Conserv. of Momentum (Applications)

Transcription

1 Conserv. of Momentum (Applications) Announcements: Next midterm a week from Thursday (3/15). Chapters 6 9 will be covered LA information session at 6pm today, UMC 235. Will do some longer examples today. Material from Chapter Wednesday Chapter Friday Chapter 9.12 Web page:

2 In all collisions, momentum is conserved: This is a vector equation. Components are conserved. Collision summary In elastic collisions, kinetic energy is conserved. In completely inelastic collisions, final velocities are the same Components:

3 Clicker question 1 Set frequency to BA Ball A strikes a stationary ball B in a 2D collision. The initial momentum of ball A,, and the final momentum of ball B,, are shown on the graph. Which vector gives the final ball A momentum? A B C D

4 Clicker question 1 Set frequency to BA Ball A strikes a stationary ball B in a 2D collision. The initial momentum of ball A,, and the final momentum of ball B,, are shown on the graph. Which vector gives the final ball A momentum? The initial momentum of the system is. By conservation of momentum, the final total momentum must be the same so or Can also do component addition: A B C D

5 Clicker question 2 Set frequency to BA Suppose the entire population of Earth gathers in one location and, at a pre-arranged signal, everyone jumps up. About a second later, 7 billion people land back on the ground. After the people have landed, the Earth's momentum is.. A: the same as it was before the people jumped. B: different than it was before the people jumped. C: impossible to know whether the Earth's momentum changed/don't know.

6 Clicker question 2 Set frequency to BA Suppose the entire population of Earth gathers in one location and, at a pre-arranged signal, everyone jumps up. About a second later, 7 billion people land back on the ground. After the people have landed, the Earth's momentum is.. A: the same as it was before the people jumped. B: different than it was before the people jumped. C: impossible to know whether the Earth's momentum changed/don't know. Same as before. Started zero, ended zero. Conservation of momentum, there are no EXTERNAL forces!

7 Kick of a Rifle (Cons. of Momentum) A 3.24 kg Winchester Super X rifle initially at rest, fires a 11.7 g bullet with a muzzle speed of 800 m/s. What is the recoil velocity of the rifle? What is the ratio of the kinetic energies of the bullet and the rifle? 0 = m r v r + m b v b v r = m b v b = kg (800m /s) = 2.89m /s m r 3.24kg K = 1 2 mv 2 = p2 2m K r K b = m b m r p is conserved, = kg 3.24kg = 0.38% K r m r = K b m b

8 A collision example A 10 g bullet hits a 2 kg block of ballistic gel and is embedded. The gel comes to rest after traveling 40 cm along the floor with a coefficient of kinetic friction of 0.5. What is the initial speed of the bullet?.01 kg 2 kg 2 kg Two sequential problems:.4 m 1. Completely inelastic collision between bullet and gel 2. An energy problem (gel kinetic energy is lost to friction)

9 The collision part of the example.01 kg The collision part: 2 kg 2 kg.4 m The collision conserves momentum: Collision is completely inelastic and the gel is initially at rest: Need v gel to get the bullet velocity

10 The energy part of the example.01 kg The work-energy part: Conservation of energy: 2 kg.4 m 2 kg No potential energy and no final kinetic energy: Work due to friction: Putting it together: So

11 The collision part again.01 kg 2 kg 2 kg Our analysis of the completely inelastic collision gave.4 m We obtained v gel from analyzing the work-energy part of the problem

12 Clicker question 3 Set frequency to BA Suppose the 0.01 kg bullet from the previous example hit the same 2 kg gel block at 400 m/s but passed through the gel, exiting the gel at 100 m/s. What speed would the gel have immediately after the bullet exits? A. 0.5 m/s B. 1.0 m/s.01 kg 2 kg C. 1.5 m/s D. 2.0 m/s E. 300 m/s

13 Clicker question 3 Set frequency to BA Suppose the 0.01 kg bullet from the previous example hit the same 2 kg gel block at 400 m/s but passed through the gel, exiting the gel at 100 m/s. What speed would the gel have immediately after the bullet exits? A. 0.5 m/s B. 1.0 m/s.01 kg 2 kg C. 1.5 m/s D. 2.0 m/s Momentum conservation: E. 300 m/s Initial gel velocity is 0:

14 An elastic example A tennis ball is put on top of a basketball and both are dropped from a height of 1 m. How high will the tennis ball go? Assume all collisions are elastic. Can use conservation of energy to get velocity after falling 1 m. becomes so Basketball hits first. What happens? If no friction losses then by conservation of energy the upward speed after bounce equals the downward speed before bounce Or, elastic collisions conserve kinetic energy. Same answer. Is momentum conserved? Only if you consider Earth as part of the system

15 An elastic example So we have a 1D elastic collision between basketball moving at v B = 4.5 m/s and tennis ball moving at v T = 4.5 m/s The collision conserves momentum: Elastic collision conserves kinetic energy: Solve this problem in the reference frame where the basketball is at rest This gives and

16 We have An elastic example and Factoring: Divide the equations to get Substitute in to get Rearrange: Solve: becomes Substitute back in to get

17 An elastic example So in the basketball rest frame we have a post collision tennis ball velocity of Masses and initial velocities: This is with respect to the basketball which is moving up at 4.5 m/s. Therefore the tennis ball speed with respect to the ground is 6.3 m/s m/s = 10.8 m/s.

18 Basketball and tennis ball Before collision, the tennis ball is moving down at 4.5 m/s After collision the tennis ball is moving up at 10.8 m/s What height will the tennis ball reach? 19 foot height from just a 3 foot drop?!

19 Elastic Collisions in 2D (Not Head-on) u 1 θ 1 v 1 v 2 A proton moving at speed 5 km/s makes an elastic collision with another proton. If we know one proton has an angle of 37 degrees, what are the velocities, v 1, v 2, θ 1 p x : p y : KE : m 1 u = m 1 v 1 cosθ 1 + m 2 v 2 cos(37) 0 = m 1 v 1 sinθ 1 m 2 v 2 sin(37) 1 2 m 1u 2 1 = 1 2 m 1v m 2 2v 2

20 Elastic Collisions in 2D (Not Head-on) v 1 m 1 u = m 1 v 1 cosθ 1 + m 2 v 2 cos(37) u 1 = 5 km/s θ 1 0 = m 1 v 1 sinθ 1 m 2 v 2 sin(37) 1 2 m 1u 2 1 = 1 2 m 1v m 2 2v 2 v 2 Divide out mass as m 1 =m 2 u 1 = v 1 cosθ 1 + v 2 cos(37) 0 = v 1 sinθ 1 v 2 sin(37) u 1 2 = v v 2 2 v 1 cosθ 1 = 5.8v 2 v 1 sinθ 1 =.6v 2 25 = v v 2 v 1 2 = 25 8v 2 + v 2 2 v 1 = 4 m/s ; v 2 = 3 m/s ; θ 1 = 53 degrees