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 bar on the car is 3000 N to the left. b ) When the vehicles are moving at constant speed, net force on each of them must be zero. Hence, referring to caravan again, the force by the tow bar on the caravan is equal in magnitude to the friction on the caravan. Hence, the force by the tow bar on the caravan is 1000 N to the right. By Newton’s third law, the force by the tow bar on the car is 1000 N to the left.    Caravan 1000 kg Friction, 1000 N Tension in tow bar, T a = 2.0 m s-2 Forward propulsion force (implied) Car 1000 kg Friction, 3000 N Tension in tow bar, T a = 2.0 m s-2 Force by A on B  Initial direction  of travel of A

D7  B Initially, the system is in equilibrium. T1 = 0.700 g Immediately after the thread is cut, Applying Newton’s 2nd Law, F = ma T 1 – 0.500 g = 0.500 a a = 2/5 g = 0.4 g                                                                                                            Initially                                  After  Note:  Questions like these come up quite regularly in exams. An object (in this case the block of 500 g) is initially at  rest, when suddenly one of the forces “disappears”. The new net force on the object immediately after the  force disappears is given by Fnet,new = ‐Fdisappeared. The force that disappears is the tension in the lower string,  which is equal to the weight of the block of 200 g.  D8  Applying Newton’s second law For X & Y: F = (m + 3m) a = 4 m a a = F/4m For X alone: f1’ = m a = m (F/4m) = F/4 Note:  The “m” that occurs in Newton’s second law is in general the mass of the object that is being accelerated.  However, in the first part of this working, the mass as indicated in Newton’s second law is actually equal to  4m.  D9  A Consider the blocks separately. B: 10 g – T = 10 (0.10 g) -------------- (1) A: T – 15 g sin 30 – f = 15 (0.10 g) -------------- (2) 10 g – 15 g sin30 – f = (10 +15) 0.10 g f = 0 N     Note:  This is a difficult question for most students. The pulley confuses them. Most students would choose  downwards as positive and then get stuck at forming the equation for the block on the slope.         T1 500 g  200 g  0.700 g 0.500 g T1 500 g  N

Part 2: Conservation of Linear Momentum/Collisions  D10  C  pi = 6.0 (5.0) + 10(-3) = 0 N s  pf = 0 N s For inelastic collision, (6.0 + 10) v = 0 F i n a l v e l o c i t y = 0 m s-1 Consider the 6.0 kg trolley, N 15020.0 )0.5(0.60   t pF D11  Let initial velocity of bullet be u, velocity of bullet after emerging from 1.2 kg block be v. Before hitting 1.2 kg block,  pi = 0.0035 u ------------------- (1) After bullet emerging from 1.2 kg block,  pf = 1.2(0.63) + 0.0035 v ------------------ (2) After bullet stuck in 1.8 kg block,  pf’ = 1.2(0.63) + (1.8 + 0.0035)(1.4) = 3.2809 ------------- (3) (a) By the principle of conse rvation of linear momentum, (2) = (3): 0.0035 v = 2.5249 v = 721 m s -1 (b) By the principle of conse rvation of linear momentum, (1) =(3): 0.0035 u = 3.2809 u = 937 m s -1 Note:  You are strongly encouraged to draw “before” and “after” diagrams to apply COLM.         v?

D12  a) (i) refer to lecture notes (ii) refer to lecture notes b) Elastic: total kinetic energy is conserved Head-on: The motions of the molecules after the collision will be along the same straight line of motions before the collision. c) For elastic collision, relative speed of separation = Relative speed of approach = 1.88 x 10 3 + 405 = 2285 m s-1. After the collision, (ii) Applying the principle of c onservation of linear momentum, taking rightwards as positive, 2 . 0 0 u (1.88 x 10 3) + 32.0u (– 405) = 2.00u vH +32.0 u vO vH +16 vO = – 4600 --------- -----------------(1) (iii) Relative speed of approach = relative speed of separation , taking rightwards as positive uH – uO = vO – vH vO – vH = 2285 -----------------------(2) (1) + (2): 17 vO = – 2315 vO = – 136 m s-1 Sub into (2): vH = – 2420 m s-1 Hence, the velocity of the oxygen molecule is 136 m s -1 to the left and the velocity of the hydrogen molecule is 2420 m s-1 to the left.   Note:  It is simpler to use relative speeds to solve the simultaneous equations, rather than the conservation of  total kinetic energy.       vH  vO

D13  a) In this case, total momentum before collision = 3mv – 2mv = mv to the right. By the principle of conservation of linear momentum (PCLM), the total momentum in a closed system is conserved. Thus at any instant, the total momentum of both the nuclei together is mv. If both the nuclei were to stop at the same instant, total momentum would be zero and the principle would be violated. b) By Principl