Calculating Velocities of Charged Balls Released from 1m Distance

[devanlevin]

2 balls are held at a distance of 1m from each other and then released, what will each of their velocities be when they are 2m from one another
m1=0.05 (kg)
q1=6*10^-6 (c)
m2=0.1 (kg)
q2=5*10^-6 (c)

what i tried was looking at each ball seperately
ball1

Energy===> Ue(initial)=Ue(final)+Ek
K(q2q1)/r(initial)=K(Q2q1)/r(final)+(1/2)mv^2
K(q2q1)/1=K(Q2q1)/2)+(1/2)mv^2
v^2=(Kq2q1)/m1=5.4
v1=2.3237m/s

V^2=(Kq2q1)/m2=2.7
v2=1.643m/s

but the answers in my textbook are
v1=1.9m/s
v2=-0.95m/s

where have i gone wrong here

 

[jackiefrost]

2 balls are held at a distance of 1m from each other and then released, what will each of their velocities be when they are 2m from one another
m1=0.05 (kg)
q1=6*10^-6 (c)
m2=0.1 (kg)
q2=5*10^-6 (c)

what i tried was looking at each ball seperately
ball1

Energy===> Ue(initial)=Ue(final)+Ek
K(q2q1)/r(initial)=K(Q2q1)/r(final)+(1/2)mv^2
K(q2q1)/1=K(Q2q1)/2)+(1/2)mv^2
v^2=(Kq2q1)/m1=5.4
v1=2.3237m/s

V^2=(Kq2q1)/m2=2.7
v2=1.643m/s

but the answers in my textbook are
v1=1.9m/s
v2=-0.95m/s

where have i gone wrong here

The (1/2)mv^2 (highlighted red, above) is the final total kinetic energy. So, it is actually (1/2*m1*v1^2)+(1/2*m2*v2^2). You then have to ask yourself how that total is split between the two masses to arrive at the individual velocities.

jf

 

[devanlevin]

so then, i say
(1/2)mv^2=(1/2*m1*v1^2)+(1/2*m2*v2^2)=27/200

v^2=9/5
v=1.34m/s

then from there i suppose i should split the velocity between the 2 as the ratio of their mass, but i still don't come to the correct answer, i see in the answer that the ratio of the velocity is the same as the ratio of the mass which is logical, but how do i get those numbers

 

[jackiefrost]

The total final kinetic energy KE is (1/2)m1v1^2 + (1/2)m2v2^2 and equals the work done by the electric field w. That work is equal to the difference between the initial and final potential energy.

KE = -w = (kq1q2/r2 - kq1q2/r1) [r2=final dist 2m, r1=initial dist 1m]

The total kinetic energy is divided according to the proportion of the individual masses to the total mass m1+m2. Note however that the ke2 of the more massive m2 is affected proportionally less than that for m1. This makes sense since the electric force acts to accelerate the less massive m1 to a proportionally higher velocity than m2 and it's those new velocities that account for the new individual kinetic energies since the masses haven't changed. That's why the following equations don't look correct (the kinetic energy of m1 using the m2/m1+m2 proportion instead of m1/m1+m2, etc...)

ke1 = -w(m2/m1+m2) [ke1 is kinetic energy of m1, etc]
ke2 = -w(m1/m1+m2)

then you can solve for v1 using ke1 and m1, etc...jf