Question
Two balls of equal mass undergo head on collision while each was moving with speed 6 m/s. If the coefficient of restitution is 1/3 , the speed of each ball after impact will be
Joshi sir comment
according to the given condition u1 = 6, u2 = -6
let after collision velocities are v1, v2
then by momentum conservation m1u1 + m2u2 = m1v1 + m2v2
and v2 - v1 = e(u2 - u1)
now solve
Read 2 Solution.
my answer comes to be 4 m/s but the correct one is 2 m/s.
SWATI KAPOOR 10 year ago
is this solution helpfull: 9 15
V2-V1/U2-U1 = 1/3
V2 - (-V2) = 1/3 *12
2V2 = 4
V2 = 2m/s
I have taken 2V2 bcoz since U1 = U2 so V1=V2 as well acc. to conservation of momentum
QWERTY 7 year ago
is this solution helpfull: 29 8