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Newton’s 3 Laws of Motion:1. Net force = 0, a = 0.2. Net force 0, a 0.

F = ma3. Action = - Reaction

Action and reaction act on 2 different bodies.

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F = ma

F= tm(v u)

a = t(v u)

F = tmv mu

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Momentum = mass velocity

P = mv F= t

mv mu

F t1

mu = Pi

mv= Pf = 0 = t

Pf Pi

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1. Why bumper ?

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2. Why seat belt ?3. Why airbag ?

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Lengthen the collision time to reduce the collision force

F t1

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If F = 0

F= tmv mu

0 = tPf Pi

Hence, Pf = Pi

Net external force

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.

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1. Cannon

Pi = 0 Pf = 0

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2. Helicopter

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2. Helicopter

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3. Newton’s cradle

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4. Rocket taking - off

F = tmv mu

e.g.

1.If : u = 0

2.V of gas = 500 ms-1

3.gas eject rate = 100 kg s-1

4.upward force on rocket by gas at take - off = ?

= tm(v u)

= (100)(500-0)= 50000 N

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Water rocket

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Class work:

P.164 3-7

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Summary :1. momentum = P = mv

F= tmv mu2.

3. F = 0, Law of Conservation of Momentum.

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Types of collision:1. Elastic collision: Momentum & kinetic

energy are conserved. (e.g. collision between atoms)

2. Inelastic collision: Momentum is conserved, k.E. is not conserved (K.E. decreased due to energy loss). (e.g. collision between cars…….)

3. 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

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u1

m1 m2

u2

v1

m1 m2

v2

Net external force = 0Pi = Pf

m1u1 + m2u2 = m1v1 + m2v2

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Class work:P.157 Check Review (3)

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P. 158 (4)

1. Mass of Tom = 60 kg = m1

2. Speed of Tom = 3 ms-1 = u1

3. Mass of Mary = 45 kg = m2

4. Speed of Mary = 2 ms-1 = u2

5. Common velocity after collision = ? = v

m1u1 + m2u2 = m1v1 + m2v2 60x3 + 45x2 = 60v + 45vV = 2.6 ms-1 along the initial direction

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P.158 (5)

u1= 2ms-1m1= 0.4 kg

u2= 0 ms-1

m2= 0.2 kg

m1+m2 = (0.4 +0.2 )kg

v = ?

m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2 )v 0.4x2 + 0.2x0 =

0.6xvv = 1.33 ms-1 along initial direction

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P. 158 (6)

1. Mass of fat boy = m1 = 70 kg

2. Initial velocity of fat boy = u1 = 0

3. Mass of thin boy = m2 = 40 kg

4. Initial velocity of thin boy = u2 = 0

5. Final velocity of thin boy = v2=1.2 ms-1

6. Final velocity of fat boy = ?

m1x0 + m2x0 = 70v1 + 40x1.2v1 = - 0.69 ms-1 in opposite direction with thin boy

By Newton’s 3rd law, the result will be the same

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1 ms-1 2 ms-1

1 ms-1? ms-1

m1u1 + m2u2 = m1v1 + m2v2 mx1+ m(-2) = mv1 + mx1

+ ve

v1 = - 2 ms-1 ball A moves backwards

Elastic collision, kinetic energy is conserved

A B

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P.158(8)1. Mass of gun=m1=3.5kg

2. Initial velocity of gun=u1=0

3. Mass of bullet=m2=10g=0.01kg

4. Initial velocity of bullet=u2=0

5. Final velocity of bullet=v2=600ms-1

6. Recoil velocity of gun = v1=? m1u1 + m2u2 = m1v1 + m2v2 3.5x0 + 0.01x0 = 3.5xv1 + 0.01x600V1 = - 1.71 ms-1

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P.158(9)

1. Mass of player = m1 =60 kg

2. Initial velocity of player = u1 = 0

3. Mass of ball = m2 = 0.4 kg

4. Initial velocity of ball = u2 =10 ms-1

5. Final velocity of ball = v2 = - 20 ms-1

6. Recoil velocity of player = ? m1u1 + m2u2 = m1v1 + m2v2 60x0 +0.4x10 = 60v1 + 0.4x(-

20)V1 = 0.2 ms-1 in opposite direction to the ball

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F =t

mv mu

P.164

2 - 6

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P.164

2. 1. Mass of car = m = 1800 kg

2. Initial velocity of car = u = 20 ms-1

3. Final velocity of car = 0

4. Collision time = 0.06s

5. Force on car = ?

F = tmv mu

= 0.060 1800(20)

= - 600 000 N

F’ = 1.10 1800(20)

= - 32700 N

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P.164

3.

F = tmv mu

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P.171

2. +1. Mass of car 1 = m1 = 2000 kg

2. Initial velocity of car 1 = u1 = 20 ms-1

3. Mass of car 2 = m2 = 1500 kg

4. 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 )vV = 5 ms-1

b) Momentum change of car 1 = m1V – m1 u1 = 2000x5 – 2000x 20 = - 30000 Ns= Momentum change of car 2

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P.171

6.1. 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

4. 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

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P.172

10, 12, 15, 17

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