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Home / Doubts and Solutions/JEE/Physics

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
Answered by Expert
Answer:
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 = begin mathsize 12px style equals space fraction numerator m subscript 1 cross times v subscript 1 space plus space m subscript 2 cross times v subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction space equals space fraction numerator 2 cross times left parenthesis negative 1.6 i right parenthesis space plus 3 cross times 2.4 i over denominator 2 plus 3 end fraction space equals space 4 over 5 i space end style
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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