that indicates presence of

(i)

Random error:

The scatter of points (about the line)

(ii)

Systematic error:

The intercept (on the t² axis)

(b)

(i)

Gradient of line of graph in Fig:

Gradient = Δy / Δx = (100 – 0) / (10.0 – 0.6) = 10.6 (cms^-2)

(ii)

Use answer to (i) to calculate acceleration of toy down slope:

s = ut + ½ at²

(½ a = gradient) So, acceleration = 2 x gradient

Acceleration = 0.212ms^‑2

 

Question 4

Ball has mass m. it is dropped onto horizontal plate as shown. Just as the ball makes contact with plate, it has velocity v, momentum p and kinetic energy E_k.

(a)

(i)

Expression for momentum p in terms of m and v:

p = mv

(ii)

Show that kinetic energy given by E_k = p² / 2m:

E_k = ½ mv²

Algebra leading to (e.g. p = mv and v = p/m)

E_k = p²/2m

(b)

Just before impact with plate, ball of mass 35g has speed 4.5ms^-1. It bounces from plate so that its speed immediately after losing contact with plate is 3.5ms^-1. Ball is on contact with plate for 0.14s.

For time that ball is in contact with plate,

(i)

Average force, in addition to weight of ball, that plate exerts on ball:

EITHER

Δp = 0.035(4.5 + 3.5) = 0.28Ns

Force = Δp / Δt (= 0.28 / 0.14) = 2.0N

OR

a = (4.5 – (-3.5)) / 0.14 = 57.1ms^-2

F = ma (= 0.035 x 575.1) = 2.0N

(ii)

Loss in kinetic energy of ball:

Loss in kinetic energy = ½ (0.035) (4.5² – 3.5²) = 0.14J

(c)

Explain whether linear momentum conserved during bounce:

The plate (and Earth) gain momentum which is equal and opposite to the change for the ball. So, momentum is conserved.

Question 5

2 forces, each of magnitude F, form a couple acting on the edge of disc of radius r, as shown.

(a)

Disc is made to complete n evolutions about axis through its centre, normal to plane of disc. Expression:

(i)

Distance moved by point on circumference of disc:

Distance = n(2πr) = 2πnr

(ii)

Work done by 1 of the 2 forces:

Work done = F x (2πnr)

(b)

Using answer to (a), show that work W done by couple producing torque T when it turns through n revolutions given by          W = 2πnT:

Total work done = 2 x F x 2πnr

But the Torque T = 2Fr

Hence, work done = 2πnT

(c)

Car engine produces torque of 470Nm at 2400 revolutions per minute. Output power of engine:

Power = Work done / time (= 470 x 2π x 2400 / 60) = 1.2x10⁵W

Question 6

Fig. 6.1 shows wavefronts incident on, and emerging from, a double slit arrangement.

Question 7

Household electric lamp rated as 240V, 60W. Filament of lamp made from tungsten and is a wire of constant radius 6.0x10^-6m. Resistivity of tungsten at normal operating temperature of lamp is 7.9x10^-7Ωm.

(a)

For lamp at its normal operating temperature:

(i)

Current in lamp:

P = VI

So, current I = P / V = 60 / 240 = 0.25A

(ii)

Show that resistance of filament is 960Ω:

Resistance, R (= V/I) = 240 / 0.25 = 960Ω

(b)

Length of filament:

R = ρL / A       (wrong formula, 0/3)

960 = (7.9x10^-7) L / (π{6.0x10^-6}²)

Length of filament, L = 0.137m

(c)

Comment on answer to (b):

Example:

The filament must be coiled / it is long for a lamp

Question 8

Thermistor has resistance 3900Ω at 0^oC and resistance 1250Ω at 30^oC. Thermistor connected into circuit of Fig in order to monitor temperature changes. Battery of e.m.f. 1.50V has negligible internal resistance and voltmeter has infinite resistance.

(a)

Voltmeter is to read 1.00V at 0^oC. Show that resistance of resistor R is 7800Ω:

V / E = R / R_tot                        or 0.5 = I x 3900

1/0/1.5 = R / (R+3900)            or 1.0 = 0.5R / 3900

R = 7800Ω                              or R = 7800Ω

(b)

Temperature of thermistor increased to 30^oC. Reading on voltmeter:

EITHER

V = 1.5 x [7800 / (7800+1250)] = 1.29V

OR

I = 1.5 / (7800+1250)              (=1.66x10^-4A)

V = IR (= (1.66x10^-4)x7800) = 1.29

(c)

Voltmeter in Fig replaced with one having resistance of 7800Ω. Reading on this voltmeter for thermistor at temperature of 0^oC:

Combined resistance of resistor R and voltmeter (= [1/7800 + 1/7800]^-1) = 3900Ω

So, reading at 0^oC (= [3900 / (3900+3900)]x1.5 ) = 0.75V