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

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

4. 1. Why bumper ?

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

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

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

8. 1. Cannon
Pi = 0
Pf = 0

9. 2. Helicopter

10. 2. Helicopter

11. 3. Newton’s cradle

12. 4. Rocket taking - off mv  mu F = t m(v  u) = t 1.If : u = 0
e.g.
1.If : u = 0 = t
m(v  u)
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 = ?
= N

13. Water rocket

14. Class work: P

15. Summary : momentum = P = mv mv  mu F= t 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. Pi = Pf m1u1 + m2u2 = m1v1 + m2v2 Net external force = 0 m1 m2 u1 u2

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

19. m1u1 + m2u2 = m1v1 + m2v2 60x3 + 45x2 = 60v + 45v 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. m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2 )v 0.4x2 + 0.2x0 = 0.6xv
P (5)
m2= 0.2 kg
u2= 0 ms-1
m1+m2 = ( )kg
m1= 0.4 kg
u1= 2ms-1
v = ?
m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2 )v
0.4x x0 = 0.6xv
v = 1.33 ms-1 along initial direction

21. m1x0 + m2x0 = 70v1 + 40x1.2 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 = ms-1 in opposite direction with thin boy
By Newton’s 3rd law, the result will be the same

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

23. m1u1 + m2u2 = m1v1 + m2v2 3.5x0 + 0.01x0 = 3.5xv1 + 0.01x600 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
3.5x x0 = 3.5xv x600
V1 = ms-1

24. m1u1 + m2u2 = m1v1 + m2v2 60x0 +0.4x10 = 60v1 + 0.4x(-20) 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 = 60v x(-20)
V1 = 0.2 ms-1 in opposite direction to the ball

25. F = t
mv  mu
P.164
2 - 6

26. P.164 2.
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 = ?
F = t
mv  mu =
0.06
0  1800(20)
= N
F’ =
1.1
0  1800(20)
= N

27. P.164 3.
F = t
mv  mu

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
2000x (-15) = ( )v
V = 5 ms-1
b) Momentum change of car 1 = m1V – m1 u1
= 2000x5 – 2000x 20 = Ns
= Momentum change of car 2

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

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