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Work, Energy And Power

Question
CBSEENPH11017363

A body of mass 4 kg moving with velocity 10 m/sec collides inelastically with another body of mass 10 kg moving with a velocity of 4m/sec. If coefficient of restitution is 0 5, then find the velocities of two after collision and loss of energy.

Solution
After collision,
Let v1 be the velocity of mass 4 kg. 
Let v2 be the velocity of mass 10 kg. 
According to the law of conservation of momentum, 
            straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals straight m subscript 1 straight v subscript 1 plus straight m subscript 2 straight v subscript 2 
Here we have,

Mass, straight m subscript 1 equals 4 kg space 
Mass,straight m subscript 2 equals 10 space kg
Initial velocity, straight u subscript 1 equals 10 space straight m divided by sec comma
Velocity after collision, straight u subscript 2 equals 4 space straight m divided by sec 
Therefore, 
           space space space 4 cross times 10 plus 10 cross times 4 equals 4 straight v subscript 1 plus 10 straight v subscript 2 
rightwards double arrow               4 straight v subscript 1 plus 10 straight v subscript 2 equals 80               ...(1) 
Coefficient of restitution is, straight e equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction 
           straight v subscript 2 minus straight v subscript 1 equals straight e left parenthesis straight u subscript 1 minus straight u subscript 2 right parenthesis 
                    equals 0.5 left parenthesis 10 minus 4 right parenthesis  
rightwards double arrow       straight v subscript 2 minus straight v subscript 1 equals 3                                 ...(2) 
Solving (1) and (2), we get 
Velocities of bodies after collision is, 
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#6 {main}</pre>   and    space space space straight v subscript 2 equals 46 over 7 straight m divided by straight s 
Total kinetic energy before collision is, 
       space space space straight K subscript straight i equals 1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 
            equals 1 half cross times 4 cross times left parenthesis 10 right parenthesis squared plus 1 half cross times 10 cross times left parenthesis 4 right parenthesis squared 
            equals 200 plus 80 equals 280 space straight J 
Total kinetic energy after collision is, 
         straight K subscript straight f equals 1 half straight m subscript 1 straight v subscript 1 superscript 2 plus 1 half straight m subscript 2 straight v subscript 2 superscript 2 
            equals 1 half cross times 4 cross times left parenthesis 25 over 7 right parenthesis squared plus 1 half cross times 10 cross times left parenthesis 46 over 7 right parenthesis squared 
             equals 241.43 space straight J 
∴   Loss of energy  = 280 - 241.43 = 38.57 J