2020 A level H2 Physics P2 Soln

1 Classified as OFFICIAL (OPEN) 2020 A Levels H2 Physics 9749/02 Paper 2 suggested solutions 1a Since the forward driving force is equal and opposite to the backwards frictional forces on the cyclist, there is zero net force on the cyclist. By Newton’s 1st law, given that there is zero resultant force on the cyclist, the cyclist continues in its state of uniform motion at constant speed in a straight line. 1b 𝑣 = √2𝐹0.5 𝐶𝐷 0.5𝜌0.5𝐴0.5 = 11.411 𝛥𝜈 𝑣 = 1 2 (𝛥𝐹 𝐹 + 𝛥𝐶𝐷 𝐶𝐷 + 𝛥𝜌 𝜌 + 𝛥𝐴 𝐴 ) = 1 2 ( 2 22 + 0.01 0.88 + 0.1 1.2 + 0.02 0.32) = 0.12305 Δ𝑣 = 0.12305 × 11.411 = 1.4041 = 1 𝑣 = 11 ± 1 m s−1 1ci Power = rate at which work is done. Power = Work done Time taken = Force × Displacement in the direction of the force Time taken Power = Force × velocity in direction of the force 1cii 𝑃 = 𝐹𝑣 = 22 × 11.411 = 251.042 = 251 W 2a It is the energy per unit mass required to bring a small test mass from infinity to that point without a change in its kinetic energy. 2b Δ𝐸P = − 𝐺𝑀𝑚 𝑟orbit − (− 𝐺𝑀𝑚 𝑟surface ) = −(6.67 × 10−11)(6.0 × 1024)(1600) ( 1 2.7 × 107 − 1 6.4 × 106) = 7.6334 × 1010 J 2ci Gravitational force provides for centripetal force 𝐺𝑀𝑚 𝑟2 = 𝑚𝑣2 𝑟 1 2 𝑚𝑣2 = 𝐺𝑀𝑚 2𝑟 Kinetic energy = 𝐺𝑀𝑚 2𝑟 2cii 𝐺𝑀𝑚 2𝑟 = (6.67 × 10−11)(6.0 × 1024)(1600) 2(2.7 × 107) = 1.19 × 1010 J 2ciii The method is incorrect. On the Earth’s surface, the satellite is not orbiting around the Earth solely due to the gravitational attraction due to the presence of Normal contact force. Hence, the kinetic energy of the satellite on the surface of Earth will not follow the equation in 2(c)(i).

2 Classified as OFFICIAL (OPEN) 3a Correct shape [M1] Uniform potential at middle of potential of R and S [A1] The total change in potential in moving from R to S is the same. However, as a conductor will have a uniform potential within the entire conductor, the potential from x to 2x should be constant (and by symmetry, the potential is in the middle of the potentials of R & S). Tips: Draw the constant potential from x to 2x first, then connect to the potentials at R and S. 3b The electric field strength is given by the negative of gradient of the potential- distance graph (i.e. potential gradient). Uniform field strength between 0 to x and 2x to 3x [B1] Field strength is 1.5 times higher than original between 0 to x and 2x to 3x [B1] zero field strength between x to 2x [B1]

3 Classified as OFFICIAL (OPEN) Since the potential gradient is now 3/2 = 1.5 times higher between 0 to x and 2x to 3x, the electric field strength in these regions should be 1.5 times of that across PQ. Within the conductor, the potential gradient is zero, hence the electric field strength is also zero. 4a Diffraction refers to the spreading of waves after they pass through a small opening or round an obstacle. Note: For diffraction to occur, the waves have to first pass through a slit or round an edge or obstacle. Avoid imprecise description su