WJEC Physics for AS Level: Revision Workbook

(ii) Statement (in words): Either : least mass if hung from end of bar Or : Wire further from mass will be slack (or equiv) when rod about to tilt [1] (If hung from left): mg × 0.2 [m] = 11.8 [N] × 1.3 [m] [1] m = 7.82 kg [1] 3 3 1 1 Total 3 6 3 12 8 Rhodri’s answers (a) The sum of the upward forces on the object is equal to the sum of the downward forces. (not enough) The sum of the clockwise moments equals the sum of the anticlockwise moments. ✓ غ (b) (i) (I) 35.316 11.76 11.8 N 35.3 N ✓ غ (II) Moments about B 11.8 × 1.3 + 35.3 × 0.6 = T A غ Ą 7 A = 36.5 N غ T A + T B = 11.8 + 35.3 ✓ ecf Ą 7 A + T B = 47.1 Ą 7 B = 47.1 - 36.5 = 10.6 N ✓ ecf (ii) Tilting most likely if mass hung from end ✓ mg × 0.2 = 11.8 × 1.3 + 10.6 × 2.6 غ mass = 22 kg غ MARKER NOTE We need force components in any direction to sum to zero. First mark not gained. Third mark not gained as it’s important to state that the clockwise and anticlockwise moments are taken about the same point. 1 mark MARKER NOTE The second mark is lost for the omission of the unit. Rhodri has given too many significant figures, but this is more likely to be penalised in a question based on experimental results. Best avoided though! 1 mark MARKER NOTE In the first equation Rhodri has written T A instead of 2.6 [m] × T A . Writing a force instead of a moment is an error of principle (and a surprisingly common one), so the first and third marks are lost. However, T A + T B = 11.8 + 35.3 is the correct force equation and is used correctly to find T B , with ecf on T A . So second and fourth marks are gained. 2 marks MARKER NOTE Only the first mark awarded for the realisation of where the mass needed to be hung. Rhodri did not realise that the tension in the far wire would be zero and (in desperation?) used the previous value. 1 mark TOTAL 5 marks / 12 Section 1 Basic physics | 21 Unit 1 Practice questions