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Physics 1501: Lecture 17, Pg 1 Physics 1501: Lecture 17 Today’s Agenda l Homework #6: due Friday l Midterm I: Friday only l Topics çChapter 9 »Momentum.

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Presentation on theme: "Physics 1501: Lecture 17, Pg 1 Physics 1501: Lecture 17 Today’s Agenda l Homework #6: due Friday l Midterm I: Friday only l Topics çChapter 9 »Momentum."— Presentation transcript:

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2 Physics 1501: Lecture 17, Pg 1 Physics 1501: Lecture 17 Today’s Agenda l Homework #6: due Friday l Midterm I: Friday only l Topics çChapter 9 »Momentum »Introduce Collisions

3 Physics 1501: Lecture 17, Pg 2 Chapter 9 Linear Momentum l Newton’s 2nd Law: Fa F = ma dvdv v p l Units of linear momentum are kg m/s. pv p = mv pv (p is a vector since v is a vector). So p x = mv x etc. l Definition: p l Definition: For a single particle, the momentum p is defined as:

4 Physics 1501: Lecture 17, Pg 3 Momentum Conservation l The concept of momentum conservation is one of the most fundamental principles in physics. l This is a component (vector) equation. çWe can apply it to any direction in which there is no external force applied. l You will see that we often have momentum conservation even when (mechanical) energy is not conserved.

5 Physics 1501: Lecture 17, Pg 4 Elastic vs. Inelastic Collisions l A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. K before = K after çCarts colliding with a spring in between, billiard balls, etc. vvivvi l 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 çCar crashes, collisions where objects stick together, etc.

6 Physics 1501: Lecture 17, Pg 5 Inelastic collision in 1-D: Example 1 l 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 : çWhat is the momentum of the bullet with speed v ? çWhat is the initial energy of the system ? çWhat is the final energy of the system ? çIs energy conserved ? v V beforeafter x

7 Physics 1501: Lecture 17, Pg 6 Example 1... Momentum is conserved in the x direction ! l 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 ! çP x,before = P x,after çmv = (M+m) V v V beforeafter x

8 Physics 1501: Lecture 17, Pg 7 Example 1... l Now consider the energy of the system before and after: l Before: l After: l So Energy is NOT conserved Energy is NOT conserved (friction stopped the bullet) However momentum was conserved, and this was useful.

9 Physics 1501: Lecture 17, Pg 8 Lecture 17, ACT 1 Inelastic Collision in 1-D ice (no friction) Winter in Storrs

10 Physics 1501: Lecture 17, Pg 9 Lecture 17, ACT 1 Inelastic Collision in 1-D v f v f = ? finally m v = 0 ice M = 2m V0V0V0V0 (no friction) initially V f = A) 0 B) V o /2 C) 2V o /3 D) 3V o /2 E) 2V o

11 Physics 1501: Lecture 17, Pg 10 Lecture 17, ACT 2 Momentum Conservation l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. çWhich box ends up moving fastest ? (a) (b) (c) (a) Box 1 (b) Box 2 (c) same 1 2

12 Physics 1501: Lecture 17, Pg 11 Inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2

13 Physics 1501: Lecture 17, Pg 12 Inelastic collision in 2-D... l There are no net external forces acting. çUse momentum conservation for both components. vv1vv1 vv2vv2 V V = (V x,V y ) m1m1 m2m2 m 1 + m 2

14 Physics 1501: Lecture 17, Pg 13 Inelastic collision in 2-D... l So we know all about the motion after the collision ! V V = (V x,V y ) VxVx VyVy 

15 Physics 1501: Lecture 17, Pg 14 Inelastic collision in 2-D... l We can see the same thing using vectors: P pp1pp1 pp2pp2 P pp1pp1 pp2pp2 

16 Physics 1501: Lecture 17, Pg 15 Comment on Energy Conservation l We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. çEnergy is lost: »Heat (bomb) »Bending of metal (crashing cars) l Kinetic energy is not conserved since work is done during the collision ! l Momentum along a certain direction is conserved when there are no external forces acting in this direction. çIn general, easier to satisfy than energy conservation.

