Solution:

1. Physics 1710—Chapter 9 Momentum & Impulse m1v1i m1v1f m2v2f 0 • Solution: P Conserved m1 v1i = - m1 v1f + m2 v2f E Conserved ½ m1 v1i2 = ½ m1 v1f2 + ½ m2 v2f2 Solve for ½ m1 v1f2 = Ef ½ m1 v1i2 = ½ m1 v1f2 + ½ m2 (m1/m2)2(v1i +v1f )2Ef = Ei [(m2 - m1)/(m2+m1)]2

2. Physics 1710—Chapter 9 Momentum & Impulse m1v1i m1v1f m2v2f 0 • Rutherford Backscattering Spectrometry: Ef = Ei [(m2 - m1)/(m2+m1)]2 = k Ei Silicon mSi = 28 u kSi = 0.56 Helium mHe = 4 u Uranium mU = 238 u kU = 0.93

3. Physics 1710—Chapter 9 Momentum & Impulse Surface Barrier Detector Amplifier +1 0 • Rutherford Backscattering Spectrometry: Pulse Height Analysis

4. Physics 1710—Chapter 9 Momentum & Impulse 0 • Rutherford Backscattering Spectrometry:

5. Physics 1710—Chapter 9 Momentum & Impulse 0 1’ Lecture 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. Force equals the time rate of change in momentum.

6. Physics 1710—Chapter 9 Momentum & Impulse 0 Impulse and Momentum d p = Fdt ∆p = ∫d p = ∫Fdt = Impulse The impulse on a body equals the change in momentum.

7. Physics 1710—Chapter 9 Momentum & Impulse 0 A B Impulse and Momentum Consider the following scenarios: Which will have the greater initial velocity? Scenario A or B?

8. Physics 1710—Chapter 9 Momentum & Impulse 0 Impulse and “Follow Through” Demonstration

9. Physics 1710—Chapter 9 Momentum & Impulse 0 ∆p = ∫Fdt ∆p=Fave∆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.

10. Physics 1710—Chapter 9 Momentum & Impulse 0 ∆p = ∫Fdt ∆p=Fave∆t For a impulse, the greater the time that the force is applied, the less will be the force. F = d p/dt

11. Physics 1710—Chapter 9 Momentum 0 Impulse and Seat Belts Seat 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.

12. Physics 1710—Chapter 9 Momentum & Impulse 0 Newton’s Second Law of Motion (What Newton actually said:) ∑F = d p/dtThe net external force is equal to the time rate of change in the linear momentum.

13. Physics 1710—Chapter 9 Momentum 0 Stopping Force ∆p = mv ∆t = s/vave = s/(v/2) Fave = ∆p/ ∆t = mv 2/(2s) Speed kills? : v 2 What about the sudden stop? :1/s

14. Physics 1710—Chapter 9 Momentum & Impulse 0 The Consider Two Bodies ⇐② ①⇒ F12= - F21 d p1/dt = - d p2/dt then ∆p1= - ∆p2 Thus, the momentum given to an ejected mass is equal and opposite to the momentum given to the ejecting mass.

15. Physics 1710—Chapter 9 Momentum & Impulse mv exhaust mv exhaust 0 Impulse Engine: Fthrust = dp/dt = - d(m v exhaust)/dt Fthrust = dp/dt = - v exhaust dm/dt

16. Physics 1710—Chapter 9 Momentum & Impulse 0 Center of Mass (CM) RCM ≡ ∑mir/ M Or RCM ≡ {∫ rdm }/ M

17. Physics 1710—Chapter 9 Momentum & Impulse 0 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.

18. Physics 1710—Chapter 9 Momentum & Impulse 0 Total Linear Momentum vCM = (1/M) ∑i mivi Thus: PCM = ∑ipi = total p aCM = dvCM /dt = (1/M) ∑i mi d vi /dt aCM = (1/M) ∑i mi d vi /dt Thus: FCM= M aCM = ∑i miai