GYSS 4E Physics Prelim P2_answer

1 The diagram shows a model car with a mass of 0.25 kg. The car is travelling at a constant speed of 1.70 m/s as it approaches the rough ramp. The average friction opposing the motion of the car down the ramp is 0.15 N. Assume air resistance is negligible. (a) Calculate the work done against friction as it moves down the rough ramp. work done = f x d = 0.15 x 0.42 = 0.0630 J [1] (b) Calculate (i) the loss in gravitational potential energy as the car moves down the the ramp, Ep = 0.25 x 10 x 0.8 = 2.00 J [1] (ii) the speed of the car at the bottom of the rough ramp. Ek = 2 + (½ x 0.25 x 1.72) – 0.063 = 2.29825 [1] 2.29825= ½ x 0.25 x v2 v2 = 18.386 v = 4.29 m/s [1] (c) Draw a free body diagram of the other forces acting on the car when it moves down the ramp, labelling the forces acting on the car. [Total: 6] weight friction contact force [2] 0.80 m 0.42 m

2 Fig. 2.1 shows a lorry accelerating along a straight road. The driving force exerted on the lorry in the forward direction is D and the total backward force acting on the lorry is B. Fig. 2.1 The gravitational field strength is 10 N/kg. (a) Compare the magnitude of the forces D and B. Give a reason for your answer. Force D is greater than force B, so that the resultant force acting on the lorry is in the forward direction, causing it to accelerate forward. [1] (b) The lorry subsequently turns a corner at a constant speed of 12 m/s. (i) Express the speed of the lorry in km/h. 12 m/s = (12/1000)/(1/3600) = 43.2 km/h [1] (ii) State and explain whether the lorry is accelerating when it turns the corner. Since the direction of the lorry’s velocity changes as it turns the corner, there is a change in its velocity. Hence the lorry is accelerating, even though its speed is constant. [1] [Total: 3]

3 Fig. 3.1 shows a uniform plank of mass 5.0 kg and length 2.0 m resting horizontally on two trestles, P and Q. Fig. 3.1 Trestle Q exerts a force of 440 N on the plank when an object of mass 65.0 kg is placed at a distance, d, from P. The gravitational field strength, g = 10 N/kg. (a) Calculate the moment due to the weight of the plank. moment = 50 x 1 = 50 Nm [1] (b) State the principle of moment. For an object to be in equilibrium, the sum of clockwise moment about the pivot is equal to sum of anti-clockwise moment about the same pivot. [1] (c) Determine the distance d, of the object from P. sum of clockwise moment = sum of anti-clockwise moment (50 x 1) + (650 x d) = 440 x 2 [1] 5 + 650d = 880 650d = 830 d = 830/650 = 1.27