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.

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

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

1. Why bumper ?

2. Why seat belt ? 3. Why airbag ?

F  t 1 Lengthen the collision time to reduce the collision 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. F= t mv  mu If F = 0 Net external force 0 = t Pf  Pi Hence, Pf = Pi

1. Cannon Pi = 0 Pf = 0

2. Helicopter

2. Helicopter

3. Newton’s cradle

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 = ? = 50000 N

Water rocket

Class work: P.164 3-7

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

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

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

Class work: P.157 Check Review (3)

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

m1u1 + m2u2 = m1v1 + m2v2 =(m1+m2 )v 0.4x2 + 0.2x0 = 0.6xv 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

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 = - 0.69 ms-1 in opposite direction with thin boy By Newton’s 3rd law, the result will be the same

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

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 3.5x0 + 0.01x0 = 3.5xv1 + 0.01x600 V1 = - 1.71 ms-1

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 = 60v1 + 0.4x(-20) V1 = 0.2 ms-1 in opposite direction to the ball

F = t mv  mu P.164 2 - 6

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) = - 600 000 N F’ = 1.1 0  1800(20) = - 32700 N

P.164 3. F = t mv  mu

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

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

P.172 10, 12, 15, 17