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# A particle of mass 2kg moving with a velocity 5i m/s collides head on with another particle of mass 3kg moving with a velocity -2i m/s .After the collision the first particle has speed of 1.6m/s in negative x direction.Find velocity of the centre of mass after the collision, velocity of the second particle after the collision ,coefficient of restitution.

Asked by m.nilu 28th September 2018, 8:44 PM
By conservation of momentum,    m1×u1 + m2 × u2 = m1 × v1 + m2 × v2 .............(1)

where u1 and u2 are initial velocities. Similarly v1 and v2 are final velocities

we are given :- m1 = 2kg,  m2 = 3kg ,  u1 = 5i m/s , u2 = -2i m/s ,  v1 = -1.6 i

substituting this data in eqn.(1), we get,   2×5i + 3×(-2i) = 2×(-1.6i) + 3×v2  .............(2)

solving eqn.(2), we get V2 = 2.4 i

velocity of centre of mass after collision =
coefficient of restitution e is given by,   v1-v2 = -e × (u1-u2) ............(3)

hence from eqn.(3),  e = ( 1.6  + 2.4  ) / ( 5 + 1.6) = 0.61
(all velocities are in x-direction, hence only magnitudes are considered to get coefficient of restitution )
Answered by Expert 29th September 2018, 2:38 PM
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