Question

Angular momentum and choice of origin: Adding angular momentum to a collision problem often feels confusing or ad hoc. What we're going to do here is verify that adding angular momentum to a collision problem isn't incompatible with anything we've done previously, by considering a very simple collision problem: One particle, moving with a velocity vo, collides with another, sticks with it, and then they move together (a) Let's start with the angular momentum of the first particle: It has mass m1 and is moving in the +x direction with velocity tox. At time t = 0 it is at x = 0, y = yo-write down the angular momentum L-r× p in terms of m1, 40, and vo-write it as a vector (because that's what it is). (b) Now the first particle undergoes a head-on collision with a second particle of mass m2 and sticks to it. Initially m2 is stationary; after the collision the particles start moving together. Which component(s) of the first particle's velocity will change as a result of this collision? (c) Compute the velocity j of the pair of particles, and keep in mind that momentum is conserved in collisions if no external force is acting. (d) Compute the angular momentum of the system after the collision. (e) If we had started the particle at yo = 0 would our calculations still give us a result consistent with angular momentum conservation? Why or why not?

Please solve only (e), and only if you're certain you're correct. Thank you.

Answer 1

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If we consider that y_0 = 0 and velocity of both particles always in x-direction and now we calculate angular momentum than L = r times P where r is perpendicular distance b/w velocity and x axis but when y_0 then r = 0 because particle moves on x axis angular momentum about x axis is zero so angular momentum before and after collision is zero so it is conserved