NJC 2022 SH1 H2 Physics Promo Paper 1 ans

[Turn over 2022 SH1 H2 Physics (9749) Promotional Examination Paper 1 Answers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B D B A B A C B D C A B A D C 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C B A D C D D C A B D A C B C Workings/Explanations 1 P − Q = P + (−Q) P is rightwards. −Q is upwards. P − Q is diagonally rightward and upward. 2 Accuracy is related to systematic error and Precision is related to random error. Having little systematic error means that the results are accurate. Why A is wrong – Removing systematic error such as zero error improves accuracy. Why B is wrong – Taking the mean value of the experiment reduces random error and improve precision and does not affect the accuracy. Why C is wrong – If mean value is close to the true value, then it is accurate. 3 2(1%) + 2% + 2% = 6% 4 The first ball bearing always has a larger speed than the second, so the distance between them keeps increasing. 5 height of object above ground – always decreasing (with decreasing gradient) until reaches constant negative gradient at terminal velocity therefore Y speed of object – starts from zero always increasing until remain constant at terminal velocity therefore Z magnitude of resultant force on obje ct – decreasing (due to air resistance increasing and weight remaining constant) until reaches zero at terminal velocity therefore X 6 Let initial velocity of elevator be u (so it is the same for the ball). At time t the ball hits the floor of elevator. For ball: 2 ball 1 2s ut gt=− (this is the displacement from its initial position) For floor: 2 floor 1 5.82s ut t= +  When they meet, taking reference from the floor as the initial position, (since ball is 2.0 m above floor, it has an initial displacement of 2.0 m) floor ball 2.0ss =+ 2211 5.8 9.81 2.022ut t ut t+  = −  + 0.51t = s 7 rate of change of momentum = resultant force = √(10000 − 9000)2 + 5002 = 1100 kg m s−2 8 impulse = change in momentum = 0.100 (−30 − 20) = −5.0 N s 9 linear momentum of 2.0 kg trolley = 2.0 (2.0) = 4.0 kg m s−1 (left)

2 therefore linear momentum of 1.0 kg trolley is 4.0 kg m s−1 (right) by Conservation of Linear Momentum since Initial Total Linear Momentum is zero therefore speed of 1.0 kg trolley = 4.0 m s−1 energy stored in spring = Total final kinetic energy = ½ (2.0)(2.0)2 + ½ (1.0)(4.0)2 = 12 J 10 The lines of action of F, the weight of the barrel and the resultant force of the ramp must meet at a common point so that resultant moment about that point is zero. The three forces will also form a closed vector triangle to ensure resultant force acting on the barrel is zero. 11 Take moments about the pin: Anti-clockwise moment = Clockwise moment ( ) + =  cos 40 20 sin 40F r r r F = 7.3 N 12 The same mass of sand displaces a smaller volume of air. Upthrust on sand is less than that for cotton. Contact force + upthrust = weight Contact forc