Next hour exam is Friday Nov 9 (next Fri) Homework is due Wed Nov 7, and is posted on Tycho. Comet

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Nancy Horton
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1 Next hour exam is Friday Nov 9 (next Fri) Homework is due Wed Nov 7, and is posted on Tycho. Comet 1

2 Billiard Balls (ignore rotation for now) elastic collision (means K is conserved): y v 1 M v 0 M v 2 θ 1 θ 2 x θ 1 can be a range of angles. What about θ 2? What about v 1 and v 2? Conservation of momentum: 2 equations Conservation of energy: 1 equation. 3 equations three unknowns, v 1, v 2 and θ 2 Challenging algebra. (special trick when both have same M.) Mv 0 =M v 1 + M v 2 v 1 v 2 θ 1 θ 2 v 2 0 = v v 2 2 for K conservation. (divide out M/2) apply Pythagoras see top angle is a right angle v 1 v 2 90 o v 0 Conserve P (divide out M) don t know top angle (yet) v 0 Congruent triangles. So all 3 angles known, and θ 1 +θ 2 = 90 ο 2

3 y v 1 M v 0 M v 2 θ 1 θ 2 x Elastic, so θ 1 +θ 2 = 90 ο and then Sin(θ 2 ) = Cos(θ 1 ) and conservation of P y gives 0 = Mv 1 sin(θ 1 ) Mv 2 cos(θ 1 ) so v 2 /v 1 = tan(θ 1 ) cons of P x gives Mv 0 = Mv 1 cos(θ 1 ) + Mv 2 sin(θ 1 ) subst for v 2 v 0 = v 1 cos(θ 1 ) + v 1 tan(θ 1 ) sin(θ 1 ) = v 1 / cos(θ 1 ) Given θ 1 (must be < 90 ο ) get other 3 variables. This depends on both masses being equal and collision must be elastic. 3

4 Section 8.4 collisions in CM Moving with CM, see P syst = Mv cm = 0 and P1 = -- P2 Example for equal masses, where v = cm v0 2 and u red = u = green v0 2 : y M v 0 /2 M v 0 /2 M v 0 /2 M v 0 /2 x Opposite momenta before and after collision. P syst = 0 before and remains so after collision. Same speeds, to conserve K. See textbook for examples with different masses Clicker Demo 4

5 A particle of mass 2m is moving to the right in projectile motion. At the top of its trajectory, an explosion breaks the particle into two equal parts. After the explosion, one part falls straight down with no horizontal motion. What is the direction of the motion of the other part just after the explosion? A. up and to the left B. stops moving C. up and to the right D. straight up E. down and to the right

6 Section 8.5 Rockets. Since rockets squirt exhaust out and decrease in mass, to keep signs straight we start with a model of a thing getting stuff shot into it and sticking. A rocket reverses this process. Have to consider change in mass and impulse: Interval Δt - thing s mass increases by ΔM System: thing (mass M) plus stuff stuck, (mass ΔM) initially, thing has v, stuff to be stuck has u finally, thing + stuff has v + Δv F Note that net, ext is EXTERNAL, not the result of accumulating ΔM. Could be gravity, for example. 5

7 Impulse completely a result of ΔM hitting and sticking to M. Involves Δu and Δv Fnet,ext Δ t = Δ P = ( M +Δ M)( v +Δv) ( Mv +ΔMu) Note ΔP is not caused directly just by ΔM and ΔP is the change in system momentum, not just M s. do some algebra (see text) and take limit Δt 0 dm dv Fnet,ext + vrel = M and rel v = u v positive v rel has stuff hitting M. For a Rocket, both dm are negative. F = dm u Thrust is th exh and v rel avoids confusion about signs. where absolute value 6

8 Alternate approach. Be in CM of rocket and bit of fuel to be burned. Rocket expels ΔM at exhaust speed u exh.with respect to rocket. Leave external forces out until later. In limit of small ΔM, speed of exhaust in CM is u exh and to conserve momentum we get ΔP rocket = ΔM u exh = F th Δt using impulse in the direction opposite to that of the exhaust F dm F Then we get th exh = u in the limit Δt0. + F where F th Net force on rocket is net,ext th points opposite the direction the exhaust is shot. To get motion of rocket going straight up dm dv F y net = Mg + Fth = Mg + uexh = M dv y 1 dm = g + uexh and M dm Mt () = M0 + t= M0 Rt(R is burn rate) acceleration is put M(t) into the equation for acceleration dv y Ru exh = M 0 Rt g 7

9 If the rocket starts at rest at t=0, we can integrate this to get which is also some features: The bigger u exh the better M 0 vy = uexh ln gt M0 Rt 1 vy = uexh ln gt 1 Rt / M0 The ratio R/M 0 is what counts Since the final mass (payload) is M 0 Rt final =M p and the fuel mass is Rt final the velocity at burn out is v = u M ln + M gt fuel p y,max exh burn-out M p so it is good to maximize M fuel / M p because ln(x) increases slowly with x, it takes a lot of fuel to work a rocket. Demo 8

Chapter 8 Conservation of Linear Momentum. Conservation of Linear Momentum

Chapter 8 Conservation of Linear Momentum Physics 201 October 22, 2009 Conservation of Linear Momentum Definition of linear momentum, p p = m v Linear momentum is a vector. Units of linear momentum are