WJEC Physics for AS Level: Revision Workbook

Q9 (a) Perpendicular component = 5.0 kN × cos 75° = 1.3 kN [1.29 kN to 3 sf] (b) 1.3 kN (c) 1.29 kN = F × sin 10° So F = sin 10° 1.29 kN = 7.4 kN [7.43 kN to 3 sf] (d) Since force components at right angles to forward direction cancel, Resultant force = forward component of 5.0 kN + forward component of F = 5.0 kN × cos 15° + 7.43 kN × cos 10° = 4.83 kN + 7.32 N = 12 kN (to 2 sf) Q10 (a) The vector sum of the forces on the object is zero. (b) The sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point. Q11 (a) at t 1 , v horiz = 15.0 m s −1 × cos 30.0° = 13.0 m s −1 v up = 15.0 m s −1 × sin 30.0° (or cos 60.0°) = 7.5 m s −1 (b) at t 2 , v horiz = 20.0 m s −1 × cos 49.5° = 13.0 m s −1 v up = –20.0 m s −1 × sin 49.5° (or cos 40.5°) = −15.2 m s −1 So, Δ v horiz = 0 And Δ v up = (–15.2 m s −1 ) − (+7.5 m s −1 ) = –22.7 m s −1 So the ball’s change in velocity is 22.7 m s −1 in the downward direction. Q12 Magnitude of − = + 12 10 2 1 2 2 v v = 15.6 m s −1 Ʌ = £ ¤ ¥ ¦ −1 tan 10 12 =39.8° ׵ Bearing = 230° to nearest degree Q13 m k kg N m kg kg m s m −1 −1 −2 = = ^ h ; E = s −2 , so k m 9 C = s 2 Now, [ T ] = s, so the first two of the given equations are clearly wrong. And [ T 2 ] = s 2 , so the third equation could be right, but the fourth is wrong. Q14 (a) π π A r 0.317 10 m − 2 3 2 × × = = 2 c m = 7.89 × 10 −8 m 2 (b) Volume = Al = 7.89 × 10 −8 m 2 × 0.85 m = 6.71 × 10 −8 m 3 (c) Volume = mass density 8.96 10 kg m 2.50 kg 3 −3 = × = 2.790 × 10 −4 m 3 l A volume 7.893 10 m 2.790 10 m − − 8 2 4 3 × × = = = 3.54 × 10 3 m = 3.54 km Q15 (a) Pressure = area of surface (magnitude of) force normal to a surface So (magnitude of) force = pressure × area = 2.5 × 10 6 Pa × π (0.100 m) 2 = 7.9 × 10 4 N to 2 sf (b) Upward force of gas on piston = weight of copper piston + downward force of air on piston. So pA = ɏ + p A A in which A is the cross-sectional area of the copper piston, l is its length and ɏ is the density of copper, so Al is the volume of the copper cylinder, and ɏ is its mass. Dividing through by A , and then substituting numerical data p = ɏ + p A = 0.10 m × 8.96 × 10 3 kg m −3 × 9.81 m s −2 + 101 × 10 3 Pa = 110 × 10 3 Pa v 1 = 12 m s −1 v 2 − v 1 θ v 2 = 10 m s −1 Unit 1 Practice questions: Answers | 185 Unit 1 Answers