1 / 9

Conservation of Momentum

https:// www.youtube.com/watch?v=vWVZ6APXM4w. Conservation of Momentum. CH6 – Section 2. Conservation of Momentum.

meara
Télécharger la présentation

Conservation of Momentum

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. https://www.youtube.com/watch?v=vWVZ6APXM4w Conservation of Momentum CH6 – Section 2

  2. Conservation of Momentum • For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. • This can be expressed by the formula: m1v1i + m2v2i = m1v1f + m2v2f (Total momentum of an isolated system remains constant)

  3. Conceptual Challenge • If a reckless ice skater collides with another skater who is standing on the ice, is it possible for both skaters to be at rest immediately after the collision? No, the only possible way for their final total momentum to be zero is if the initial total momentum is also zero. This could happen only if both skaters initially have the same magnitude of momentum but opposite directions.

  4. Inside Story • Read the INSIDE STORY ON SURVIVING A COLLISION in your textbook on p. 207

  5. Sample problem 1 • A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

  6. Sample Problem 1 cont. • State what is GIVEN: • m1 = 76 kg (mass of boater) • m2 = 45 kg (mass of boat) • v1i = 0 (initial velocity of boater) • v2i = 0 (initial velocity of boat) • v1f = 2.5 m/s to the right Establish a direction convention Right is positive Left is negative

  7. Sample Problem 1 cont. • Solution: Because the total momentum of an isolated system remains constant, the total initial momentum of the boater and the boat will be equal to the total final momentum of the boater and the boat, and thus we can solve this problem by using the Conservation of Momentum formula: m1v1i + m2v2i = m1v1f + m2v2f

  8. Sample Problem 1 cont. Because the boater and the boat are initially at rest, the total initial momentum of the system is equal to zero. Therefore: 0 = m1v1f + m2v2f Rearrange the formula to solve for final velocity of the boat. v2f= -(m1/m2) v1f v2f= -(76kg/45kg) 2.5m/s v2f=-4.2 m/s to the left

  9. Highlights of CH6 - Section 2 • In all interactions between isolated objects, momentum is conserved. • In every interaction between two isolated objects, the change in momentum of the first object is equal to and opposite the change in momentum of the second object. • If the momentum of one object increases after a collision, then the momentum of the other object in the situation must decrease by an equal amount.

More Related