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Physics 207, Lecture 12, Oct. 15 Agenda: Finish Chapter 9, start Chapter 10 • Chapter 9: Momentum & Collisions Momentum conservation in 2D Impulse Assignment: HW5 due Wednesday HW6 posted soon Physics 207: Lecture 12, Pg 1 Impulse & Linear Momentum Transition from forces to conservation laws Newton’s Laws Conservation Laws Conservation Laws Newton’s Laws They are different faces of the same physics phenomenon for special “cases” Physics 207: Lecture 12, Pg 2 Lecture 12, Exercise 1 Momentum Conservation Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks. In which case does the box ends up moving fastest ? No external force then, vectorially, COM A. B. C. Box 1 Box 2 Same 1 2 Physics 207: Lecture 12, Pg 3 Lecture 12, Exercise 1 Momentum Conservation Which box ends up moving fastest ? Examine the change in the momentum of the ball. In the case of box 1 the balls momentum changes sign and so its net change is largest. Since momentum is conserved the box must have the largest velocity to compensate. (A) Box 1 1 (B) Box 2 (C) same 2 Physics 207: Lecture 12, Pg 4 A perfectly inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). V v1 m1 + m2 m1 m2 v2 before after If no external force momentum is conserved. Momentum is a vector so px, py and pz Physics 207: Lecture 12, Pg 5 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 Physics 207: Lecture 12, Pg 6 Elastic Collision in 1-D m2 m1 before v1b v2b x m1 m2 after v1a v2a Physics 207: Lecture 12, Pg 7 Force and Impulse (A variable force applied for a given time) Gravity: usually a constant force to an object 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. Physics 207: Lecture 12, Pg 8 Force and Impulse (A variable force applied for a given time) J reflects momentum transfer t t p J F dt (dp / dt )dt dp F Impulse J = area under this curve ! (Transfer of momentum !) t t Impulse has units of Newton-seconds ti tf Physics 207: Lecture 12, Pg 9 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 same area F t t big, F small t t t t small, F big Physics 207: Lecture 12, Pg 10 Average Force and Impulse F Fav F Fav t t big, Fav small t t t t small, Fav big Physics 207: Lecture 12, Pg 11 Example from last time A 2 kg cart initially at rest on frictionless horizontal surface is acted on by a 10 N horizontal force along the positive x-axis for 2 seconds what is the final velocity? F is in the x-direction F = ma so a = F/m = 5 m/s2 v = v0 + a t = 0 m/s + 2 x 5 m/s = 10 m/s (+x-direction) but mv = F t [which is the area with respect to F(t) curve] Physics 207: Lecture 12, Pg 12 Lecture 12, 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 A. B. C. D. light F heavy heavier lighter same can’t tell Physics 207: Lecture 12, Pg 13 Lecture 12, 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 ? (A) heavier F (B) light lighter F (C) same heavy Physics 207: Lecture 12, Pg 14 Boxers: Physics 207: Lecture 12, Pg 15 Back of the envelope calculation t J F dt Favg t (2) varm~7 m/s (3) Impact time t ~ 0.01 s (1) marm~ 7 kg Impulse J = p ~ marm varm ~ 49 kg m/s F ~ J/t ~ 4900 N (1) mhead ~ 6 kg ahead = F / mhead ~ 800 m/s2 ~ 80 g ! Enough to cause unconsciousness ~ 40% of fatal blow Physics 207: Lecture 12, Pg 16 Woodpeckers During "collision" with a tree ahead ~ 600 - 1500 g How do they survive? • Jaw muscles act as shock absorbers • Straight head trajectory reduces damaging rotations (rotational motion is very problematic) Physics 207: Lecture 12, Pg 17 Energy What do we mean by an isolated system ? What do we mean by a conservative force ? If a force acting on an object act for a period of time then we have an Impulse change (transfer) of momentum What if