Physics 1501 Lecture 17

Transcription

1 Physics 50: Lecture 7 Today s Agenda Homework #6: due Friday Midterm I: Friday only Topics Chapter 9» Momentum» Introduce Collisions Physics 50: Lecture 7, Pg Newton s nd Law: Chapter 9 Linear Momentum Definition: For a single particle, the momentum p is defined as: (p is a vector since v is a p = mv vector). So p x = mv x etc. F = ma = m dv d = (mv dt dt ) dp F = dt Units of linear momentum are kg m/s. Physics 50: Lecture 7, Pg Page

2 Momentum Conservation F EXT dp dp = = 0 F EXT = 0 dt dt The concept of momentum conservation is one of the most fundamental principles in physics. This is a component (vector) equation. We can apply it to any direction in which there is no external force applied. You will see that we often have momentum conservation even when (mechanical) energy is not conserved. Physics 50: Lecture 7, Pg 3 Elastic vs. Inelastic Collisions A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. K before = K after Carts colliding with a spring in between, billiard balls, etc. v i A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. K before K after Car crashes, collisions where objects stick together, etc. Physics 50: Lecture 7, Pg 4 Page

3 Inelastic collision in -D: Example A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : What is the momentum of the bullet with speed v? What is the initial energy of the system? What is the final energy of the system? Is energy conserved? v V x before after Physics 50: Lecture 7, Pg 5 Example... Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction Momentum is conserved in the x direction! P x,before = P x,after mv = (M+m) V! v = M + m $ # &V " m % v V x before after Physics 50: Lecture 7, Pg 6 Page 3

4 v =! M + m$ # & " m % V Example... Now consider the energy of the system before and after: Before: After: E B = mv = m! M + m $ # & V =! M + m $ # &( M + m)v " m % " m % E A = ( M + m )V So! m $ E A = # &E B " M + m % Energy is NOT conserved (friction stopped the bullet) However momentum was conserved, and this was useful. Physics 50: Lecture 7, Pg 7 Lecture 7, ACT Inelastic Collision in -D Winter in Storrs ice (no friction) Physics 50: Lecture 7, Pg 8 Page 4

5 Lecture 7, ACT Inelastic Collision in -D M = m initially m V 0 v = 0 ice (no friction) finally v f =? V f = A) 0 B) V o / C) V o /3 D) 3V o / E) V o Physics 50: Lecture 7, Pg 9 Lecture 7, ACT Momentum Conservation Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box bounces back, while the ball hitting box gets stuck. Which box ends up moving fastest? (a) Box (b) Box (c) same Physics 50: Lecture 7, Pg 0 Page 5

6 Inelastic collision in -D Consider a collision in -D (cars crashing at a slippery intersection...no friction). v V m m + m m v before after Physics 50: Lecture 7, Pg Inelastic collision in -D... There are no net external forces acting. Use momentum conservation for both components. P x,a = Px,b m v = m + m P y,a = Py,b m v = m + m m ( )V x V x = ( m + m ) v m ( )V y V y = ( m + m ) v v V = (V x,v y ) m m + m m v Physics 50: Lecture 7, Pg Page 6

7 Inelastic collision in -D... So we know all about the motion after the collision! V = (V x,v y ) θ V x V y m Vx = m + m v V y = ( ) m m m v ( + ) V y tan! = = = V x m v m v p p Physics 50: Lecture 7, Pg 3 Inelastic collision in -D... We can see the same thing using vectors: P p p P θ p p tan! = p p Physics 50: Lecture 7, Pg 4 Page 7

8 Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Energy is lost:» Heat (bomb)» Bending of metal (crashing cars) Kinetic energy is not conserved since work is done during the collision! Momentum along a certain direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy than energy conservation. Physics 50: Lecture 7, Pg 5 Elastic Collisions Elastic means that energy is conserved as well as momentum. This gives us more constraints. We can solve more complicated problems!! Billiards (-D collision). The colliding objects have separate motions after the collision as well as before. Before After Start with a simpler -D problem. Physics 50: Lecture 7, Pg 6 Page 8

9 Elastic Collision in -D before m m v,b v,b x after m m v,a v,a Physics 50: Lecture 7, Pg 7 Elastic Collision in -D m m Conserve P X before v,b v,b m v,b + m v,b = m v,a + m v,a x after Conserve Energy v,a v,a / m v,b + / m v,b = / m v,a + / m v,a Suppose we know v,b and v,b We need to solve for v,a and v,a Should be no problem equations & unknowns! Physics 50: Lecture 7, Pg 8 Page 9

10 Elastic Collision in -D After some moderately tedious algebra, we can derive the following equations for the final velocities, m! m m + m m m + m v = ( ) v + ( ) v A B B m m + m m! m m + m v = ( ) v + ( ) v A B B Physics 50: Lecture 7, Pg 9 Example - Elastic Collision Suppose I have identical bumper cars. One is motionless and the other is approaching it with velocity v. If they collide elastically, what is the final velocity of each car? Note that this means, m = m = m v B = 0 Physics 50: Lecture 7, Pg 0 Page 0

11 Example - Elastic Collision Let s start with the equations for conserving momentum and energy, mv,b = mv,a + mv,a () / mv,b = / mv,a + / mv,a () ( v v ) = v + v A A A A + v 0 Initially, v = 0. After the collision, v = 0. To satisfy equation (), v A = v B. v = A A v A =0, v A =0 or both Both: contradicts () If v A =0 : v A = v B from () or () no collision! Physics 50: Lecture 7, Pg Lecture 7, ACT 3 Elastic Collisions I have a line of 3 bumper cars all touching. A fourth car smashes into the others from behind. Is it possible to satisfy both conservation of energy and momentum if two cars are moving after the collision? All masses are identical, elastic collision. A) Yes B) No Before After? Physics 50: Lecture 7, Pg Page

12 Example of -D elastic collisions: Billiards If all we know is the initial velocity of the cue ball, we don t have enough information to solve for the exact paths after the collision. But we can learn some useful things... Physics 50: Lecture 7, Pg 3 Billiards Consider the case where one ball is initially at rest. p b p a v cm F P a before the final direction of the red ball will depend on where the balls hit. after Physics 50: Lecture 7, Pg 4 Page

13 Billiards We know momentum is conserved: p b = p a + P a p b = (p a + P a ) = p a + P a + p a P a We also know that energy is conserved: b + p p a = m m P a m p b = p a + P a Comparing these two equations tells us that: p a P a = 0 and must therefore be orthogonal! Or one momentum must be zero. p a p b P a Physics 50: Lecture 7, Pg 5 Billiards The final directions are separated by 90 o. p a p b v cm F P a before after Active Figure Physics 50: Lecture 7, Pg 6 Page 3

14 Lecture 7 ACT 4 Pool Shark Can I sink the red ball without scratching? Ignore spin and friction. A) Yes B) No C) More info needed Physics 50: Lecture 7, Pg 7 Billiards. More generally, we can sink the red ball without sinking the white ball fortunately. Physics 50: Lecture 7, Pg 8 Page 4

15 Billiards However, we can also scratch. All we know is that the angle between the balls is 90 o. Physics 50: Lecture 7, Pg 9 Page 5