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HarleyM
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Homework Statement
Two frictionless pucks are placed on a level surface as shown, at an initial distance of 20m.
Puck 1 has a mass of 0.8 kg and a charge of + 3x10^-4 while puck 2 has a mass of 0.4 kg and a charge of +3 x10^-4.
The initial velocity of puck 1 is 12 m/s [E] and the initial velocity of puck 2 is 8 m/s [W]. Find the minimum separation of the two pucks; the min distance between the two pucks
Homework Equations
Pt=Pt'
M1V1+M2V2=(M1+M2)V
Ek=1/2mv2
The Attempt at a Solution
Conservation of momentum is used to find the velocity of closest approach, since it must be constant for both.
Pt=Pt'
M1V1+M2V2=(M1+M2)V
(0.8)(12)+(0.4)(8)=(1.2)V (Should I set East or west as +, and therefore should 1 velocity be negative?)
V=10.7 m/s when pucks reach point of closest approach
Ek1+Ek2= Ee +Ek
1/2M1v12 +1/2M2v22=(kq1q2/r) + 1/2(M1+M2)V
1/2(0.8)(12)2+1/2(0.4)(8)2= [(9x10^9)*(3x10-4)2/r] + [1/2(1.2)*(10.7)]
(70.4-6.42)r=(9x109)(3x10-4)2
r= 12.66 m
SO my question is, is 12.7 m the closest they reach or would it be 20-12.7?
I think I put all the variables into the equations correctly and accounted for conservation of energy and momentum, any help would be great. I am quite confused.
Thanks everyone and merry Christmas, happy holidays and a happy winter solstice