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Momentum and Impulse

Momentum and Impulse. Principles of Physics. Momentum. - property of an object related to its mass and velocity. - “mass in motion” or “inertia in motion” p = momentum (vector) p = mv m = mass (kg) v = velocity (m/s) (vector) units for momentum are kg m/s

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Momentum and Impulse

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  1. Momentum and Impulse Principles of Physics

  2. Momentum - property of an object related to its mass and velocity. - “mass in motion” or “inertia in motion” p = momentum (vector) p = mv m = mass (kg) v = velocity (m/s) (vector) units for momentum are kg m/s **If an object is not moving, it has no momentum

  3. Momentum of an individual object p = mv Example: Calculate the momentum of a 250 kg cart with a velocity of 25 m/s. p = mv p = 250 kg (25 m/s) p = 6250 kg m/s

  4. Momentum of a system p = m1v1 + m2v2 + m3v3 + … Example: A 300 kg car is travelling to east at 45 m/s and a 500 kg truck is travelling west at 30 m/s. Calculate the total momentum of the system. p = m1v1 + m2v2 p = 300 kg(45 m/s) + 500 kg(-30 m/s) p = -1500 kg m/s

  5. Conservation • Conservation – to keep constant even though changes occur • whatever you had before an event you will still have after the event

  6. Conservation of Momentum The total momentum of a system is the same before and after a collision ptotal before = ptotal after

  7. Conservation of Momentum ptotal before = ptotal after ptotal before= m1v1i + m2v2i + m3v3i + … **remember v has direction ptotal after = m1v1f + m2v2f + m3v3f + … Newton’s Cradle Demo:

  8. Types of Collisions Elastic - bounce • objects hit and bounce off from each other Inelastic – stick • multiple objects hit and stick together or • one objects separates into 2 or more (explosion)

  9. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Write down given information Givens: m1 = 0.050 kg m2 = 0.050 kg v1i= 0.20 m/s v2i= 0.10 m/s v1f = 0.08 m/s v2f = ?

  10. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Determine the momentum of the cue ball before and after the collision Givens: m1 = 0.050 kg m2 = 0.050 kg v1i= 0.20 m/s v2i= 0.10 m/s v1f = 0.08 m/s v2f = ? p1i = m1v1i = 0.050 kg(0.20 m/s) = 0.01 kg m/s p1f = m1v1f = 0.050 kg(0.08 m/s) = 0.004 kg m/s

  11. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Determine the momentum of the 8 ball before the collision Givens: m1 = 0.050 kg m2 = 0.050 kg v1i= 0.20 m/s v2i= 0.10 m/s v1f = 0.08 m/s v2f = ? p2i = m2v2i = 0.050 kg(0.10 m/s) = 0.005 kg m/s

  12. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Set up a conservation equation pbefore = pafter p1i + p2i = p1f+ p2f

  13. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Substitute momentum values into the conservation equation and solve for the momentum of the 8 ball after the collision pbefore = pafter p1i + p2i = p1f+ p2f 0.01 kg m/s + 0.005 kg m/s = 0.004 kg m/s + p2f p2f = 0.011 kg m/s

  14. Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity? Determine the final velocity of the 8 ball p2f = m2v2f 0.011 kg m/s = 0.050 kg (v2f) v2f = 0.22 m/s

  15. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Write down given information Givens: mb = 0.015 kg mbb = 5.085 kg vbi= ? vbbi= 0 m/s vbf = 1 m/s vbbf = 1 m/s

  16. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Determine the momentum of the block before and after the collision p2i= m2v2i = 5.085 kg(0 m/s) = 0 kg m/s Givens: m1= 0.015 kg m2= 5.085 kg v1i= ? v2i= 0 m/s v1f= 1 m/s v2f= 1 m/s p2f= m2 v2f = (5.085 kg) (1 m/s) = 5.085 kg m/s

  17. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Determine the momentum of the bullet after the collision Givens: m1= 0.015 kg m2= 5.085 kg v1i= ? v2i= 0 m/s v1f= 1 m/s v2f = 1 m/s p1f= (m1)v1f = (0.015 kg) (1 m/s) = 0.015 kg m/s

  18. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Set up a conservation equation pbefore = pafter p1i+ p2i = p1f + p2f

  19. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Substitute the momentum values and determine the momentum of the bullet before the collision pbefore = pafter p1i+ p2i = p1f + p2f p1i+ 0 = 0.015 kg m/s + 5.085 kg m/s p1i = 5.1 kg m/s

  20. Example 2 A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet. Determine the initial velocity of the bullet p1i= m1v1i 5.1 kg m/s = 0.015 kg(v1i) v1i = 340 m/s

  21. http://www.youtube.com/watch?NR=1&v=wCXqcBpURZE&feature=endscreenhttp://www.youtube.com/watch?NR=1&v=wCXqcBpURZE&feature=endscreen

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