17 Physics 1501: Lecture 17, Pg 16 Elastic Collisions l Elastic means that energy is conserved as well as momentum. l This gives us more constraints. çWe can solve more complicated problems !! çBilliards (2-D collision). çThe colliding objects have separate motions after the collision as well as before. l Start with a simpler 1-D problem. BeforeAfter

18 Physics 1501: Lecture 17, Pg 17 Elastic Collision in 1-D v 1,b v 2,b before x m1m1 m2m2 v 1,a v 2,a after m1m1 m2m2

19 Physics 1501: Lecture 17, Pg 18 Elastic Collision in 1-D v 1,b v 2,b v 1,a v 2,a before after x m1m1 m2m2 Conserve P X m 1 v 1,b + m 2 v 2,b = m 1 v 1,a + m 2 v 2,a Conserve Energy 1 / 2 m 1 v 2 1,b + 1 / 2 m 2 v 2 2,b = 1 / 2 m 1 v 2 1,a + 1 / 2 m 2 v 2 2,a Suppose we know v 1,b and v 2,b We need to solve for v 1,a and v 2,a Should be no problem 2 equations & 2 unknowns ! m 1 v 1,b + m 2 v 2,b = m 1 v 1,a + m 2 v 2,a

20 Physics 1501: Lecture 17, Pg 19 Elastic Collision in 1-D l After some moderately tedious algebra, we can derive the following equations for the final velocities,

21 Physics 1501: Lecture 17, Pg 20 Example - Elastic Collision l Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v 1. If they collide elastically, what is the final velocity of each car ? Note that this means, m 1 = m 2 = m v 2B = 0

22 Physics 1501: Lecture 17, Pg 21 Example - Elastic Collision l Let’s start with the equations for conserving momentum and energy, mv 1,B = mv 1,A + mv 2,A (1) 1 / 2 mv 2 1,B = 1 / 2 mv 2 1,A + 1 / 2 mv 2 2,A (2) Initially, v 2 = 0. After the collision, v 1 = 0. To satisfy equation (1), v 2A = v 1B. v 1A =0, v 2A =0 or bothBoth: contradicts (2) If v 2A =0 : v 1A = v 1B from (1) or (2) no collision !

23 Physics 1501: Lecture 17, Pg 22 Lecture 17, ACT 3 Elastic Collisions l 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) YesB) No Before After?

24 Physics 1501: Lecture 17, Pg 23 Example of 2-D elastic collisions: Billiards l 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...

25 Physics 1501: Lecture 17, Pg 24 Billiards l Consider the case where one ball is initially at rest. ppappa ppbppb F PPaPPa beforeafter the final direction of the red ball will depend on where the balls hit. v v cm

26 Physics 1501: Lecture 17, Pg 25 ppP l We know momentum is conserved: p b = p a + P a l We also know that energy is conserved: l Comparing these two equations tells us that: Billiards pPpP p b 2 = (p a + P a ) 2 = p a 2 + P a 2 + 2 p a  P a m2 P m2 p m2 p 2 a 2 a 2 b  pP p a  P a = 0 and must therefore be orthogonal! Or … one momentum must be zero. ppappa ppbppb PPaPPa Ppp 2 a 2 a 2 b 

27 Physics 1501: Lecture 17, Pg 26 Billiards l The final directions are separated by 90 o. ppappa ppbppb F PPaPPa beforeafter v v cm Active FigureFigure

28 Physics 1501: Lecture 17, Pg 27 Lecture 17 – ACT 4 Pool Shark l Can I sink the red ball without scratching ? Ignore spin and friction. A) YesB) NoC) More info needed

29 Physics 1501: Lecture 17, Pg 28 Billiards. l More generally, we can sink the red ball without sinking the white ball – fortunately.

30 Physics 1501: Lecture 17, Pg 29 Billiards l However, we can also scratch. All we know is that the angle between the balls is 90 o.


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