In this problem, the messes of the two bodies are equal, so let the mass of each body be m. We will now write down the expressions for the kinetic energies of both the bodies separately.
(i) Mass of first body = m
Velocity of first body = v
So, K.E. of first body `= (1)/(2) mv^(2) " "` ...(1)
(ii) Mass of second body = m
Velocity of second body = 3v
So, K.E. of second body `= (1)/(2)m(3v)^(2)`
`= (1)/(2)m xx 9V^(2)`
`= (9)/(2)mv^(2)" "` ...(2)
Now, to find out the ration of kinetic energies of the two bodies, we should divide equation (1) by equation (2), so that :
`(K.E. "of first body")/(K.E. " of second body") = ((1)/(2)mv^(2))/((9)/(2)mv^(2))`
or `(K.E. "of first body")/(K.E. " of second body") = (1)/(9)" "` ...(3)
Thus, the ratio of the kinetic energies is 1 : 9.
We can also write down the equation (3) as follows :
K.E. of second body = `9 xx` K.E. of first body
That is, the kinetic energy of second body is 9 times the kinetic energy of the first body. It is clear from this example that when the velocity (or speed) of a body is "tripled" (from v to 3v), then its kinetic energy becomes "nine times".