we consider this force acting over a distance: Can we identify another useful quantity? Physics 207: Lecture 12, Pg 18 Energy Fy = m ay and let the force be constant y(t) = y0 + vy0 t + ½ ay t2 y = y(t)-y0= vy0 t + ½ ay t2 vy (t) = vy0 + ay t t = (vy - vy0) / ay = vy / ay So y = vy0 vy / ay + ½ ay (vy/ay)2 = (vy vy0 - vy02 ) / ay + ½ (vy2 - 2vy vy0+vy02 ) / ay 2 ay y = (vy2 - vy02 ) Finally: If falling: may y = ½ m (vy2 - vy02 ) -mg y = ½ m (vy2 - vy02 ) Physics 207: Lecture 12, Pg 19 Energy -mg y= ½ m (vy2 - vy02 ) -mg (yf – yi) = ½ m ( vyf2 -vyi2 ) A relationship between y displacement and y speed Rearranging ½ m vyi2 + mgyi = ½ m vyf2 + mgyf We associate mgy with the “gravitational potential energy” Physics 207: Lecture 12, Pg 20 Energy Notice that if we only consider gravity as the external force then then the x and z velocities remain constant To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyf where vi2 = vxi2 +vyi2 + vzi2 ½ m v2 terms are referred to as kinetic energy Physics 207: Lecture 12, Pg 21 Energy If only “conservative” forces are present, the total energy (sum of potential, U, and kinetic energies, K) of a system is conserved. Emech = K + U Emech = K + U = constant K and U may change, but E = K + Umech remains a fixed value. Emech is called “mechanical energy” Physics 207: Lecture 12, Pg 22 Another example of a conservative system: The simple pendulum. Suppose we release a mass m from rest a distance h1 above its lowest possible point. What is the maximum speed of the mass and where does this happen ? To what height h2 does it rise on the other side ? m h1 h2 v Physics 207: Lecture 12, Pg 23 Example: The simple pendulum. What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum. y y=h1 y= 0 Physics 207: Lecture 12, Pg 24 Example: The simple pendulum. What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh1 at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 y=0 h1 v Physics 207: Lecture 12, Pg 25 Example: The simple pendulum. To what height h2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point. E = mgh1 = mgh2 or h1 = h2 y y=h1=h2 y=0 Physics 207: Lecture 12, Pg 26 Lecture 12, Exercise 3 Conservation of Mechanical Energy A block is shot up a frictionless 40° slope with initial velocity v. It reaches a height h before sliding back down. The same block is shot with the same velocity up a frictionless 20° slope. On this slope, the block reaches height A. B. C. D. E. 2h h h/2 Greater than h, but we can’t predict an exact value. Less than h, but we can’t predict an exact value. Physics 207: Lecture 12, Pg 27 Lecture 12, Example The Loop-the-Loop … again Ub=mgh U=mg2R To complete the loop the loop, how high do we have to let the release the car? Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) Use fact that E = U + K = constant ! Car has mass m h? R (A) 2R (B) 3R (C) 5/2 R Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is v gR (D) 23/2 R Physics 207: Lecture 12, Pg 28 Lecture 12, Example The Loop-the-Loop … again Use E = K + U = constant mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R (C) v gR h? R (A) 2R (B) 3R (C) 5/2 R (D) 23/2 R Physics 207: Lecture 12, Pg 29 Inelastic collision in 1-D: Example 1 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. What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved? x v V before after Physics 207: Lecture 12, Pg 30 Inelastic collision in 1-D: Example 1 mv What is the momentum of the bullet with speed v ? 