Answer:
The object with a higher mass has three times the kinetic energy of the less massive object.
We know that K.E.= 1/2 mv²
mass of first object = 1kg
speed = v
So K.E. = 1/2mv²
= 1/2 × (1) × v²
=
[tex] \: \: \: \: \: \: \: \: \: = \frac{1}{2} {v}^{2} \\ [/tex]
Now the mass of second body = 3kg
speed = v
So K.E. = 1(3)v²/2
=
[tex] \: \: \: \: \: \: \: \: \: = \: \: \frac{3}{2} {v}^{2} \\ [/tex]
now
K.E of first / KE of 2nd
[tex] \frac{1}{2} {v}^{2} \div \frac{3}{2} {v}^{2} \\ [/tex]
[tex]\frac{ \frac{ke \: of \: first \: }{ - - - - - - - - - - -} }{ \frac{ke \: of \: 2nd}{} } = \frac{ \frac{1}{ - -} }{ 3 } \\ \\ so \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ 3(ke \: of \: first \:) = ke \: of \: 2nd[/tex]
