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Answers to JC2 Preliminary Examination Paper 3 (H2 Physics)
Suggested Solutions:
No. / Solution1(a) / Negative acceleration is when the acceleration of a body is opposite to direction of travel assigned to be in the positive direction but it can also mean the body is slowing down. Deceleration means the process of slowing down (speed decreases).
1(b)(i) /
ms−1
1(b)(ii) /
s
Therefore, distance apart = 1.23 m.
1(b)(iii) / Large angle gives a longer time of flight for a given projection speed and hence juggler has more time to manipulate balls.
Large angle keeps (horizontal) range small and within the natural distance between the juggler’s hands.
1(b)(iv) / In reality, air resistance would lead to a smaller range leading to the ball not landing within juggler’s left hand. Therefore, speed of projection should be greater.
2(a) / The intensity of a progressive wave is the energy per unit time passing normally through a unit area perpendicular to the direction of propagation of the wave.
2(b) / Diffraction refers to the spreading of light waves when they pass through each of the apertures or obstacle of the diffraction grating.
The width of the apertures has to be comparable to the wavelength of the light waves for diffraction to be observable.
2(c)(i) / For the 588.99 nm wavelength,
Since , we have
Therefore, the number of orders observable is 3.
2(c)(ii) / The maximum angular separation occurs at the highest order.
For the 588.99 nm wavelength,
For the 589.59 nm wavelength,
Maximum angular separation = 62.176°– 62.066°
= 0.11°
3(a)(i) /
Ω
3(a)(ii) / Energy supplied
= Pt
= (1200)(60)
= 7.20 × 104 J
3(b)(i) /
Since 3 heating elements are used in series, 3R = 48 Ω, R = 16 Ω.
3(b)(ii) / Since there is an inverse relationship between resistance R and cross-sectional area A (), when A decreases, resistance of the nichrome wires will increase.
The power of the toaster is then inversely related to the resistance (), when the resistance increases, the power of the toaster will decrease.
4(a) / Magnetic flux density is defined as the force acting on a straight conductor of unit length, carrying unit current, placed perpendicularly to the magnetic field.
SI unit: Tesla
4(b)(i) /
4(b)(ii) / N
4(c)(i) / As wire cuts the magnetic field, it experiences a change in magnetic flux. According to Faraday’s law, an e.m.f. is induced across wire.
4(c)(ii) /
ms−1
5(a) / Photoelectric effect is the emission of electrons from a metal surface when it is exposed to electromagnetic radiation of sufficiently high frequency.
5(b)(i) /
5(b)(ii) / Power = Intensity × area of illumination
s−1
5(b)(iii) / Light of higher wavelength has lower energy. When photons of lower energy are absorbed by electrons in the rhodium metal, the maximum K.E. of photoelectrons emitted is lower.
Since the stopping potential depends on the maximum K.E. of photoelectrons emitted, the stopping potential needed is lower.
6(a)(i) / rads−1
Highest tension occurs when toy satellite is at lowest point of a vertical circle.
2.3 N
6(a)(ii) / There is a gain in G.P.E. but the K.E. remains constant. Hence, work has to be done by the motor (external force).
6(a)(iii) / Work done = mgh = (0.200) (9.81)(0.30) ≈ 0.589 J
6(b)(i) / A geostationary satellite must
- be in the same plane as the equator
- revolves from west to east
- revolves with a period of 24 hours
6(b)(ii) / Both the spacecraft and astronaut are falling towards the centre of Earth with the same acceleration.
Hence, the normal contact force on the astronaut by the spacecraft is zero.
6(b)(iii)1. / Resultant force on satellite = Gravitational force on satellite
6(b)(iii)2. /
6(b)(iv)
6(b)(v) / Atmospheric resistance causesto decrease, and so radius of orbit decreases.increases and satellite’s speed increases.
7(a) / When a liquid is boiling, the separation of its particles increases. There is thus an increase in microscopic potential energy.
There is no change in temperature and hence no change in average microscopic kinetic energy.
Therefore, the internal energy of the substance increases.
There is work done against atmosphere during this expansion.
Since W is negative and is positive, by reference to first law of thermodynamics, Q has to be positive.
Therefore, thermal energy must be supplied.
7(b)(i) / Specific latent heat of vaporisation of a liquid is the quantity of heat (thermal energy) required to convert per unit massof the liquid, at its boiling point,into gas at the same temperature.
7(b)(ii) /
7(b)(iii) / Current = A
7(b)(iv)1. / There is thermal energy loss H to the surroundings for the duration of 5.00 minutes.
H is the same if experiment is done with two different values of P. Hence, H can be eliminated using two equations.
7(b)(iv)
2. / ----- (1)
----- (2)
Solving, we have
Jkg−1
7(c)(i) / Some of the water is vaporised by the solar energy and escape into the atmosphere.
7(c)(ii) / Surface area of canal = m2
Power P incident on canal = W
Mass rate of vaporisation = kgs−1
Volume rate of vaporisation = m3s−1
Rate of supply from reservoir = 2.7 + 0.1868
= 2.8868
m3s−1
8(a)(i) /
8(a)(ii) / Nuclear fission reaction.
8(b)(i) / Nuclear binding energy of a nucleus is the minimum amount of energy required to break the nucleus into its constituent particles. It is also the energy released when a nucleus is formed.
8(b)(ii) / Energy released
MeV
Therefore, the binding energy per nucleon of Krypton-90
MeV
8(b)(iii) / Energy released in one hour due to nuclear reactions
J
Number of nuclear reactions required
Mass of Uranium-235 consumed per hour
g
8(c)(i) / Isotopes are nuclides with the same number of protons but different number of neutrons.
8(c)(ii)1. / Since the radiation cannot be stopped by the alumimum sheet, it can be either or radiation. However, radiation is unaffected by magnetic field since it is uncharged. Therefore the unknown radiation is radiation.
8(c)(ii)2. / The radiation will be slowed down by the aluminum sheet. The magnetic force on the particles will decrease which results in a smaller radius of curvature. Hence, the point of entry is at point X.
8(c)(iii)1. /
s−1
8(c)(iii)2. /
days
8(c)(iii)3. /