Question

A particle moves at 5 i m/s. If its mass is 2g, then its magnitude of momentum is​

Answer 1

Answer:

Explanation:

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 )

Answer 2

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 )

Explanation: