HCI 03 Dynamics Solutions (Self-Review Questions)

1 2023 Dynamics Tutorial - Suggested Solutions for Self-Review Questions Self-Review Questions Part 1: Newton’s Laws, Inertia, Force, Momentum, Impulse S1 C Net force is equal to the rate of change of momentum (Newton’ s second law). S2 u = 0 and v = 270 km h1 = 270 / 3.6 = 75 m s1. Hence, Δv = v – u = 75 m s1. Δp = m Δv = 20,000 x 75 = 1,500,000 kg m s1 and Δt = 2.0 s. Favg = Δp / Δt = 1,500,000 / 2.0 = 750,000 N = 750 kN. S3 C Favg = Δp / Δt = 40 / 5 = 8 N. Hence, the impulse is Δ p = (8 N) x (5 s) = 40 N s. The impulse is also equal to the area under the F-t graph. Hence, Δ p = 40 N s = ½ (5 + 3) x = 4x  x = 40 / 4 = 10 N. S4 C The two forces in an action-reaction pair must be of the same type (gravitational) and due to the 2 bodies interacting (the man and the Earth). S5 a) Consider all 3 blocks as one body. By Newton’s second law, taking rightwards as positive, F = m a 30 N = (2.0 + 4.0 + 3.0) a a = (30 / 9.0) m s 2 Applying Newton’s 2nd Law separately on each block, net force acting on A = (2.0) (30/9.0) = 6.7 N to the right, net force acting on B = (4.0) (30/9.0) = 13 N to the right, net force acting on C = (3.0) (30/9.0) = 10 N to the right. b)(i) Consider only block C and use that the net force on C is 10 N. Since the only force acting on C is the force exerted by B on C, F B on C = 10 N to the right. b)(ii) Consider only block A and use that the net force on A is 6.7 N . Since only two forces are acting on block A, 3 0 + (  F B on A) = 6.7 FB on A = 30  6.7 = 23.3 N FB on A is 23.3. N to the left. A 30 N FB on A FB on C A B 30 N C C

2 Part 2: Conservation of Linear Momentum / Collisions S6 D Linear momentum is always conserved in a closed system if no net external force acts on the system. Kinetic energy is conserved only for elastic colli sions. But total energy is conserved for all types of collisions (again for a closed system). S7 C Taking to the right as positive, the initial total linear mome ntum is 20 + (-12) = 8 N s. By the principle of conservation of linear momentum, the final linear momentum is also 8 N s. Thus, (-2) + pY’ = 8 N s  pY’ = 8 + 2 = 10 N s. S8 D “Moves off together” => all the masses share the same velocity after collision. Hence, this is a perfectly inelastic collision. Applying PCOLM, 2 m (5.0) = (2m + 4m) v v = 1.7 m s 1 S9 a) At velocity = 0, acceleration = 0 m s-2, thus the net force acting on the helicopter = 0 N b) In equilibrium, upward force produced by rotor = weight of helicopter = 5500 (9.81) = 53 955 N = 54.0 kN c) By Newton’s third law, magnitude of force exerted by rotor on air = m