Conservation of Momentum in 1D and 2D Collisions - Lecture 13

1. Lecture 13 Goals: • Chapter 9 • Employ conservation of momentum in 1 D & 2D • Examine forces over time (aka Impulse) • Chapter 10 • Understand the relationship between motion and energy Assignments: • HW5, due tomorrow • For Wednesday, Read all of Chapter 10

2. Momentum is conserved if no external force

3. Inelastic collision in 1-D: Example • A block of massMis 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 final speed V. • In terms of m, M,andV : What is the momentum of the bullet with speed v ? x v V before after

4. P After P Before v V x before after Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ? • Key question: Is there a net external x-dir force ? • If not, then momentum in the x-direction is conserved!

5. Momentum is a vector

6. V q m1 + m2 after A perfectly inelastic collision in 2-D • Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). v1 m1 m2 v2 before • If no external force momentum is conserved. • Momentum is a vector so px, py and pz

7. p1 A perfectly inelastic collision in 2-D • x-dir px : m1 v1 = (m1 + m2 ) Vcos q • y-dir py : m2 v2 = (m1 + m2 ) Vsin q p V v1 m1 + m2 q m1 p2 m2 v2 before after

8. Exercise Momentum is a Vector (!) quantity • Yes • No • Yes & No • Too little information given • A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction • In regards to the block landing in the cart is momentum conserved?

9. Before After Elastic Collisions • Elastic means that the objects do not stick. • There are many more possible outcomes but, if no external force, then momentum will always be conserved • Start with a 1-D problem.

10. Billiards • Consider the case where one ball is initially at rest. after before paq pb Paf F The final direction of the red ball will depend on where the balls hit.

11. Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) • Conservation of Momentum • x-dir Px : m vbefore = m vafter cos q + m Vafter cos f • y-dir Py :0 = m vafter sin q + m Vafter sin f after before pafterq pb Pafterf F

12. Explosions: A collision in reverse • A two piece assembly is hanging vertically at rest at the end of a 2.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. The time of the “explosion” is short compared to the swing of the string. • Does the tension in the string increase or decrease after the explosion? After Before

13. Explosions: A collision in reverse • A two piece assembly is hanging vertically at rest at the end of a 2.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 10 kg is ejected horizontally at 30 m/s. • Decipher the physics: 1. The green ball recoils in the –x direction (3rd Law) and, because there is no net force in the x-direction the x-momentum is conserved. 2. The motion of the green ball is constrained to a circular path…there must be centripetal (i.e., radial acceleration) After Before

14. Explosions: A collision in reverse • A two piece assembly is hanging vertically at rest at the end of a 20.0 m long massless string. The mass of the two pieces are 60 & 20 kg respectively. Suddenly you observe that the 20 kg mass is suddenly ejected horizontally at 30 m/s. • Cons. of x-momentum Px before= Px after = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s Tbefore = Weight = (60+20) x 10 N = 800 N SFy = m ay (radial) = M V2/r = T – Mg T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N After Before

15. Impulse (A variable force applied for a given time) • Gravity: At small displacements a “constant” force • Springs often provide a linear force (-kx) towards its equilibrium position (Chapter 10) • Collisions often involve a varying force F(t): 0  maximum  0 • We can plot force vs time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the time of the collision.

16. t t ti tf Force and Impulse (A variable force applied for a given time) • J a vector that reflects momentum transfer F ImpulseJ= area under this curve ! (Transfer of momentum !) Impulse has units ofNewton-seconds

17. F t Force and Impulse • Two different collisions can have the same impulse since J depends only on the momentum transfer, NOT the nature of the collision. same area F t t t t big, F small t small, F big

18. F t Average Force and Impulse Fav F Fav t t t t big, Favsmall t small, Fav big

19. F F heavy light Exercise Force & Impulse • heavier • lighter • same • can’t tell • Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ?

20. Boxing: Use Momentum and Impulse to estimate g “force”

21. Back of the envelope calculation (1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s  Question: Are these reasonable? Impulse J= p ~ marm varm ~ 49 kg m/s  Favg ~ J/t ~ 4900 N (1) mhead ~ 7 kg ahead = F / mhead ~ 700 m/s2 ~ 70 g ! • Enough to cause unconsciousness ~ 40% of a fatal blow • Only a rough estimate!

22. Ch. 10 : Kinetic & Potential energies • Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of masses • Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy • Mechanical energy, EMech, is defined to be the sum of U and K.

23. Lecture 13 Assignment: • HW6 up soon • For Wednesday: Read all of chapter 10