Question

the energy of a body is increased by 21% what is the percentage increase in the linear momentum of the body

Answer 1

Fractional increase in K.E. =[ K.E. (final) - K.E.(initial)] / K.E. (initial) = (V^2 - v^2)/ v^2 = (V^2/v^2) - 1 = 21/100
i.e.
V^2/v^2 = 121/100 i.e. V/v = 11/10 ----- (i) , where V is final velocity andv is initial velocity
Now Fractional increase in linear momentum = [p(final) - p (initial)]/ p (initial) = (mV - mv)/mv = V/v -1 = 11/10 -1 = 1/10          (by (i))
Therefore, % increase in p = 1/10 *100 = 10%
Ans = 10%