1 / 21

Understanding Momentum and Impulse: Principles of Physics

This guide explores the concepts of momentum and impulse, two fundamental principles in physics. Momentum is defined as the product of an object's mass and velocity, often referred to as "mass in motion." We calculate momentum using the formula p = mv, where p is momentum, m is mass in kilograms, and v is velocity in meters per second. Key topics include conservation of momentum, calculations involving collisions (both elastic and inelastic), and examples demonstrating these principles in practical scenarios. Learn the significance of momentum's direction and how it remains conserved during interactions.

chandler-wiley
Télécharger la présentation

Understanding Momentum and Impulse: Principles of Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related