1a) The rate of change of momentum of a body is directly proportional to the net external force acting on the body and takes place in the direction of the force.
b) The tension is not zero and act opp to W , hence resultant force is less than mg , acc is less than g
ii) Write two equation for the bucket A and B , solve to obtain expression for acc . Note the direction positive is down for bucket A and up for bucket B .
iii) Sub the mass in the expression to obtain a = 1.6 ms-2 . Use equation of motion to obtain time taken to travel 4 m. time = 2.2s
2i) resistance is infinite from 0 to 0.6 V. Resistance is negligible from 0.6 V onwards.
iii) A curve of decreasing gradient and starting from origin.
b) R = pL/A but V=AL , subs to obtain R=pL2/V
ii) Using the formula above, fractional change in resistance is equal to twice the fractional change in length. The fractional change in length = 0.052 (5.2%) . The fractional change in resistance = 0.104 .The new length is 1.104 of the initial length
b) Draw two lines one directly from source to X , the other to mirror and then to X. ii) At X, constructive interference occur implies phase diff = 0 . Which also means other angles like 360 deg and 720 deg.
c) Use the young double slit formula to calc wavelength.The fringe separation is read from the fig. 3.3 as 0.45 mm , wavelength is 6x10-7 m.
ii) if one of the slit is blocked , the intensity arriving will be 1/4 (amplitude A) of the maximum intensity at constructive interferece (2A)
4 a) Minimum frequency of EM radiation that can emit photoelctrons from the material.
b) As the pd becomes less negative, some of the more energetic photoelectrons will be able to reach the collector and constitute the photoelectric current.
c) eVs = 1/2 mv2
v = 8.6 x 105 ms-1
d) At the intensity is halved , the photo-current will be halved but the KE remain the same hence cut off voltage the same. Indicate which is original and which is the answer.
5a)one kg yields 45x 106 J, hence by proportionality 28 kg will yield 1260 MJ
b) Power = VI and power = energy / time , hence time = energy / VI= 8.1 Hour.
c) 20 kg yields 2.8 MJ hence 1260 MJ needs 9000 kg .
cii) mass of car is about 1000kg m , the mass of the batteries are a few times greater than the mass of the car. Hence mor energy is required to carry the batteries in the car.
d) At constant speed , force of engine = Drag force. Work done by engine = 0.25 ( 1260 MJ ) Work done = F d where F = 580 N , d= 543 km
e) Power = kv3
power=k(0.8v)3
dividing first and second equation
Power /new power = 2, new power is half of the old power .
6. Summation of force is zero and summation of torque is zero.
i) At equilibrium , draw vector triangle that has arrows flowing from one to the next and ending at the starting point.
ii) Solve by sine rule the ratio T1/T2
Since tensions only depends on the angle , the weight will not change the ratio.
c) Taking moments about O, Wd1=Fd2 Where d2 = 42 cm and d1 = 18.3 cm.
F = 77 N
di) Pressure increases linearly with depth of seawater.
ii) Draw F1 shorter at the top and F2 longer at the base.
iii) Resultant = F2 -F1
= hpg A =1640N
2 Upthrust = weight
1640 = mg
m= 167 kg
3. Average density = mass /vol
= 167 /120 x0.19 = 733 kgm-3
iv) The grad is steepest at the earliest time 0.45s. Gradient is the speed.
7. a) he amount of charge which passes through a given cross section of a conductor is 1 second when a constant current of 1 ampere is flowing.
bi) Q = Ne
N = Q/e= 2.0 x 1020
bii) Q = IT
T= Q/I = 32/1.5 x 104=2.1 x 10-3 s
ci) The energy converted by a source from other forms of energy into electrical energy when a unit charge passes through it.
cii) R eff = 0.57 R
2 V= IR
I = E/0.57
R=1.57 E/R
di) V= IR2
V=(E/R1+R2)R2
diii) As the temp of the thermistor increases , its resistance decreases. Hence potential difference across thermistor decreases.This lead to corresponding increase in pd across 100 ohm as the two resistors are connected series.
ei) I = 2.0 A
ii) pd across R = 3V
r= 1.5 ohm.
8a) vertically downwards,
ii) the mass will move in parabolic path towards the ground.
bi) horizontally leftwards
bii) decelerate to a rest and then accelerated along the same line of motion in the opp direction.
c) the molecule will experience a clockwise rotation about its centre , with no translational movement .
di) see def.
dii) draw to scale : Ans 53 mT
e) force F = BIL
torque = Fxd= BIab = BIA Since B and I are constants , torque is directly proportional to area.
fi) r = 8.1 x 10-3 m
fii) F/L = BI= 3.3 x 10-3 Nm-1
