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Collisions

Collisions. Review. Momentum is a quantity of motion. p = mv A change in momentum is called impulse. Impulse = p = mv Impulses are caused by forces applied over time. p = Ft. Collisions.

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Collisions

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  1. Collisions

  2. Review • Momentum is a quantity of motion. • p = mv • A change in momentum is called impulse. • Impulse = p = mv • Impulses are caused by forces applied over time. • p = Ft

  3. Collisions • Collision – an isolated event in which 2 or more bodies exert strong forces over short periods of time against each other. • Momentum is conserved in all collisions (assuming no outside force). • “sticky” collision – objects stick together after colliding and move as one. • “non-sticky” collision – objects bounce apart after colliding.

  4. Collisions • A train car with a mass of 3.00x104 kg traveling north at 1.5 m/s collides and couples with a 3.20x104-kg train car going south at 0.80 m/s. • What is the velocity of the coupled cars after the collision?

  5. mass = 3.00x104 kg mass = 3.20x104 kg North Collisions • Sketching the situation can help. 1.5 m/s -0.80 m/s p1 = m1v1 p2 = m2v2 p1 = (3.00x104 kg)(1.5 m/s) p2 = (3.20x104 kg)(-0.80 m/s) p1 = 4.5x104 kg*m/s p2 = -2.6x104 kg*m/s ptotal= (4.5x104 kg*m/s) + (-2.6x104 kg*m/s) = 1.9x104 kg*m/s

  6. North Collisions mtotal = 6.20x104 kg ptotal= 1.9x104 kg*m/s p = mv v = p/m v = (1.9x104 kg*m/s) / (6.20x104 kg) = 0.31 m/s north

  7. Collisions • A 20.0-kg ball is moving north at 4.00 m/s toward a 5.00-kg ball that is stationary. After they collide, the 20.0-kg ball is moving at 2.40 m/s north. • What is the 5.00-kg ball’s velocity after the collision? • Neglect friction.

  8. North Collisions Before Collision m1 = 20.0 kg m2 = 5.00 kg 4.00 m/s p1 = m1v1 p2 = m2v2 p1 = (20.0 kg)(4.00 m/s) p2 = (5.00 kg)(0 m/s) p1 = 80.0 kg*m/s p2 = 0 kg*m/s ptotal= (80.0 kg*m/s) + (0 kg*m/s) = 80.0 kg*m/s

  9. North Collisions After Collision m1 = 20.0 kg m2 = 5.00 kg 2.40 m/s ??? ptotal= 80.0 kg*m/s p2 = (80.0 kg*m/s) – (48.0 kg*m/s) = 32.0 kg*m/s p1 = (20.0 kg)(2.40 m/s) v2 = p2 / m2 p1 = 48.0 kg*m/s v2 = (32.0 kg*m/s) / (5.00 kg) = 6.40 m/s

  10. Collisions • A 0.0085-kg bullet strikes a stationary wooden block with a mass of 40.0 kg on a frictionless surface. After the impact, the block-bullet combo is moving at 0.20 m/s. • What was the bullet’s original velocity?

  11. Don’t know! Collisions Before Collision m1 = 0.0085 kg 40.0 kg ??? p1 = m1v1 p2 = m2v2 p2 = (40.0 kg)(0 m/s) p2 = 0 kg*m/s ptotal = Don’t know yet!

  12. Collisions After Collision mtotal = 40.0085 kg = 40.0 kg 0.20 m/s ptotal = mtotalv = (40.0 kg)(0.20 m/s) = 8.0 kg*m/s

  13. Collisions Before Collision m1 = 0.0085 kg 40.0 kg ??? ptotal = 8.0 kg*m/s v1 = p1 / m1 v1 = (8.0 kg*m/s) / (0.0085 kg) v1 = 940 m/s

  14. Recoil (Collision in Reverse) • Whenever one object pushes another object away, momentum is still conserved. • Think of it like a collision in reverse.

  15. Recoil • A 60.-kg student on ice skates stands at rest on a frictionless frozen pond holding a 10.-kg brick. He throws the brick east with a speed of 18 m/s. • What is the resulting velocity of the student?

  16. Recoil Before Throwing m1 = 60. kg m2 = 10. kg ptotal = 0 kg*m/s

  17. After Throwing m1 = 60. kg m2 = 10. kg Recoil ptotal = 0 kg*m/s ??? 18 m/s p1 + p2 = ptotal p2 = m2v2 p1 + (180 kg*m/s) = 0 kg*m/s p2 = (10. kg)(18 m/s) p1 = -180 kg*m/s p2 = 180 kg*m/s v1 = p1 / m1 = (-180 kg*m/s) / (60. kg) v1 = -3.0 m/s

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