HCI 03 Dynamics Solutions (Lecture Examples)

Hwa Chong Institution (College) H2 Physics C1 2023 1 Example 1a: Change in Momentum A 110 g billiard ball rebounds off a wall, with velocities as shown in the (top view) diagram. The ball moves on a horizontal plane. Evaluate the change in momentum of the ball. Answer ()fi fipp p m v v    Draw the vector triangle. 221.2 1.2 2(1.2)(1.2)cos120v    22 10.110 1.2 1.2 2(1.2)(1.2)cos120 0.23 kg m s (downward)pm v       Alternative Method Horizontally, ,,() 0.110(1.2sin30 1.2sin30 ) 0 Hx f x ipm v v      Vertically, ,, 1 () 0.110[1.2cos30 ( 1.2cos30 )] 0.23 kg m s Vy f y ipm v v         10.23 kg m s (downward)Vpp    1.2 m s-1 1.2 m s-1 Δv 30° 30° 1.2 m s1 1.2 m s1 110 g 30° 30°

Hwa Chong Institution (College) H2 Physics C1 2023 2 Example 1b: Rate of Change in Momentum The time interval of contact between the billiard ball and the wall is 0.100 s. Find the average net force acting on the wall during the collision. Answer: In example 1a, we found 10.23 kg m sp  By Newton’s 2nd Law, wall on ball 0.23 2.3 N (downward)0.100 pF t   By Newton’s 3rd Law, wall on ball ball on wallFF  So the average net force acting on the wall by the ball during the collision is 2.3 N directed upward (perpendicularly into the wall). Example 2 (N89/II/8 modified) In order to stop a car of mass 1500 kg travelling at 30 m s1, the driver applies his brakes so that F, the total stopping force, increases steadily to a maximum and then decreases to zero as shown in the figure. Calculate (a) the momentum of the car when it is travelling at 30 m s1, (b) the impulse due to the braking force, (c) the magnitude of the average stopping force, <F>, (d) the value of Fmax. Answer: (a) Momentum of car, 31(1500)(30) 45 10 kg m spm v    (b) Impulse due to the braking force = change in linear momentum of car, carp = ifpp = 1500(0 – 30) = 45 x 103 kg m s1 (c) 345 10 20 carp xF t   = 2.3 x 103 N Note: negative sign => force acts to oppose car’s motion. It is hence called “stopping force”. (d) carp = area under F-t graph 45 x 103 = (0.5)(20)Fmax Fmax = 4500 N 1.2 m s1 1.2 m s1 110 g 30° 30° 20 10 Stopping force F Fmax time / s <F> = 2.3 x 103 N

Hwa Chong Institution (College) H2 Physics C1 2023 3 Example 3 A helicopter of mass M and weight W rises with vertical acceleration, a, due to the upward thrust U generated by its rotor. The crew and passengers of total mass m and total weight w, exerts a combined force R on the floor of the helicopter. Draw an appropriate free-body diagram, and write down an equation for the motion of (a) the helicopter, (b) the crew and passengers, (c) helicopter, crew and passengers Answer: (a) helicopter (b) crew and passengers (c) helicopter, crew, passengers Example 4 Two blocks A and B, of masses 2M and 4M, respectively, are pushed along a smooth horizontal surface by a force of F as shown in the diagram. What is the magnitude of the force exerted by block A on block B during the acceleration? Answer: Consider A + B: Consider B: Resultant upwards force on helicopter = U – W – R Hence, by Newton’s Second Law, U – W – R = Ma Resultant upwards force on crew and passengers = R’ – w Hence, by Newton’s Second Law, R’ – w = ma (1) U – W – R = Ma (2) R’ – w = ma (1) + (2): U – (W+w) = (M+m)a U W U w W + w R’ R A F B By Newton’s second law: FA on B = 4Ma = 4M ( F 6M) = 2 3 F By Newton’s second law: F = (2M + 4M)a a = F 6M F A on B B 4M F B 4M A 2M Note: the weight and normal contact force, despite equal in value, should be included in FBD for completeness. F B 4M A 2M F B 4M A 2M (Extension) Which case has a larger contact force between the two blocks?

Hwa Chong Institution (College) H2 Physics C1 2023 4 Example 5 A 75 kg skier is accelerating at 2.6 m s2 down a slope at an angle θ = 30°. Find the friction f and the normal contact force N acting on the skier. Step 1 Choose/Identify the system, in this case the skier, and draw a simple box to represent it. Step 2 Draw the free-body diagram. Step 3 For two-dimensional pr oblems, resolve forces into two convenient, mutually perpendicular directions to simplify analysis. - We choose the components along and perpendicular to the slope. - The frictional force is parallel to the slope. - The normal contact force is perpendicular to the slope. - The weight has components along and perpendicular to the slope. Step 4 Apply Newton’s second law to the 2 perpendicular directions: Parallel to the slope, taking downwards along the slope as positive Direction // to slope: () sin (75)(9.81)sin30 (75)(2.6) 173 N netFm a Wf M a f f       Perpendicular to the slope, taking upwards away from the slope as positive Direction  to slope: (0 ) cos (75)(9.81)cos30 637 N netF NM g      Don’t know how to resolve the weight into the 2 components? Watch this: θ Friction, f Normal contact force, N Weight, W W cos  W sin  a  +y +x

Hwa Chong Institution (College) H2 Physics C1 2023 5 Example 6 (Man in the lift) An 80 kg man weighs himself by standing on a weighing scale inside a lift. What does the scale read if the lift (a) is at rest, (b) moves with an upward acceleration of 1.8 m s2, (c) the lift is moving upwards with a constant velocity of 2.2 m s1, (d) the lift slows from its velocity in (c) to rest at rate of 1.9 m s2. Answer: Note that the weighing scale actually does not measure the man’ s weight. It measures the normal contact force pushing down on it. As we do not have any other information about the weighing scale, we are unable to choose the weighing scale as the system and draw its FBD. Instead we will determine the normal contact force on the scale by using its third-law partner. Since the no rmal contact force on the scale is due to the man, its third-law partner is the normal contact force on the man by the scale. We will solve the question by choosing the man as the system and drawing the FBD of the man. (a) Consider the FBD of the man: (c) This outcome is true for (c) as well since moving at constant speed also implies that the net force is zero. (b) The lift moves with an upward acceleration a Net force upwards () () (80)(9.81 1.8) 928.8 N netFm a Nm gm a Nm ga      Mass reading 928.8 9.81 94.7 kg  (d) The lift slows to rest at rate of a () () (80)(9.81 1.9) 632.8 N netFm a mg N ma Nm ga      Mass reading 632.8 9.81 64.5 kg  This YouTube video sheds more light on apparent weight and true weight: https://www.youtube.com/watch?v=AbNJv1VNWu8 Weight, W Normal Contact Force, N The man is at rest. Net force = 0 N = mg By Newton’s Third Law, The scale reading reads mg. Weighing scale reads the true weight of the man.

Hwa Chong Institution (College) H2 Physics C1 2023 6 Example 7: A hovering jetpack with nozzles of total cross-sectional area A expels water at constant velocity v relative to the jetpack. Show that the force exerted by the expelled water on the jetpack is 2FA v , where  is the density of the water jet. Answer: Since the jetpack is hovering, the speed of water relative to the jetpack is the same as the speed of water observed by a ground observer. Hence in a short time interval t, the column of water ejected from the jetpack is as shown. Consider the cylindrical colum