CAPE PHYSICS OTHER ISLANDS June Paper 2 (2008)

■-m

(2008) PAPER 2 -Qu.l

(a) (i) «

(b)

<ii»

174
 (2008) PAPER 2-Qu.l
Paper 2 OTHER CAPE
^ (ii)
(2008)The graph shows an initial acceleration
ISLANDSwhich decreases with timePHYSICS
(since the
slope of the graph decreases). The acceleration eventually reaches zero (slope
zero) at 2.8 seconds and beyond. This is the time at which terminal velocity is
reached.
The terminal velocity is 4.12 ms-1.

(iii)

Average acc = —
At
3.25-2.85 0.7-0.5 = 2.0 ms-2

(b) (i)

Y - mE 67trV(
= kg ms-2 irr1 rrr’s unit of K = kg m_1 s-1 (viscosity)

(ii)

_ 5 x 10~3 x9.81 67tl.0xl0-3 x4.12 = 0.632 kg nr1 s-1

r

From the equation,

V - m§
1
67tkr

if r' - 2r and m is constant, then y ' = jvt,

i.e., the terminal velocity will be half of what it was.

08) Paper 2 (2008) OTHER ISLANDS CAPE PHYSICS

(2008) PAPER 2 -Qu.2

Any membrane that is made to vibrate will cause the medium (air in this case) around it
to vibrate and so cause sound. The frequency of the sound produced in the case of a
string depends on the tension in the string, the length of the string and the mass per unit
length of the string.

A, = — = 1.84 m

(iii) For n antinodes there will be n half wavelengths. Son(f)

=L
Thermometer Advantage Disadvantage
Liquid in glass Not very accurate
Reads temperature
directly
Thermocouple Best for varying Does not read
temperature temperature directly
Constant Volume gas Very accurate for wide Inconvenient to carry
range

(b) (i) To keep the volume constant, adjust the right-hand side tube up and
down until the level in the left side tube (of the ruler) is back to a
specified mark on the ruler.

(ii) (a) When the bulb is in pure melting ice the right arm is adjusted so that the
left arm comes back to the designated mark on the ruler. The
difference in mercury levels is h0.

(b) At 100°C, again the difference in mercury levels will be h 100.

(c) At temperature t°C, the difference in mercury levels will be h (. In all

cases, the left hand side must be brought back to the designated
standard mark.
 Paper 2 (2008) OTHER ISLANDS CAPE PHYSICS

(2008) PAPER 2-Qu.3

(c) (i) t= 16-8 " 5'° x 100 20.0-5.0

= 1L8X100
15
= 78.7°C

(ii) The pressure of the gas in the bulb will be Atomospheric

Pressure + Ahpg

P = (0.76 + 0.168) 13600x9.81 =

1.24 x 105 Pa
 CAPE PHYSICS OTHER ISLANDS June Paper 2 (2008) Paper 2 2(

(2008) PAPER 2-Qu.4

4 (a) (i) If the body is moving in a circle, then the direction is continuously changing.
So even if the speed is constant, the velocity changes because direction
changes (velocity is a vector, speed is a scalar). Acceleration is rate of change
of velocity, so if velocity is changing, then there must be an acceleration.
v2
a=—
r
(ii) This acceleration is always directed towards the centre of the circle.

(iii) Work done is force multiplied by distance moved in direction of force. The
centripetal force is directed towards the centre of the circle whereas the
velocity (direction of motion at any instant) is tangential. So, F is
perpendicular to the direction of motion. So no work is done by that force.

(b) (i)

mg

(ii) (a) Resolving vertically,

T Cos 30 mg
1x9.81
T = 11.3 N
Cos 30

180
Paper 2 (2008) OTHER ISLANDS CAPE PHYSICS

(2008) PAPER 2-Qu.4

(b) Finding the radius of the circle,

Sin 30 = —
0.5
r = 0.25 m
„ mv 2

So-------= T Sin 30
r
113 1
v2 = —x-x0.25 = 1.4125 1 2
v = 1.19 ms-1

(c) The mass will move in a parabolic path with an initial vertical
velocity of zero.

Looking from the side.

s = ut + ^at2 1.5
= -j(9.81)t2

9.81 t =
0.55 s

181
 CAPE PHYSICS OTHER ISLANDS June Paper 2 (2008) 1 Paper 2 (20

(2008) PAPER 2 -Qu.5

(a) (i) Role of diffraction:

For each slit, spreading of the waves takes place.

Same X on both sides.

Diffraction takes place because the size of the slits is about the size of the
wavelength of the light waves.

Role of Interference:
(b)
These wave fronts now have a chance to meet and so interfere with each
other. In directions in which the wave fronts meet inphase, there is
constructive interference and the directions in which the wave fronts meet out
of phase there will be destructive interference.

Each colour diffracts by a different amount depending on the wavelength as

given by

d Sin 0 = n/L

The bigger the wavelength, the greater the diffraction.

Direction of constructive interference.

(ii) From the formula d Sin 0 = n^

nX
or Sin 0 =

182
Paper 2 (2008) OTHER ISLANDS CAPE PHYSICS

(2008) PAPER 2-Qu.5

The shorter the wavelength, the smaller the 0. Blue has the shortest
wavelength of the three, so it will be diffracted least. C will be red, since red
has the longest wavelength and B will be yellow.
nX-
(iii) From the formula Sin 9 = —
d

For zero order, n = 0,

0 will be zero for all X .

So all the colours will overlap at O.

(b) (i) d =---------r m = 1.67 x 10-6 m

6xl05
nX
Sin 0 = — < 1 using longer X 590 nm d
So n < —
X
<_l_xl0^_
"" 6x5.90xl0-7
<2.8
So max n is 2nd order.

(ii) When n = 2,
2 x 5.89 x IQ"7
Sin0j
1.67 xlO-
6
0
1 44.86°
Also Sin 0^ 2x5.90xl0~7
1.67 xlO"
0 6
2
So angular separation is 0.10°.

183
CAPE PHYSICS OTHER ISLANDS June Paper 2 (2008) Paper 2 (2008#

(2008) PAPER 2-Qu.6

(i) Au => The increase in internal energy of the gas. (iii) i S

Q is the heat supplied to the gas.

W is the work done ON the gas by the surroundings.

(ii) At constant pressure,

Q = Au + w for expansion at constant pressure.

Q is heat supplied nCp AT

In this case, heat supplied goes to increase the internal energy (hence
temperature) plus do work on the surroundings.

If the volume is kept constant then W = 0, and so all the energy supplied
nCyAT goes to increase internal energy alone (no change in volume) so less
energy is needed to raise the temperature of one mole of the gas by one
kelvin. Hence Cp > Cv.

(iv) ■

(b) (i) Net work done is given by area enclosed by loop.

Net work done = P x V

= 1.01 x 105 x 0.0225
= 2.27 x 103 J

(ii) Pl=2^
TI T2
T2 = 2Tj = 2 x 273 = 546 K

184
Paper 2 (2008) OTHER ISLANDS CAPE PHYSICS

(2008) PAPER 2-Qu.6

(iii) From 1 — > 2 no change in volume.

So AQ = Au = nCvAT

= lx-^Rx(546-273)

= 3400 J

From 2 —> 3
AQ = Au + PAV =
nCvAT + PAV

= 1 x |R x (1092 - 546) + 2P x V
= 6800 +2(2.27+ 103)
= 11.34xl03 J

(iv) Efficiency =
Useful work done 100
x
Total energy supplied 1
2.27 x 103 100