MOMENTUM!

All objects have MASSConsider the following;A flying aeroplaneA stationary carHas MOMENTUMHas no MOMENTUMAny moving object has MOMENTUM

Momentum Definedp = m vp = momentum m = massv = velocity The product of mass and velocitymomentum = mass x velocity

Example 1 Calculate the momentum of the bus of mass of 9 000 kg moving with velocity of 16 ms-1.Momentum : mass x velocityMomentum : 9 000 kg x 16 ms-1 Momentum : 144 000 kg ms-1

Example 2 Calculate the momentum of a car of mass of 1 800 kg moving with velocity of 80 ms-1.Momentum : mass x velocityMomentum : 1 800 kg x 80 ms-1 Momentum : 144 000 kg ms-1

Something big and slow could have the same momentum as something small and fast.

MOMENTUM in collisionBefore DuringcollisionAfter u1u2v1v2m1m2m1m2Total Momentum Before collision m1 x u1m2 x u2m1 u1 + m2 u2 Total Momentum After collision m1 v1 m1 v1 + m2 v2 m2 v2

MOMENTUM in collisionTotal Momentum Before collision m1 u1 + m2 u2 Total Momentum After collision m1 v1 + m2 v2 From the PRINCIPLE OF CONSERVATION OF MOMENTUMTotal Momentum = Before collision IN THE ABSENCE OF AN EXTERNAL FORCE, THE TOTAL MOMENTUM OF A SYSTEM REMAINS UNCHANGEDTotal Momentum After collision m1 u1 + m2 u2 = m1 v1 + m2 v2

ELASTIC COLLISIONBefore DuringcollisionAfter u1u2v1v2m1m2m1m2Total Momentum Before collision m1 x u1m2 x u2m1 u1 + m2 u2 Total Momentum After collision m1 v1 + m2 v2 The objects move separately after the collisionm1 x v1m1 x v2m2

INELASTIC COLLISIONBefore DuringcollisionAfter u1u2vm1m2m1m2Total Momentum Before collision m1 x u1m2 x u2m1 u1 + m2 u2 Total Momentum After collision m1 v + m2 v m1 x vm2 x v(m1 + m2)v Object moving together after collision

INELASTIC COLLISIONTotal Momentum Before collision m1 u1 + m2 u2 Total Momentum After collision m1 v + m2 v

m1 u1 + m2 u2 = (m1 + m2)v Object moving together after collision

EXPLOSIONBefore DuringcollisionAfter u = 0v2m1m2m1m2Total Momentum Before collision m1 x um2 x um1 u + m2 u Total Momentum After collision m1 x v1m2 x v2(m1 + m2)u m1v1 + m1v2v1m1v1 + m1 (-v2)(m1 + m2)0 = 0 kgms-1 m1v1 - m1 v2

EXPLOSIONTotal Momentum Before collision (m1 + m2 )u Total Momentum After collision m1 v1 + m2 v2

0 = m1v2 - m2v2 Object attached together before explosion but after the explosion the object move separately at opposite direction

Example of explosionJump out of a boatRecoil of a gun when it fires a bullet

MASTERY PRACTICE 2.4mA = 500 g = 0.5 kguA = 2 ms-1mB = 400 g = 0.4 kguB = 0 ms-1vB = 1 ms-1

Will trolley A collide with trolley B for second time?Yes. Because velocity of trolley A is greater than trolley B as they move along the same direction.vA = 2 ms-1No. 1

MASTERY PRACTICE 2.4mr = 5 kgur = 0 ms-1vr = ? ms-1

mB = 50 g = 0.05 kguB = 0 ms-1vB = 80 ms-1

Why recoil velocity of a rifle is much less than the velocity of the bullet?vr = 8 ms-1No. 3Because the final momentum of the rifle is equal to the final momentum of the bullet. This was stated by the Principle of Conservation of momentum.

MASTERY PRACTICE 2.4m1 = 50 kgu1 = 0 ms-1v1 = ? ms-1

m2 = 60 kgu2 = 0 ms-1v2 =2 ms-1

What happens to their motion when they collide?vr = -2.4 ms-1No. 4The two boys move oppositely because the final momentum of the heavier boy is equal to the final momentum of the lighter boy as they move oppositely.

Application of Principle of Conservation of momentumTHE LAUNCHING OF ROCKET1. Fuel burns explosively in the combustion chamber3. Momentum of the hot = Momentum of the rocket gases downward forward4. Rocket can be launched to the space2. Hot gases expelled at high speed have a large momentum downwards.

JET ENGINEAir is sucked into combustion chamberAir is mixed with kerosene vapour and burnt-produce hot exahust gas at high temperatureHot gases expand & expelled from the back at high speedMomentum of hot gases = Momentum of jet backward forwardApplication of Principle of Conservation of momentum

Application of Principle of Conservation of momentumBOATMomentum of water = Momentum of boat backward forward

Sample Problem 4beforeafter3 kg15 kg10 m/s-16 m/s-13 kg15 kg4.5 m/s-1v

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