1. Solution: m 1 v 1i m 1 v 1f m 2 v 2f P Conserved m 1 v 1i = - m 1 v 1f + m 2 v 2f E Conserved ½ m 1 v 1i 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f 2 Solve for Solve for ½ m 1 v 1f 2 = E f ½ m 1 v 1i 2 = ½ m 1 v 1f 2 + ½ m 2 (m 1 /m 2 ) 2 (v 1i +v 1f ) 2 E f = E i [(m 2 - m 1 )/(m 2 +m 1 )] 2 Physics 1710—Chapter 9 Momentum & Impulse

2. Rutherford Backscattering Spectrometry: m 1 v 1i m 1 v 1f m 2 v 2f E f = E i [(m 2 - m 1 )/(m 2 +m 1 )] 2 = k E i Silicon m Si = 28 u Uranium m U = 238 u Helium m He = 4 u k Si = 0.56 k U = 0.93 Physics 1710—Chapter 9 Momentum & Impulse

3. Rutherford Backscattering Spectrometry: Amplifier Surface Barrier Detector Pulse Height Analysis +1 Physics 1710—Chapter 9 Momentum & Impulse

4. Rutherford Backscattering Spectrometry: Physics 1710—Chapter 9 Momentum & Impulse

5. 1’ Lecture Impulse is the time integrated force. Impulse is the time integrated force. The motion of a system of point particles is a combination of motion of the center of mass (CM) and the motion about the CM. The motion of a system of point particles is a combination of motion of the center of mass (CM) and the motion about the CM. Force equals the time rate of change in momentum. Force equals the time rate of change in momentum. Physics 1710—Chapter 9 Momentum & Impulse

6. Impulse and Momentum d p = F dt ∆p = ∫d p = ∫F dt = Impulse The impulse on a body equals the change in momentum. Physics 1710—Chapter 9 Momentum & Impulse

7. Impulse and Momentum Consider the following scenarios: Consider the following scenarios: Which will have the greater initial velocity? Scenario A or B? AB Physics 1710—Chapter 9 Momentum & Impulse

8. Impulse and “Follow Through” Demonstration Physics 1710—Chapter 9 Momentum & Impulse

9. ∆p = ∫F dt ∆p =F ave ∆t For a given force, the greater the time that the force is applied, the greater will be the impulse and, thus, the change in momentum. Physics 1710—Chapter 9 Momentum & Impulse

10. ∆p = ∫F dt ∆p =F ave ∆t For a impulse, the greater the time that the force is applied, the less will be the force. F = d p/dt Physics 1710—Chapter 9 Momentum & Impulse

11. Impulse and Seat Belts Seat Belts ( and air bagsSeat Belts ( and air bags and crumple zones) increase the stopping time ∆t. If ∆p is the same in two instants the impulse will be the same. The case with the longer ∆t will exhibit the smaller average force.If ∆p is the same in two instants the impulse will be the same. The case with the longer ∆t will exhibit the smaller average force. Physics 1710—Chapter 9 Momentum

12. Newton’s Second Law of Motion (What Newton actually said:) ∑F = d p/dt The net external force is equal to the time rate of change in the linear momentum. Physics 1710—Chapter 9 Momentum & Impulse

13. Stopping Force ∆ p = mv ∆t = s/v ave = s/(v/2) F ave = ∆ p/ ∆t = mv 2 /(2s) Speed kills? : v 2 What about the sudden stop? :1/s Physics 1710—Chapter 9 Momentum

14. The Consider Two Bodies ⇐② ①⇒ F 12 = - F 21 d p 1 /dt = - d p 2 /dt then ∆p 1 = - ∆p 2 Thus, the momentum given to an ejected mass is equal and opposite to the momentum given to the ejecting mass. Physics 1710—Chapter 9 Momentum & Impulse

15. Impulse Engine: F thrust = dp/dt = - d(m v exhaust )/dt F thrust = dp/dt = - d(m v exhaust )/dt F thrust = dp/dt = - v exhaust dm/dt F thrust = dp/dt = - v exhaust dm/dt Physics 1710—Chapter 9 Momentum & Impulse exhaust mv exhaust

16. Center of Mass (CM) R CM ≡ ∑m i r/ M Or R CM ≡ {∫ rdm }/ M Physics 1710—Chapter 9 Momentum & Impulse

17. The center of mass (CM) of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the resultant external force on the system. Physics 1710—Chapter 9 Momentum & Impulse

18. Total Linear Momentum v CM = (1/M) ∑ i m i v i Thus: P CM = ∑ i p i = total p a CM = d v CM /dt = (1/M) ∑ i m i d v i /dt a CM = (1/M) ∑ i m i d v i /dt Thus: F CM = M a CM = ∑ i m i a i Physics 1710—Chapter 9 Momentum & Impulse