Physics (AS & A Level) · May / June 2016 · Paper 21 · Question 3
9702
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A ball of mass 150 g is at rest on a horizontal floor, as shown in Fig. 3.1. ball floor Fig. 3.1
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3(a)(i)[1]
force (= mg = 0.15 × 9.81) = 1.5 (1.47) N [1]
3(a)(ii)[1]
resultant force (on ball) is zero so normal contact force = weight or the forces are in opposite directions so normal contact force = weight or normal contact force up = weight down [1]
3(b)(i)[1]
(resultant) force proportional/equal to rate of change of momentum [1]
3(b)(ii)[3]
change in momentum = 0.15 × (6.2 + 2.5) (= 1.305 N s) magnitude of force = 1.305 / 0.12 = 11 (10.9) N or (average) acceleration = (6.2 + 2.5) / 0.12 (= 72.5 m s–2) (C1) magnitude of force = 0.15 × 72.5 = 11 (10.9) N (A1) (direction of force is) upwards/up [3]
3(b)(iii)[2]
there is a change/gain in momentum of the floor this is equal (and opposite) to the change/loss in momentum of the ball so momentum is conserved [2]
or change of (total) momentum of ball and floor is zero (M1) so momentum is conserved (A1) or (total) momentum of ball and floor before is equal to the (total) momentum of ball and floor after (M1) so momentum is conserved (A1)
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