- #1
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
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