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Newton’s 3 Laws of Motion :

Newton’s 3 Laws of Motion :. Net force = 0, a = 0. Net force  0, a  0. F = ma. Action = - Reaction Action and reaction act on 2 different bodies. mv  mu. m(v  u). (v  u). F=. a =. F =. t. t. t. F = ma. 1. mv  mu. P f  P i. F=. =. F . t. t. t.

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Newton’s 3 Laws of Motion :

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  1. Newton’s 3 Laws of Motion: • Net force = 0, a = 0. • Net force  0, a  0. • F = ma • Action = - Reaction • Action and reaction act on 2 different bodies.

  2. mvmu m(vu) (vu) F= a = F = t t t F = ma

  3. 1 mvmu PfPi F= = F  t t t Momentum = mass  velocity mu = Pi P = mv mv= Pf = 0

  4. 1. Why bumper ?

  5. 2. Why seat belt ? 3. Why airbag ?

  6. 1 F  t Lengthen the collision time to reduce the collision force

  7. mvmu Pf Pi F= 0 = t t When the net force acting on a body is zero, the momentum of the body(or system) must be constant, called theLaw of Conservation of Momentum. If F = 0 Net external force Hence, Pf = Pi

  8. 1. Cannon Pi = 0 Pf = 0

  9. 2. Helicopter

  10. 2. Helicopter

  11. 3. Newton’s cradle

  12. mvmu m(vu) F = = t t 4. Rocket taking - off e.g. 1.If : u = 0 2.V of gas = 500 ms-1 3.gas eject rate = 100 kg s-1 = (100)(500-0) 4.upward force on rocket by gas at take - off = ? = 50000 N

  13. Water rocket

  14. Class work: P.164 3-7

  15. mvmu F= t Summary : • momentum = P = mv 2. 3. F = 0, Law of Conservation of Momentum.

  16. Types of collision: • Elastic collision: Momentum & kinetic energy are conserved. (e.g. collision between atoms) • Inelastic collision: Momentum is conserved, k.E. is not conserved (K.E. decreased due to energy loss). (e.g. collision between cars…….) • Explosion: Momentum is conserved, k.E. is not conserved ( K.E.increased from chemical energy). (e.g. bombs) Note: Momentum is conserved in all cases if net external F = 0

  17. m1 m2 u1 u2 m1 m2 v1 v2 Net external force = 0 Pi = Pf m1u1 + m2u2 = m1v1 + m2v2

  18. Class work: P.157 Check Review (3)

  19. P. 158 (4) • Mass of Tom = 60 kg = m1 • Speed of Tom = 3 ms-1 = u1 • Mass of Mary = 45 kg = m2 • Speed of Mary = 2 ms-1 = u2 • Common velocity after collision = ? = v m1u1 + m2u2 = m1v1 + m2v2 60x3 + 45x2 = 60v + 45v V = 2.6 ms-1 along the initial direction

  20. P.158 (5) m2= 0.2 kg u2= 0 ms-1 m1+m2 = (0.4 +0.2 )kg m1= 0.4 kg u1= 2ms-1 v = ? m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2)v 0.4x2 + 0.2x0 = 0.6xv v = 1.33 ms-1 along initial direction

  21. P. 158 (6) • Mass of fat boy = m1 = 70 kg • Initial velocity of fat boy = u1 = 0 • Mass of thin boy = m2 = 40 kg • Initial velocity of thin boy = u2 = 0 • Final velocity of thin boy = v2=1.2 ms-1 • Final velocity of fat boy = ? m1x0 + m2x0 = 70v1 + 40x1.2 v1 = - 0.69 ms-1 in opposite direction with thin boy By Newton’s 3rd law, the result will be the same

  22. + ve B A 1 ms-1 2 ms-1 ? ms-1 1 ms-1 m1u1 + m2u2 = m1v1 + m2v2 mx1+ m(-2) = mv1 + mx1 v1 = - 2 ms-1 ball A moves backwards Elastic collision, kinetic energy is conserved

  23. P.158(8) • Mass of gun=m1=3.5kg • Initial velocity of gun=u1=0 • Mass of bullet=m2=10g=0.01kg • Initial velocity of bullet=u2=0 • Final velocity of bullet=v2=600ms-1 • Recoil velocity of gun = v1=? m1u1 + m2u2 = m1v1 + m2v2 3.5x0 + 0.01x0 = 3.5xv1 + 0.01x600 V1 = - 1.71 ms-1

  24. P.158(9) • Mass of player = m1 =60 kg • Initial velocity of player = u1 = 0 • Mass of ball = m2 = 0.4 kg • Initial velocity of ball = u2 =10 ms-1 • Final velocity of ball = v2 = - 20 ms-1 • Recoil velocity of player = ? m1u1 + m2u2 = m1v1 + m2v2 60x0 +0.4x10 = 60v1 + 0.4x(-20) V1 = 0.2 ms-1 inopposite direction to the ball

  25. mvmu F = t P.164 2 - 6

  26. P.164 2. 0 1800(20) 01800(20) mvmu F = = F’ = t 0.06 1.1 • Mass of car = m = 1800 kg • Initial velocity of car = u = 20 ms-1 • Final velocity of car = 0 • Collision time = 0.06s • Force on car = ? = - 600 000 N = - 32700 N

  27. P.164 3. mvmu F = t

  28. P.171 2. + • Mass of car 1 = m1 = 2000 kg • Initial velocity of car 1 = u1 = 20 ms-1 • Mass of car 2 = m2 = 1500 kg • Initial velocity of car 2 = u2 = - 15 ms-1 a) Let common velocity after collision = v m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2)v 2000x20 + 1500(-15) = (2000+1500 )v V = 5 ms-1 b) Momentum change of car 1 = m1V – m1 u1 = 2000x5 – 2000x 20 = - 30000 Ns = Momentum change of car 2

  29. P.171 6. + • Mass of trolley = m1=0.5 – 0.0002 kg • 2. initial speed of trolley + bullet = u1 = u2 =0 • 3. Mass of bullet = m2 =0.0002 kg • final speed of bullet = v2 = 60 ms-1 a) Let final speed of trolley = v1 m1u1 + m2u2 = m1v1 + m2v2 0 = 0.49998v1 + 0.0002x60 V1 = 0.024 ms-1 b) 0

  30. P.172 10, 12, 15, 17

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