PHY 107 1st Edition Lecture 20 Outline of Last Lecture I. Linear MomentumII. Collision and Impulse Outline of Current LectureIII. Collision and ImpulseCurrent LectureCollision and Impulse:- A massive (stationary) target: m2 >> m1 → m1m2 << 1 v1f ~ -v1i v2f ~ 2(m1m2)v1i Body 1 (small mass): bounces back along the incoming path with its speed practically unchanged Body 2 (large mass): moves forward with very small speed- A massive projectile: m1 >> m2 → m2m1 << 1 v1f ~ v1i v2f ~ 2v2iThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Body 1 (large mass): keeps on going, scarcely slowed by the collision Body 2 (small mass): charges ahead at twice the speed of body 1 - Collision in 2 dimensions: Consider two masses in the xy-plane; masses m1 and m2 The linear momentum of the system is conserved´p1 i + ´p2 i = ´p1 f + ´p2 f If the collision is elastic, then kinetic energy is conserved K1i + K2i = K1f + K2f Assume m2 is stationary and after the collision m1 and m2 move at angles of θ1 and θ2 with the initial direction of motion of m1 Conservation of momentum and kinetic energy: o x-axis: m1v1i = m1v1fcosθ1 + m2v2fcosθ2o y-axis: 0 = m1v1fsinθ1 + m2v2fsinθ2½ m1v1i2 = ½ m1v2f2 + ½ m2v2f2 - The rocket: A rocket of mass M and speed v ejects mass backwards at a constant rate – the ejected material is expelled at constant speed vrel relative to the rocket. The rocket loses mass and accelerates forward. Use linear momentum to determine speed, v, of the rocket as a function of time:p(t) = p(t + dt) → Mv = (dM)(v – vrel) + (M – dM)(v + dv) v – vrel is the velocity of the ejected gasses with respect to observer O who measures the rocket’s speed v Gas velocity in terms of vrel (with respect to the rocket) Mdv = vreldM → Ma = vrelR (first rocket equation) For the rocket, a is not constant but increases with time because M decreasesVf – vi = vrel
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