1. Principles of Physics

2. - property of an object related to its mass and velocity. - “mass in motion” or “inertia in motion” p = momentum (vector) p = mvm = 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 = m 1 v 1 + m 2 v 2 + m 3 v 3 + … 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 = m 1 v 1 + m 2 v 2 p = 300 kg(45 m/s) + 500 kg(-30 m/s) p = -1500 kg m/s

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

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

7. p total before = m 1 v 1i + m 2 v 2i + m 3 v 3i + … **remember v has direction p total after = m 1 v 1f + m 2 v 2f + m 3 v 3f + … Newton’s Cradle Demo:

8. 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? Givens: m 1 = 0.050 kg m 2 = 0.050 kg v 1i = 0.20 m/s v 2i = 0.10 m/s v 1f = 0.08 m/s v 2f = ? Write down given information

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? Givens: m 1 = 0.050 kg m 2 = 0.050 kg v 1i = 0.20 m/s v 2i = 0.10 m/s v 1f = 0.08 m/s v 2f = ? Determine the momentum of the cue ball before and after the collision p 1i = m 1 v 1i = 0.050 kg(0.20 m/s) = 0.01 kg m/s p 1f = m 1 v 1f = 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? Givens: m 1 = 0.050 kg m 2 = 0.050 kg v 1i = 0.20 m/s v 2i = 0.10 m/s v 1f = 0.08 m/s v 2f = ? Determine the momentum of the 8 ball before the collision p 2i = m 2 v 2i = 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 p before = p after p 1i + p 2i = p 1f + p 2f

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 p before = p after p 1i + p 2i = p 1f + p 2f 0.01 kg m/s + 0.005 kg m/s = 0.004 kg m/s + p 2f p 2f = 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 p 2f = m 2 v 2f 0.011 kg m/s = 0.050 kg (v 2f ) v 2f = 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. Givens: m b = 0.015 kg m bb = 5.085 kg v bi = ? v bbi = 0 m/s v bf = 1 m/s v bbf = 1 m/s Write down given information

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 p 2i = m 2 v 2i = 5.085 kg(0 m/s) = 0 kg m/s p 2f = m 2 v 2f = (5.085 kg) (1 m/s) = 5.085 kg m/s Givens: m 1 = 0.015 kg m 2 = 5.085 kg v 1i = ? v 2i = 0 m/s v 1f = 1 m/s v 2f = 1 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 p 1f = (m 1 ) v 1f = (0.015 kg) (1 m/s) = 0.015 kg m/s Givens: m 1 = 0.015 kg m 2 = 5.085 kg v 1i = ? v 2i = 0 m/s v 1f = 1 m/s v 2f = 1 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. p before = p after p 1i + p 2i = p 1f + p 2f Set up a conservation equation

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. p before = p after p 1i + p 2i = p 1f + p 2f p 1i + 0 = 0.015 kg m/s + 5.085 kg m/s p 1i = 5.1 kg m/s Substitute the momentum values and determine the momentum of the bullet before the collision

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 p 1i = m 1 v 1i 5.1 kg m/s = 0.015 kg(v 1i ) v 1i = 340 m/s

21.  http://www.youtube.com/watch?NR=1&v=wC XqcBpURZE&feature=endscreen http://www.youtube.com/watch?NR=1&v=wC XqcBpURZE&feature=endscreen