HCI 03 Dynamics Solutions (Discussion Questions)

2023 Dynamics Tutorial - Suggested Solutions for Discussion Questions Part 1: Newton’s Laws, Inertia, Force, Momentum, Impulse  D1  D  Constant velocity implies a = 0, Fnet = 0. Hence, the first part of the graph should be zero. Constant deceleration implies a = constant, Fnet = constant, so the last part must be constant.  Comment:  Students tend to get confused or distracted by the term “deceleration” and leap to the conclusion that the  graph must be sloping downwards which is actually how the veloc ity is changing and not the acceleration or  net force.   D2  D  Constant force implies uniform acceleration. Applying equations of motion: v2 = u2 + 2as. The railway carriage starts from rest, so u = 0 Thus v = (2as)  p = m(2as), i.e. p  s Note: Slight error in the answer option. lim ௦→଴ ௗ௣ ௗ௦ ൌ0 (i.e. limit of the gradient) should approach infinity as s approaches zero. Problem‐solving skills:  A repeat of skills taught in kinematics regarding the use of an equation relating the 2 axes is deployed here.   Common mistake:  Students may choose A, thinking they are tested on force = rate of change of momentum.   Extension:  Try to plot momentum‐time graph.  D3  A  u + v = v    v = v + (u)   Using the sine rule,   24 sin120 sin30 v     v = 48 sin 120  Alternatively, use the cosine rule.  By Newton’s 2 nd Law, or impulse‐momentum relationship       (0.11)(48sin120 ) 180 N0.025 pm vF tt   Comment:  Students who are not as strong in their maths or vectors can choose to solve the components separately.  Taking rightwards and upwards as positive:   horizontally: Δvx = ‐24 cos 60o – 24 = ‐36 m s‐1;   vertically: Δvy = 24 sin 60o = 20.8 m s‐1;   total change in momentum is Δp = m Δv = m √36ଶ ൅2 0 . 8ଶ = 0.11 x 41.6 = 4.57 N s;   average force is <F> = Δp/Δt = 4.57 / 0.025 = 183 N = 180 N (2 s.f.)       v = 24 m s-1 u = 24 m s-1 60 120 30 v = v + (-u)

D4 At the instant of collision, if the collision is not head-on (line joining center of A and B is not along the initial direction of travel of A) the contact force between them will have a component perpendicular to the initial direction of travel of A. This means that there will be a change in momentum of sphere B in the perpendicular direction and the initial velocity of sphere B will no longer be along the direction of travel of A. D5 At terminal velocity, 1 (0 ) (3.0)(9.81) 0.60 49.1 m s netF mg R v v           At  112 m sv  , 2 () (3.0)(9.81) 0.60(12) 3.0 7.41 m s netFm a mg R ma a a             D6 a)  As the light tow bar is under t ension, it exerts forces of equal magnitude on the caravan as well as the car. The tension in the tow bar pulls the caravan forward and the car back.  Applying Newton’s second law on the caravan, taking rightwards as positive, F net = ma T – 1000 = (1000) (2.0) T = 3000 N  The force by the tow bar on the caravan is 3000 N to the right .  Hence the force by the tow b