NJC_2024_H1_Physics_Prelim_P2_Ans

[Turn over Section A Answer all the questions in this section. 1 An object is launched at a speed of 30 m s -1 with an angle of 60 ° from the ground as shown in Fig. 1.1. Ignore air resistance. (a) Show the time taken for the object to reach its maximum height is 2.65 s. [1] Consider vertical motion: ݑ௬ = 30݊݅ݏ݊݅ݏ 60 (upwards), ܽ= 9.81 (downwards), ݒ௬ = 0 at maximum height Using ݒ௬ =ݑ௬ +ݐܽtaking upwards positive: 0 = 30݊݅ݏ݊݅ݏ 60 + (−9.81)ݐ ݐ= ଷ଴௦௜௡௦௜௡ ଺଴ ଽ.଼ଵ = 2.648 ≈ 2.65 s (shown) (b) Calculate the time taken for the object to hit the ground. Time taken = 2 × 2.65 = 5.30 s time = ………………… s [1] (c) Hence or otherwise, calculate the horizontal distance travelled by the object from the point of launch to the point it first hits the ground. Consider horizontal motion: ݏ௫ = (ݑݏ݋ܿ ݏ݋ܿ ߠ )×ݐ= (30ݏ݋ܿݏ݋ܿ 60 ) × 5.30 = 79.5 or 80 m horizontal distance = ………………… m [2]

2 (d) On Fig. 1.2 below, sketch the variation with time t of the vertical component of the velocity vy from the time it leaves the ground to the time it hits the ground. [2] Straight line negative gradient (1 mark) Start at (0, 26) and end at (5.3, -26) (1 mark)

3 [Turn over (e) On Fig. 1.3, sketch (i) the variation of the horizontal component of the velocity with time of the object for the duration of time in flight. Label this line A. [1] Horizontal speed (ݑݔ)= 30ݏ݋ܿݏ݋ܿ 60 = 15 m s-1 is constant. (ii) the variation of the horizontal component of the velocity with time of the object for the duration of time in flight if air resistance is not negligible. Label this line B. [1] Horizontal speed decreases from 15 m s-1 to zero in a time of less than 5.3 s. In the presence of air resistance, horizontally net force = −ݒ݇ where ݇ is a constant ܽ݉= −ݒ݇  ܽ= − ቀ ௞ ௠ቁݒ  gradient of curve B = acceleration proportional to −ݒ [Total: 8] A B

4 2 An aeroplane of mass 1.5 x 10 5 kg moves horizontally with constant velocity. The forces exerted on the aeroplane are as shown in the Fig. 2.1 below. Fig. 2.1 (a) Calculate the values of the lift and the drag. Moves horizontally  vertical forces are balanced, hence Lift = weight = 1.5 x 105 x 9.81 = 1.47 x 106 N Moves horizontally with constant velocity  horizontal forces are also balance, hence Drag = Thrust = 0.60 x 106 N Lift = ….........…….… N [1] Drag = ….........…….… N [1] (b) (i) Define torque of a couple. The product of one force in the couple and the perpendicular distance between the two forces in the couple. [1] (iv) The horizontal separation of the lines of action of lift and weight is 0.80 m. Using your answer to (ii) and (iii), determine the vertical separation of the lines of action of the thrust and drag. Refer to Fig. 2.1. The lift and weight are equal in magnitude, opposite in dire