Physics 207: Lecture 11, Pg 1 Lecture 11 Goals: Assignment: l Read through Chapter 10 l MP HW5, due Wednesday 3/3 Chapter 9: Momentum & Impulse Chapter 9: Momentum & Impulse  Understand what momentum is and how it relates to forces  Employ momentum conservation principles  In problems with 1D and 2D Collisions  In problems having an impulse (Force vs. time)  Chapter 8: Use models with free fall

Physics 207: Lecture 11, Pg 2 Problem 7.34 Hint Suggested Steps l Two independent free body diagrams are necessary l Draw in the forces on the top and bottom blocks l Top Block  Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding) l Bottom Block  Forces: 1. normal to bottom surface 2. normal to top block interface 3. rope tension (to the left) 4. weight (2 kg) 5. friction with top block 6. friction with surface N Use Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.

Physics 207: Lecture 11, Pg 3 Locomotion: how fast can a biped walk?

Physics 207: Lecture 11, Pg 4 How fast can a biped walk? What about weight? (a)A heavier person of equal height and proportions can walk faster than a lighter person (b)A lighter person of equal height and proportions can walk faster than a heavier person (c)To first order, size doesn’t matter

Physics 207: Lecture 11, Pg 5 How fast can a biped walk? What about height? (a)A taller person of equal weight and proportions can walk faster than a shorter person (b)A shorter person of equal weight and proportions can walk faster than a taller person (c)To first order, height doesn’t matter

Physics 207: Lecture 11, Pg 6 How fast can a biped walk? What can we say about the walker’s acceleration if there is UCM (a smooth walker) ? Acceleration is radial ! So where does it, a r, come from? (i.e., what external forces are on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways

Physics 207: Lecture 11, Pg 7 Orbiting satellites v T = (gr) ½

Physics 207: Lecture 11, Pg 8 Geostationary orbit

Physics 207: Lecture 11, Pg 9 Geostationary orbit l The radius of the Earth is ~6000 km but at km you are ~42000 km from the center of the earth. l F gravity is proportional to r -2 and so little g is now ~10 m/s 2 / 50 n v T = (0.20 * ) ½ m/s = 3000 m/s At 3000 m/s, period T = 2  r / v T = 2  / 3000 sec = = sec = s/ 3600 s/hr = 24 hrs l Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) l Great for communication satellites (1 st pointed out by Arthur C. Clarke)

Physics 207: Lecture 11, Pg 10 Impulse & Linear Momentum l Transition from forces to conservation laws Newton’s Laws  Conservation Laws Conservation Laws  Newton’s Laws They are different faces of the same physics NOTE: We have studied “impulse” and “momentum” but we have not explicitly named them as such Conservation of momentum is far more general than conservation of mechanical energy

Physics 207: Lecture 11, Pg 11 Collisions are a fact of life

Physics 207: Lecture 11, Pg 12 Forces vs time (and space, Ch. 10) l Underlying any “new” concept in Chapter 9 is (1) A net force changes velocity (either magnitude or direction) (2) For any action there is an equal and opposite reaction l If we emphasize Newton’s 3 rd Law and look at the changes with time then this leads to the Conservation of Momentum Principle

Physics 207: Lecture 11, Pg 13 Example 1 A 2 kg block, initially at rest on frictionless horizontal surface, is acted on by a 10 N horizontal force for 2 seconds (in 1D). What is the final velocity? l F is to the positive & F = ma thus a = F/m = 5 m/s 2 v = v 0 + a  t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction) Notice: v - v 0 = a  t  m (v - v 0 ) = ma  t  m  v = F  t If the mass had been 4 kg … now what final velocity? - + F (N) time (s)

Physics 207: Lecture 11, Pg 14 Twice the mass l Same force l Same time Half the acceleration (a = F / m ’ ) l Half the velocity ! ( 5 m/s ) F (N) Time (sec) Before

Physics 207: Lecture 11, Pg 15 Example 1 l Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity). l Here, mass times the velocity always gives the same value. (Always 20 kg m/s.) F (N) Time (sec) Area under curve is still the same ! Force x change in time = mass x change in velocity

Physics 207: Lecture 11, Pg 16 Example 1 l There many situations in which the sum of the products “mass times velocity” is constant over time l To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s) l A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” F (N) Time (sec) Area under curve : “impulse” With: m  v = F avg  t

Physics 207: Lecture 11, Pg 17 Force curves are usually a bit different in the real world

Physics 207: Lecture 11, Pg 18 Example 1 with Action-Reaction l Now the 10 N force from before is applied by person A on person B while standing on a frictionless surface l For the force of A on B there is an equal and opposite force of B on A F (N) Time (sec) A on B B on A M A x  V A = Area of top curve M B x  V B = Area of bottom curve Area (top) + Area (bottom) = 0

Physics 207: Lecture 11, Pg 19 Example 1 with Action-Reaction M A  V A + M B  V B = 0 M A [V A (final) - V A (initial) ] + M B [V B (final) - V B (initial) ] = 0 Rearranging terms M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V (Remember: frictionless surface)

Physics 207: Lecture 11, Pg 20 Example 1 with Action-Reaction M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V Define MV to be the “momentum” and this is conserved in a system if and only if the system is not acted on by a net external force (choosing the system is key) Conservation of momentum is a special case of applying Newton’s Laws

Physics 207: Lecture 11, Pg 21 Applications of Momentum Conservation 234 Th 238 U Alpha Decay 4 He v1v1 v2v2 Explosions Radioactive decay: Collisions

Physics 207: Lecture 11, Pg 22 Impulse & Linear Momentum l Newton’s 2 nd Law: Fa F = ma l This is the most general statement of Newton’s 2 nd Law pv p ≡ mv pv (p is a vector since v is a vector) So p x = mv x and so on (y and z directions) l Definition: For a single particle, the momentum p is defined as:

Physics 207: Lecture 11, Pg 23 Momentum Conservation l Momentum conservation (recasts Newton’s 2 nd Law when F = 0) is an important principle l It is a vector expression (P x, P y and P z ).  And applies to any situation in which there is NO net external force applied (in terms of the x, y & z axes).

Physics 207: Lecture 11, Pg 24 Momentum Conservation l Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved l Momentum is a vector quantity and we can independently assess its conservation in the x, y and z directions (e.g., net forces in the z direction do not affect the momentum of the x & y directions)

Physics 207: Lecture 11, Pg 25 Exercise 1 Momentum is a Vector (!) quantity A. Yes B. No C. Yes & No D. Too little information given l A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction l In regards to the block landing in the cart is momentum conserved?

Physics 207: Lecture 11, Pg 26 Exercise 1 Momentum is a Vector (!) quantity 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? l x-direction: No net force so P x is conserved. l y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) P y is not conserved (system is block and cart only) 5.0 m 30 ° 7.5 m 10 kg 2 kg

Physics 207: Lecture 11, Pg 27 Inelastic collision in 1-D: Example 2 l 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 speed V. In terms of m, M, and V : What is the momentum of the bullet with speed v ? v V beforeafter x

Physics 207: Lecture 11, Pg 28 Inelastic collision in 1-D: Example 2 What is the momentum of the bullet with speed v ?  Key question: Is x-momentum conserved ? v V beforeafter x BeforeAfter

Physics 207: Lecture 11, Pg 29 Example 2 Inelastic Collision in 1-D with numbers ice (no friction) Do not try this at home! 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?

Physics 207: Lecture 11, Pg 30 Exercise 2 Momentum Conservation A. Box 1 B. Box 2 C. same l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.  Which box ends up moving fastest ? 1 2