Lecture 12 Chapter 9: Momentum & Impulse Goals: Chapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time) Chapter 10 Understand the relationship between motion and energy Define Potential & Kinetic Energy Develop and exploit conservation of energy principle Assignment: HW6 due Wednesday 3/3 For Tuesday: Read all of chapter 10 1

Momentum Conservation Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (usually when forces act over a short time) It is a vector expression so must consider Px, Py and Pz if Fx (external) = 0 then Px is constant if Fy (external) = 0 then Py is constant if Fz (external) = 0 then Pz is constant

Inelastic collision in 1-D: Example A block of mass M is 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, and V : What is the momentum of the bullet with speed v ? x v V before after

Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ? Key question: Is x-momentum conserved ? P Before P After v V before after x

Exercise Momentum is a Vector (!) quantity 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? Yes No Yes & No Too little information given

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) 2 kg 5.0 m 30° 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? 10 kg 7.5 m

Exercise Momentum is a Vector (!) quantity 1) ai = g sin 30° = 5 m/s2 2) d = 5 m / sin 30° = ½ ai Dt2 10 m = 2.5 m/s2 Dt2 2s = Dt v = ai Dt = 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. i j N 5.0 m mg 30° 30° 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 7.5 m y x

Home Exercise Inelastic Collision in 1-D with numbers Do not try this at home! ice (no friction) Before: A 4000 kg bus, twice the mass of the car, moving at 30 m/s impacts the car at rest. What is the final speed after impact if they move together?

Home exercise Inelastic Collision in 1-D v = 0 ice M = 2m V0 (no friction) initially vf =? finally 2Vo/3 = 20 m/s

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

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 ) V cos q y-dir py : m2 v2 = (m1 + m2 ) V sin q

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. Before After

Billiards Consider the case where one ball is initially at rest. after before pa q pb vcm Pa f F The final direction of the red ball will depend on where the balls hit.

Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) 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 before after pafter q pb Pafter f F

Force and Impulse (A variable force applied for a given time) Gravity: At small displacements a “constant” force t Springs often provide a linear force (-k x) 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.

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

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. F t same area F t t t t big, F small t small, F big

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

Exercise 2 Force & Impulse 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 ? F light heavy heavier lighter same can’t tell

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

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 ~ 6 kg  ahead = F / mhead ~ 800 m/s2 ~ 80 g ! Enough to cause unconsciousness ~ 40% of a fatal blow Only a rough estimate!

Chapter 10: Energy (Forces over distance) We need to define an “isolated system” ? We need to define “conservative force” ? Recall, chapter 9, force acting for a period of time gives an impulse or a change (transfer) of momentum What if a force acts over a distance: Can we identify another useful quantity?

For Tuesday: Read all of chapter 10 Lecture 12 Assignment: HW6 due Wednesday 3/3 For Tuesday: Read all of chapter 10