A body of mass 2kg moving with velocity 3 m/s collides head on with 3kg moving with a velocity of 4m/s in opposite direction. After collision the two bodies stick together. Find the loss of energy in the process.

Solution

Given,
Mass of first body, $straight m subscript 1 space equals space 2 kg$
Initial velocity of the body, $straight u subscript 1 equals 3 straight m divided by straight s$
Mass of the second body, $space straight m subscript 2 equals 3 kg$

Initial velocity of the body, $straight u subscript 2 equals negative 4 straight m divided by straight s$

Let, v be the velocity of composite mass after collision.
Therefore by conservation of momentum,
$straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v$
$rightwards double arrow$ $2 cross times 3 plus 3 left parenthesis negative 4 right parenthesis space equals space left parenthesis 5 right parenthesis straight v$
$rightwards double arrow$ $straight v equals negative 1.2 space straight m divided by straight s$
Now kinetic energy of system before collision is,
$1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 equals 1 half cross times 2 left parenthesis 3 right parenthesis squared plus 1 half cross times 3 left parenthesis negative 4 right parenthesis squared$
$equals 9 plus 24 equals 33 straight J$
Kinetic energy of the sytem after collision is,
$1 half left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v squared equals 1 half cross times 5 cross times left parenthesis 1.2 right parenthesis squared equals 3.6 straight J$