1 1 2 mv v mv 2 2 1 What is the final energy of the system ? (m M )V 2 2 aaaa Is momentum conserved (yes)? mv M 0 (m M )V What is the initial energy of the system ? Examine Ebefore-Eafter 1 1 1 1 m 1 m mv 2 [( m M )V]V mv 2 (mv) v mv 2 1 2 2 2 2 mM 2 mM ( v No! before ) V after x Is energy conserved? Physics 207: Lecture 12, Pg 31 Example – Fully Elastic Collision Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v1. If they collide elastically, what is the final velocity of each car ? Identical means m1 = m2 = m Initially vGreen = v1 and vRed = 0 COM mv1 + 0 = mv1f + mv2f v1 = v1f + v2f COE ½ mv12 = ½ mv1f2 + ½ mv2f2 v12 = v1f2 + v2f2 v12 = (v1f + v2f)2 = v1f2 +2v1fv2f + v2f2 2 v1f v2f = 0 Soln 1: v1f = 0 and v2f = v1 Soln 2: v2f = 0 and v1f = v1 Physics 207: Lecture 12, Pg 32 Lecture 12, Exercise for home Elastic Collisions I have a line of 3 bumper cars all touching. A fourth car smashes into the others from behind. Is it possible to satisfy both conservation of energy and momentum if two cars are moving after the collision? All masses are identical, elastic collision. (A) Yes (B) No (C) Only in one special case v Before v1 v2 After? Physics 207: Lecture 12, Pg 33 Physics 207, Lecture 12, Oct. 15 Agenda: Finish Chapter 9, start Chapter 10 • Chapter 9: Momentum & Collisions Momentum conservation in 2D Impulse Assignment: HW5 due Wednesday HW6 posted soon Finish Chapter 10, Start 11 (Work) Physics 207: Lecture 12, Pg 34 Lecture 12, Exercise for home Elastic Collisions COM mv = mv1 + mv2 so v = v1 + v2 COE ½ mv2 = ½ mv12 + ½ mv22 v2 = (v1 + v2)2 = v12 + v22 v1 v2 = 0 (A) Yes (B) No (C) Only in one special case Before After? Physics 207: Lecture 12, Pg 35 Example of the U - F relationship For a Hooke’s Law spring, U(x) = (1/2)kx2 Notice that the derivative gives Hooke’s Law Fx = - d ( (1/2)kx2) / dx Fx = - 2 (1/2) kx Fx = -kx Physics 207: Lecture 12, Pg 36 Equilibrium Example Spring: Fx = 0 => dU / dx = 0 for x=0 The spring is in equilibrium position In general: dU / dx = 0 for ANY function establishes equilibrium U stable equilibrium U unstable equilibrium Physics 207: Lecture 12, Pg 37 Hooke’s Law Springs: Another kind of potential energy Physics 207: Lecture 12, Pg 38 Elastic vs. Inelastic Collisions A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. Kbefore = Kafter Carts colliding with a perfect spring, billiard balls, etc. vi A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter Car crashes, collisions where objects stick together, etc. Physics 207: Lecture 12, Pg 39 Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Mechanical energy is lost: Heat (friction) Bending of metal and deformation Kinetic energy is not conserved since negative work is by a non-conservative force done during the collision ! Momentum along a certain direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy than energy conservation. Physics 207: Lecture 12, Pg 40 Example of 2-D Elastic collisions: Billiards If all we are given is the initial velocity of the cue ball, we don’t have enough information to solve for the exact paths after the collision. But we can learn some useful things... Physics 207: Lecture 12, Pg 41 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. Physics 207: Lecture 12, Pg 42 Billiards: All that really matters is conservation of energy and momentum COE: ½ m vb2 = ½ m va2 + ½ m Va2 x-dir COM: m vb = m va cos q + m Vb cos f y-dir COM: 0 = m va sin q + m Vb sin f after before pa q pb vcm F Active Figure The Pa f final directions are separated by 90° : q – f = 90° Physics 207: Lecture 12, Pg 43 Lecture 12 – Exercise 4 Pool Shark Can I sink the red ball without scratching (sinking the cue ball) ? (Ignore spin and friction) (A) Yes (B) No (C) More info needed Physics 207: Lecture 12, Pg 44 Lecture 12 – Exercise 4 Pool Shark Can I sink the red ball without scratching (sinking the cue ball) ? (Ignore spin and friction) (A) Yes (B) No (C) More info needed Physics 207: Lecture 12, Pg 45