Lecture-08-08[1] - Outline Outline c Momentum c Momentum...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Outline Outline c Momentum c Momentum and Impulse c Conservation of Linear Momentum c Collisions Chapter-9 Impulse, Momentum & Conservation of Momentum….!!!! Momentum "momentu Definition: m" p mv ≡ = a a ; ; ; x x y y z z p mv p mv p mv = = = nd ( ) 2 Law: dv d mv dp F ma m dt dt dt = = = = a a a a a 2 1 2 1 ( ) t x x x x t p p p F t dt Δ =- = ∫ Momentum is a vector quantity. Force changes momentum. 2 2 1 mv p mv v c γ ≡ =- a a a 2 2 1 1 v c γ =- Momentum and Impulse Profile of the force during a collision. Microscopic view of a “bounce”. Impulse 2 1 "impulse Defi " niti ( on: ) x t x t J F t dt ≡ = ∫ x avg J F t = Δ x x f i x p J p p J Δ = = + (area under force curve) (area under force curve) fx ix x ix fx fx ix p p J p p v v m m = + = + = = + Quiz!! 6 Non Quiz!! Quiz!! Example: A Bouncing Ball A 100 g rubber ball is dropped from a height of 2.0 m onto a hard floor. The floor exerts a force on the ball as shown above. How high does the ball bounce? 2 2 1 2 1 2 2 2 2(9.80 m/s )(-2.0m) 6.26 m/s y y y v v g y g y v g y =- Δ = - Δ = - Δ = - = - 1 1 y y 2 2 J = F (t)dt= (base)(altitude) (300 N)(0.008 s) 1.20 Ns = = ∫ 2 1 (0.1 kg)( 6.26 m/s)+(1.20 Ns)=0.574 kg m/s y y y p p J = + =- 2 2 / (0.574 kg m/s)/(0.1 kg)=5.74 m/s y y v p m = = 1 1 2 2 2 2 2 3y 2y 2y 2 2 v =v 2 0 so y= v (5.74 m/s) (9.80 m/s ) 1.68 m / g y g- Δ = Δ = = Newton’s Second Law for a system of particles ( Reminder!!! ) • From the previous slide: • Here is a resultant force on particle i • According to the Newton’s Third Law, the forces that particles of the system exert on each other (internal forces) should cancel : • Here is the net force of all external forces that act on the system (assuming the mass of the system does not change) ∑ = i i com F a M a a net com F a M a a = i F a net F a tot net dP F dt = a a Conservation of Momentum 1 2 on 1 2 1 on 2 2 on 1 ( ) ( ) ( ) ( ) ( ) x x x x x d p F dt d p F F dt = = = - 1 2 1 2 2 on 1 2 on 1 [( ) ( ) ] ( ) ( ) ( ) ( ) x x x x x x d p p d p d p dt dt dt F F + = + =- = 1 2 ( ) ( ) constant x x p p + = 1 2 Before 1 2 After [( ) ( ) ] =[( ) ( ) ] x x x x p p p p + + The Law of Conservation of Momentum (derivable from Newton’s Laws and the definition of momentum.) In an isolated system, momentum is conserved. dP dt = a A System of N Particles The N=3 Case: For every pair of particles, the action/reaction pairs F j on k and F k on j are equal and opposite force vectors. In addition, each particle may be subjected to possible external forces F ext on k from agents outside the system....
View Full Document

This note was uploaded on 02/21/2009 for the course PHY A10 taught by Professor Tawfiq during the Fall '08 term at University of Toronto- Toronto.

Page1 / 20

Lecture-08-08[1] - Outline Outline c Momentum c Momentum...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online