2022 A Level H2 Physics P2 Soln

2022 A Levels H2 Physics 9749/02 Paper 2 suggested solutions 1a     22 1 22 1 1 7.2 0.72 s10 2 4 0.1204 9.1385 N m 0.72 2 1 0.22 9.1385 100 7.2 0.6 N m (1 s.f.) 9.1 0.6 N m T mT k mk T k m T k m T k T t T t k k                                       1bi From Fig. 1.2, 1 0 1.4 m sv  and 0 0.16 mx        0 0 1 22 2 0 0 1.4 0.16 8.75 rad s 8.75 0.16 12.3 m s v x a x              1bii Taking downwards as positive, the initial displacement will be positive (assuming the mass is pulled down), and upon release, the velocity will be upwards (hence negative. 2ai Vector is a quantity that has both magnitude and direction. 2aii Force or velocity or momentum etc. 2bi Component of weight along the slope = W sinθ = 16 sin35° = 9.18 N 2bii When going down the slope at constant speed, Fnet =ma = 0 Frictional force is equal to the component of weight along the slope = 9.18 N. On the way up the slope at constant speed, Fnet = ma = 0 P – frictional force – component of weight along slope = 0 Hence P = 9.18 x 2 = 18.4 N

2ci Impulse on X = change in momentum of X = mvf – mvi = 0.22 (3.5) – 0.22 (2.6) = 0.198 N s. 2cii Impulse on Y = – impulse on X = – 0.198 N s Hence Impulse on Y = mvf – mvi – 0.198 = 0.40 (vf) – 0.40 (3.3) vf = 2.81 m s-1. 2ciii Distance travelled by X = area under graph from 1.8 seconds to 2.0 s. = (0.5) (2.6+3.5) (0.20) = 0.61 m 3a For a mass in a gravitational field, the gravitational force is in the direction of the field. For a charge particle in an electric field, the electric force is either in the direction of the field for a positive charge or opposite for a negative charge. In both cases, the direction of the force is perpendicular to both its initial direction of motion. This causes an acceleration only in the direction of the force but no component in the direction of its initial velocity. Its component of velocity parallel to its initial direction of motion thus remains unchanged while its component of velocity parallel to the force increases from zero. This results in the object moving in a parabolic path. 3bi Gravitational force is always attractive in nature, thus the direction of the gravitational force FG is towards each proton. Since both protons have a positive charge, hence the electric force FE is a repulsive force and the direction is away from each proton. 3bii No. The protons are of extremely small mass, hence their gravitational force of attraction is many orders of magnitude smaller than their electic force of repulsion. 3ci No. The electric field strength due to positive charge P points away from P (rightwards) and the electric field strength due to negative charge R points towards R (rightwards). Being in the same direction, they add up vectorially thus the resultant electric field strength between P and R points towards the right. 3cii Yes. The electric potential due t