2. 1© Manhattan Press (H.K.) Ltd. Newton’s First Law of Motion Linear momentum and Newton’s Second Law of Motion Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion Newton’s Third Law of Motion Newton’s Third Law of Motion

3. 2 © Manhattan Press (H.K.) Ltd. Newton’s First Law of Motion 3.1 Newton’s Laws of Motion (SB p. 104) Newton’s First Law of Motion: An object will remain at rest or move along a straight line with constant speed, unless it is acted upon by a net force. Object resists changes to its state of rest or motion. inertia Inertia of object :  its mass (independent of its velocity) Go to Common Error

4. 3 © Manhattan Press (H.K.) Ltd. Newton’s First Law of Motion 3.1 Newton’s Laws of Motion (SB p. 105) 1. Use seat belt / head-rest 2. Like being thrown away when turning around corner

5. 4 © Manhattan Press (H.K.) Ltd. Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion (SB p. 105) 1. Linear momentum Momentum = Mass x Velocity p = mv vector Unit – kg m s -1 or N s

6. 5 © Manhattan Press (H.K.) Ltd. Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion (SB p. 106) 2. Newton’s Second Law of Motion Newton’s Second Law of Motion: The rate of change of momentum of an object is directly proportional to the net force acted on it, and the motion occurs along the direction of the force. Rate of change of momentum ∝ Net force

7. 6 © Manhattan Press (H.K.) Ltd. Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion (SB p. 106) 2. Newton’s Second Law of Motion

8. 7 © Manhattan Press (H.K.) Ltd. Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion (SB p. 106) 2. Newton’s Second Law of Motion Unit – newton (N) F = kma 1 N = k (1 kg) (1 m s -2 ) = k kg m s -2 k = 1 F = ma

9. 8 © Manhattan Press (H.K.) Ltd. Linear momentum and Newton’s Second Law of Motion 3.1 Newton’s Laws of Motion (SB p. 106) 2. Newton’s Second Law of Motion Force = Rate of change of momentum impulse = Area under F-t curve Go to Example 1 Example 1 Go to More to Know 1 More to Know 1 Go to Example 2 Example 2 Go to Example 3 Example 3

10. 9 © Manhattan Press (H.K.) Ltd. Newton’s Third Law of Motion 3.1 Newton’s Laws of Motion (SB p. 109) E.g water rocket: water pushed out from the bottle, opposite force exerted on the bottle Newton’s Third Law of Motion: When two bodies interact, they exert equal but opposite forces on each other. They are called action and reaction. push it upwards

11. 10 © Manhattan Press (H.K.) Ltd. Newton’s Third Law of Motion 3.1 Newton’s Laws of Motion (SB p. 110) E.g. Ball is being hit by bat Magnitude of F 1 = Magnitude of F 2 = F Action and reaction are equal in magnitude but opposite in direction. They act on different bodies.

12. 11 © Manhattan Press (H.K.) Ltd. Newton’s Third Law of Motion 3.1 Newton’s Laws of Motion (SB p. 110) E.g. Ball is released Magnitude of F 1 = Magnitude of F 2 M >> m, a 1 >> a 2

13. 12 © Manhattan Press (H.K.) Ltd. End

14. 13 © Manhattan Press (H.K.) Ltd. It is incorrect to think that a force can always cause a motion. Only a net force can do so. The net force is the vector sum of all forces acting on an object. If the forces cancel each other, there is no net force acted on the object. Return to Text 3.1 Newton’s Laws of Motion (SB p. 104)

15. 14 © Manhattan Press (H.K.) Ltd. Return to Text 3.1 Newton’s Laws of Motion (SB p. 106)

16. 15 © Manhattan Press (H.K.) Ltd. Q: Q:A body of mass 3 kg experiences a force F which varies with time t as shown in the figure. What is the increase in momentum of the body after 8 s? Solution 3.1 Newton’s Laws of Motion (SB p. 107)

17. 16 © Manhattan Press (H.K.) Ltd. Solution: Return to Text 3.1 Newton’s Laws of Motion (SB p. 107)

18. 17 © Manhattan Press (H.K.) Ltd. Q: Q:A plumb-line is hung from the roof of a train carriage. The bob is vertically above a mark on the wall as shown when the train is stationary. Draw figures to show the position of the bob relative to the mark when (a) the train accelerates uniformly. (b) the train moves with uniform velocity. (c) the train retards uniformly. Explain your figures. Solution 3.1 Newton’s Laws of Motion (SB p. 108)

19. 18 © Manhattan Press (H.K.) Ltd. Solution: 3.1 Newton’s Laws of Motion (SB p. 108) (a) The net force acting on the bob is the horizontal component of tension T in the string, i.e. T cosθ. ∴ T cosθ = ma where m is the mass of the bob. Since T cosθ is in the same direction as ma, the bob is swung to the back of the mark. (b) Since v = constant, acceleration (a) = 0. From T cosθ = ma cosθ = (m/T)a ∴ cosθ = 0, θ = 90° The bob is vertically above the mark.

20. 19 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Return to Text 3.1 Newton’s Laws of Motion (SB p. 108) (c) When acceleration = – a (or retardation) By Force = Mass × Acceleration T cosθ = m (– a) T cosθ = –ma T cosθ is in the opposite direction to the direction of motion. Therefore, the bob is swung to the front of the mark.

21. 20 © Manhattan Press (H.K.) Ltd. Q: Q:A space research rocket stands vertically on its launching pad. Prior to ignition, the mass of the rocket and its fuel is 1.9 × 10 3 kg. On ignition, gas is ejected from the rocket at a speed of 2.5 × 10 3 m s −1 relative to the rocket, and fuel is consumed at a constant rate of 7.4 kg s −1. Find the thrust of the rocket and hence explain why there is an interval between ignition and lift-off. Let acceleration of free-fall (g) be 10 m s −2. Solution 3.1 Newton’s Laws of Motion (SB p. 109)

22. 21 © Manhattan Press (H.K.) Ltd. Solution : Return to Text 3.1 Newton’s Laws of Motion (SB p. 109)  Thrust of rocket = (2.5  10 3 )  7.4 = 1.85  10 4 N Initial weight of rocket = mg = (1.9  10 3 )  10 = 1.9  10 4 N which is greater than the thrust of the rocket. Therefore, the rocket is not lifted immediately at ignition. As the fuel is burnt, the weight of the rocket decreases. When the weight of the rocket is less than 1.85  10 4 N, the rocket will be lifted. Hence, there is an interval between ignition and lift-off.