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 pas