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- 1. 2.4 Analysing Momentum
- 2. Learning Outcome : Define the momentum of an object Define momentum (p) as the product of mass ( m) and velocity (v), e. p=mv State the principle of conversation of momentum Describe applications of conservation of momentum.
- 3. Momentum is a commonly used term in sports A team that has the
- 4. The momentum of an object is the product of the mass and the velocity of the object
- 5. Situation 1 Car A and car B have the same mass but they move with different velocities. Which car possess more momentum?
- 6. Situation 2 The lorry and the car move with the same velocity but they have different masses. Which vehicle possess more momentum ?
- 7. The principle of conservation of momentumstates that in a system make out ofobjects that react (collide or explode), thetotal momentum is constant if no externalforce is acted upon the system.Sum of Momentum Before Reaction = Sum of Momentum After Reaction
- 8. Formula of Principle of Conservation of Momentum
- 9. Example : Both objects are same direction before collision A Car A of mass 600 kg moving at 40 ms-1 collides with a car B of mass 800 kg moving at 20 ms-1 in the same direction. If car B moves forwards at 30 ms-1 by the impact, what is the velocity, v, of the car A immediately after the crash? m1 = 600kg m2 = 800kg u1 = 40 ms-1 u2 = 20 ms-1 v1 = ? v2 = 30 ms -1
- 10. Answer : According to the principle of conservation of momentum, m1u1 + m2u2 = m1v1 + m2v2 (600)(40) + (800)(20) = (600)v1 + (800)(30) 40000 = 600v1 + 24000 600v1 = 16000 v1 = 26.67 ms-1
- 11. Example 2 : Both objects are in opposite directionbefore collision A 0.50kg ball traveling at 6.0 ms-1 collides head-on with a 1.0 kg ball moving in the opposite direction at a speed of 12.0 ms-1. The 0.50kg ball moves backward at 14.0 ms-1 after the collision. Find the velocity of the second ball after collision. m1 = 0.5 kg m2 = 1.0 kg u1 = 6.0 ms-1 u2 = -12.0 ms-1 v1 = -14.0 ms-1 v2 = ? (IMPORTANT: velocity is negative when the object move in opposite siredtion)
- 12. Answer :According to the principle of conservation ofmomentum, m1u1 + m2u2 = m1v1 + m2v2(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v 2 -9 = - 7 + 1v 2 v2 = -2 ms-1
- 13. Elastic collision is the collision where the kineticenergy is conserved after the collision.Total Kinetic Energy before Collision= Total Kinetic Energy after Collision Additional notes: -In an elastic collision, the 2 objects separated right after the collision, and -the momentum is conserved after the collision.
- 14. Inelastic collision is the collision where thekinetic energy is not conserved after thecollision. Additional notes: -In a perfectly elastic collision, the 2 objects attach together after the collision, and -the momentum is also conserved after the collision.
- 15. Example 3 : Perfectly Inelastic Collision A lorry of mass 8000kg is moving with a velocity of 30 ms-1. The lorry is then accidentally collides with a car of mass 1500kg moving in the same direction with a velocity of 20 ms-1. After the collision, both the vehicles attach together and move with a speed of velocity v. Find the value of v.
- 16. According to the principle of conservation ofmomentum, m1u1 + m2u2 =( m1+ m2) v (8,000)(30) + (1,500)(20) = (8,000)v+ (1,500)v 270,000 = 9500v v = 28.42 ms-1 Answer: (IMPORTANT: When 2 object attach together, they move with same speed.)
- 17. Application of Conservation of Momentum : Jet Engine •The hot gas is forced through the engine to turn the turbine blade, which turn the compressor. • High-speed hot gases are ejected from the back with•Air is taken in from the front and is high momentum. This compressed by the compressor. produces an equal and•Fuel is injected and burnt with the opposite momentum to push compressed air in the combustion the jet plane forward. chamber.

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