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Momentum and Energy in Collisions

Momentum and Energy in Collisions. Starter. A 2kg car moving at 10m/s strikes a 2kg car at rest. They stick together and move to the right at ___________m/s. Definition of Momentum. The symbol p stands for momentum. Momentum is the product of mass and

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Momentum and Energy in Collisions

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  1. Momentum and Energy in Collisions

  2. Starter A 2kg car moving at 10m/s strikes a 2kg car at rest. They stick together and move to the right at ___________m/s.

  3. Definition of Momentum The symbol p stands for momentum. Momentum is the product of mass and velocity. p = mv

  4. Examples of calculating momentum A 2000kg car is moving at 30m/s. What is the momentum of the car? p = mv = (2000kg)(30 m/s) = 60,000 kg m/s A .1 kg bullet has a momentum of 50 kg m/s. How fast is it moving? v = p/m = 50/.1 = 500 m/s

  5. More Examples: A 10kg rock and a 2 kg rock have the same momentum of 100 kg m/s. What is the speed of each rock? Answer: For the 10kg rock: 100 = 10v, 0r v = 10m/s. For the 2 kg rock, 100 = 2v, or v = 50 m/s.

  6. The Vector Nature of Momentum Momentum is a vector – it points in the same direction as the velocity. In one dimension, momentum pointing to the right is positive. Momentum pointing to the left is negative.

  7. Example: Find the momentum of each ball. Be careful of the signs! Answer: For the 3kg ball, p = 3(20) = 60 kg m/s For the 10 kg ball, p = 2(-10) = -20 kg m/s

  8. Newton’s 2nd Law in terms of Momentum Favg = maavg = mDv/Dt = (pf – pi)/ Dt = Dp/Dt FavgDt = Dp Impulse = Change in momentum

  9. Impulse Area = FavgDt = Impulse = Dp

  10. Example This force is applied to a 3kg particle moving at 4m/s. 1. What is the impulse? Impulse = area = ½bh = ½(3)(3) =4.5Ns 2. How fast is the particle moving after 4 seconds? Impulse = Dp I = mvf – mvi 4.5 = 3vf – 3(4) vf= 5.5 m/s

  11. Conservation of Momentum Momentum is Conserved for Collisions Total momentum = Total momentum before the collision after the collision Pbefore = Pafter

  12. Types of Collisions • Elastic ( Energy and Momentum are conserved) • Inelastic ( Only momentum is conserved) • Note: Momentum is Always Conserved for any collision.

  13. Completely Inelastic Collisions • When two objects hit and stick together. • Or, the reverse of this – when one object breaks apart into two objects. Momentum is Conserved Total momentum = Total momentum before the collision after the collision Pbefore = Pafter

  14. Example A cannon ( mass = 500kg) fires a cannon ball ( m = 50kg) at 40m/s. How fast does the cannon move after it fires the cannon ball? Before: Pi = 0 After: Pf = mballvball + mcannonvcannon Pi = Pf 0 = mballvball + mcannonvcannon (-mballvball )/mcannon =vcannon = (-50)(40)/500 = -4 m/s

  15. Example A car mass = 1kg moving at 3m/s hits another 1kg car and they stick together. How fast are they moving after they stick together? Pi = mvi = 1(3) = 3 Pf = 2mv = 2v 2v = 3, v = 1.5 m/s

  16. Example A car mass = 10kg moving at 2m/s hits another 15kg car moving to the left at 3m/s and they stick together. How fast are they moving after they stick together? Pi = m1v1i + m2v2i = 10(2) + 15(-3) = -25 Pf = m1v1f + m2v2f = (m1 + m2 )vf = 25vf -25 = 25vf vf = -25/25 = -1 m/s

  17. Starter A 2kg car moving at 24m/s strikes a 10kg car at rest. They stick together and move to the right at ___________m/s.

  18. 2 particle,1-D Elastic Collisions Momentum is conserved: m1v1i + m2v2i = m1v1f + m2v2f Energy is Conserved : v1i + v1f = v2i + v2f This gives you 2 equations and 2 unknowns.

  19. Example A 10kg ball moving to the right at 3m/s strikes a 5kg ball at rest. Find the velocity of each ball after the collision. m1v1i + m2v2i = m1v1f + m2v2f 10(3) + 0 = 10v1f + 5v2f OR (1) 30 = 10v1f + 5v2f v1i + v1f = v2i + v2f OR (2) 3 + v1f = v2f

  20. CONTINUED……… (1) 30 = 10v1f + 5v2f (2) 3 + v1f = v2f The problem now is to solve two equations and two unknowns. Sub (2) into (1) : (1) 30 = 10v1f + 5 ( 3 + v1f ) 30 = 10v1f + 15 + 5v1f , 15 = 15v1f , v1f = 1 m/s Then (2) : 3 + v1i = v2f , 3 + 1 = 4 = v2f

  21. Alternate Form You should be able to show that from : m1v1i+ m2v2i = m1v1f + m2v2f and v1i+ v1f = v2i + v2f It follows that: v1f= v1i [(m1 - m2 )/(m1+ m2)] + v2i [ 2m2 /(m1 + m2 )] v2f= v1i [ 2m1/(m1 + m2 )] + v2i [(m2 – m1 )/(m1 + m2)]

  22. Special Case 1 m1 = m2 v1f = v2i v2f= v1 Velocity Exchange Special Case 2 v2i = 0 v1f = v1i [(m1 - m2 )/(m1 + m2 )] v2f= v1i [ 2m1 /(m1 + m2 )]

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