Lecture 15: Momentum Conservation in 1D & 2D

1. Lecture 15 • Goals • Employ conservation of momentum in 1 D & 2D • Introduce Momentum and Impulse • Compare Force vs time to Force vs distance • Introduce Center-of-Mass Note: 2nd Exam, Monday, March 19th, 7:15 to 8:45 PM

2. Comments on Momentum Conservation • More general than conservation of mechanical energy • Momentum Conservation occurs in systems with no net external forces (as a vector quantity)

3. Explosions: A collision in reverse • A two piece assembly is hanging vertically at rest at the end of a 20 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? • If the time of the explosion is short then momentum is conserved in the x-direction because there is no net x force. This is not true of the y-direction but this is what we are interested in. After Before

4. Explosions: A collision in reverse • A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg mass 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 external 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

5. Explosions: A collision in reverse Before • A two piece assembly is hanging vertically at rest at the end of a 20 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 acy = M V2/r = T – Mg T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N After

6. 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?

7. Exercise Momentum is a Vector (!) quantity • x-direction: No net force so Px is conserved. • y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) py is not conserved (system is block and cart only) Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart What is the final velocity of the cart? 2 kg 5.0 m 30° 10 kg 7.5 m

8. Exercise Momentum is a Vector (!) quantity j i 1) ai = g sin 30° = 5 m/s2 2) d = 5 m / sin 30° = ½ aiDt2 10 m = 2.5 m/s2Dt2 2s = Dt v = aiDt= 10 m/s vx= v cos 30° = 8.7 m/s • x-direction: No net force so Px is conserved • y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it lands in the cart. Initial Final Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s V’x = 1.4 m/s N 5.0 m mg 30° 30° 7.5 m y x

9. Impulse (A variable external force applied for a given time) • Collisions often involve a varying force F(t): 0  maximum  0 • We can plot force vs time for a typical collision. The impulse, I, of the force is a vector defined as the integral of the force during the time of the collision. • The impulse measures momentum transfer

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

11. Force and Impulse F t • Two different collisions can have the same impulse since I 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

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

13. Exercise Force & Impulse F F heavy light • 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 ?

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

15. A perfectly inelastic collision in 2-D • If no external force momentum is conserved. • Momentum is a vector so px, py and pz are conseved V v1 m1 + m2 q m1 m2 v2 before after • x-dir px : m1 v1 = (m1 + m2 ) Vcos q • y-dir py : m2 v2 = (m1 + m2 ) Vsin q

16. 2D Elastic Collisions Before After • Perfectly elastic means that the objects do not stick and, by stipulation, mechanical energy is conservsed. • There are many more possible outcomes but, if no external force, then momentum will always be conserved

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

18. Billiards: Without external forces, conservation of both momentum & mech. energy after before pafterq pb Pafterf F • 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 If the masses of the two balls are equal then there will always be a 90° angle between the paths of the outgoing balls

19. Center of Mass • Most objects are not point-like but have a mass density and are often deformable. • So how does one account for this complexity in a straightforward way? Example • In football coaches often tell players attempting to tackle the ball carrier to look at their navel. • So why is this so?

20. System of Particles: Center of Mass (CM) + + m2 m1 m1 m2 • If an object is not held then it will rotate about the center of mass. • Center of mass: Where the system is balanced ! • Building a mobile is an exercise in finding centers of mass. mobile

21. System of Particles: Center of Mass RCM m2 m1 r2 r1 y x • How do we describe the “position” of a system made up of many parts ? • Define the Center of Mass (average position): • For a collection of N individual point-like particles whose masses and positions we know: (In this case, N = 2)

22. Momentum of the center-of-mass is just the total momentum • Notice • Impulse and momentum conservation applies to the center-of-mass

23. Sample calculation: RCM = (12,6) (12,12) 2m m m (0,0) (24,0) • Consider the following mass distribution: XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters XCM= 12 meters YCM= 6 meters

24. A classic example • There is a disc of uniform mass and radius r. However there is a hole of radius a a distance b (along the x-axis) away from the center. • Where is the center of mass for this object?

25. System of Particles: Center of Mass • For a continuous solid, convert sums to an integral. dm where dmis an infinitesimal mass element (see text for an example). r y x

26. Recap • Thursday, Review for exam • For Tuesday, Read Chapter 10.1-10.5