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 direction and acting at two different points so these two forces formed a couple. The torque is anti-clockwise. Torque due to lift and weight = 1.47 x 106 x 0.80 Since the aeroplane is not rotating, the torque due to thrust and drag must be clockwise and same magnitude as the torque due to lift and weight.

5 [Turn over (0.60 x 106)(y) = 1.47 x 106 x 0.80  y = ଵ.ସ଻ ଴.଺଴ × 0.80 = 1.96  2.0 m 3 (a) State the principle of conservation of linear momentum. In the absence of a resultant external force acting on a system of interacting objects, the total momentum of the system is constant. (b) Fig. 3.1 shows two discs, A and B, on a frictionless table collide head -on. Disc A has a mass of 0.36 kg and disc B has a mass of 0.18 kg. Before colliding, disc A has a velocity of 0.40 m s1 and disc B a velocity of 0.10 m s-1 in the opposite direction. On colliding they stick together. Fig. 3.1 Calculate (i) the velocity of the discs after the collision. From momentum conservation and taking velocity to the right as positive: (0.36)(0.40) + (0.18)(0.10) = (0.36 + 0.18)(v) v = 0.233 m s-1 velocity = …………….………… m s -1 [2] direction = ……………………………… [1] (ii) the kinetic energy lost during the collision expressed as a percentage of the initial kinetic energy of the two discs. Total kinetic energy before collision = ଵ ଶ (0.36)(0.40)ଶ + ଵ ଶ (0.18)(0.10)ଶ = 0.029 7 Total kinetic energy after collision = ଵ ଶ (0.36 + 0.18)(0.233)ଶ = 0.014 658 0.233 rightwards

6 Percentage lost = (଴.଴ଶଽ଻ି଴.଴ଵସ଺ହ଼) ଴.଴ଶଽ଻ × 100% = 51% [3] [Total: 8] 4 A company rents out tower cranes of many different sizes. A tower crane is illustrated in Fig. 4.1. This type of tower crane is called a flat -top tower crane because the jib and counter jib are horizontal. A crane can be constructed to different arrangements of height, jib and counter -jib length, and balancing load. The size of the base can be varied to cope with different maximum loads lifted by the crane. Note: The masses of the loads in Fig. 4.1 and in Table 4.1 are given in tonnes (t). One tonne is 1000 kg. Fig. 4.1 Distance x is the fixed distance. This is a different distance for each different crane arrangement. Distance y is variable and changes as the load L is moved in and out from the tower, along the jib. Table 4.1 lists information for four different crane arrangements. x y counter jib jib tower load L 16.0 t balancing load force due to mass of structure width w 22.0 t 22.0 t

7 [Turn over The maximum load L in tonnes that can be lifted for different distances y from the centre of the tower for each arrangement is also shown. Table 4.1 crane arrangement total length of jib and counter jib / m distance x to 16.0 t balancing load / m Maximum load L at different distances y / t y = 30 m y = 52 m y = 75 m A 95.0 17.3 8.48 4.31 2.60 B 75.0 19.4 9.79 5.15 − C 75.0 21.1 10.81 5.77 − D 55.0 22.3 11.53 − − (a) (i) Calculate the weight of the 16.0 t balancing load. Weight = 16.0 x 1000 x 9.81 = 1.57 x 105 N weight = ………………………. unit ……….. [2] (ii) Using the data in Table 4.1, explain why there is no detail provided for crane D when y = 52 m. Referring to Fig 4.1 and the data for crane D in table 4.1, Maximum value of y  55.0 22.3 = 32.7 m. Hence, no data for y = 52 m [1] (b) (i) Show, for crane A, that the load and the balancing load given in the table can never put the crane into equilibrium. Taking moments about the mid-point of the tower: Anti-clockwise moment due to 16.0 t balancing load = 1.57 x 105 x 17.3 = 2.72 x 106 Nm When y = 30 m, clockwise moment due to load = (8.48)(1000)(9.81)(30) = 2.50 x 106 Nm When y = 52 m, clockwise moment due to load = (4.31)(1000)(9.81)(52) = 2.20 x 106 Nm When y = 75 m, clockwise moment due to load = (2.60)(1000)(9.81)(75) = 1.91 x 106 Nm Clockwise moment due to the load is always less than anticlockwise moment due to the balancing load, hence the two loads cannot put the crane into equilibrium. [3]

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9 [Turn over (ii) When in use, crane A is in equilibrium. Suggest how this is achieved. The weight of the crane structure and the weight of the 22.0 t base will provide the necessary moment to keep the crane in equilibrium. [2] (c) The width w of the base of a crane is important in providing stability. For crane C , the foundations of the base are two identical large cubic concrete masses, each of mass 22.0 t. These masses are firmly attached to the crane. The total mass of the crane structure is 17.0 t and the force due to the mass of the crane acts through the centre of the legs. The balancing load is 16.0 t and is 21.1 m from the centre of the tower. By taking moments about point P in Fig. 4.1, determine, for zero load, the minimum possible value of w before the