Wednesday, September 27, 2017

homework and exercises - Use the relative velocity formula to find v2f in terms of v1f?




Q: A 0.150 kg glider is moving to the right (+x) on a frictionless, horizontal air track with a speed of 0.80 m/s. It has an elastic collision with a 0.300 kg glider moving to the left (x) with a speed of 2.20 m/s.


a.) What is the initial momentum of each glider? Express the momentum in terms of unit vectors.


b.) Use the relative velocity formula to find v2f in terms of v1f.


c.) Use the relative velocity result to solve conservation of momentum to find the velocity (magnitude and direction) of each glider after the collision.



I've figured part (a) using the definition of momentum: p=mv:


1st glider: p1=m1v1=(0.15 kg)(+0.8ˆx m/s)=+0.12ˆx kg m/s


2nd glider: p2=m2v2=(0.3 kg)(2.20ˆx m/s)=0.66ˆx kg m/s


Parts (b) and (c) are what have me confused at the moment. I'm not positive I have the equations for relative velocity right nor how to solve for v2f in terms of v1f. My book lists this an equation that can be gotten from manipulation of a kinetic energy equation: v1iv2i=(v1fv2f). Is this the relative velocity formula? would just isolating v2f in this equation be solving for v2f in terms of v1f?




Answer



I have no clue whatsoever as to what the "relative velocity formula" is, but I think that the idea here is to find a relation between the vfi (assuming f means final).


That comes straight out of momentum conservation (using s for starting and f for final and that the problem is in one dimension): pf=psm1vf1+m2vf2=m1vs1+m2vs2

which has only two unknowns, namely the vfi.


(Kinetic) energy conservation would lead to: Ef=Esm1(vf1)2+m2(vf2)2=m1(vs1)2+m2(vs2)2

which again has the two same unknowns.


So, two equations with two unknowns...


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