A spring is held compressed so that its stored energy is `2.4 J`. Its ends are in contact with masses `1 g and 48 g` placed on a frictionless table. When the spring in released, the heavier mass will acquire a speed of:

Options

A. `(2.4)/(49) ms^(-1)`
B. `(2.4 xx 48)/(49) ms^(-1)`
C. `(10^(3))/(7)cms^(-1)`
D. `(10^(6))/(7)cms^(-1)`

Answer

C

Explanation

`(1)/(2)m_(1)v_(1)^(2) + (1)/(2) m_(2) v_(2)^(2) = 2.4`
or `m_(1)v_(1)^(2) + m_(2) v_(2)^(2) = 4.8`…(i)
Now `m_(1)v_(1)= m_(2)v_(2)` or `v_(1) = 48v_(2)`
Using (i) `(1)/(1000) (48v_(2))^(2) + (48)/(1000) v_(2)^(2) = 4.8`
or `v_(2) = (10)/(7) m//s = (10^(2))/(7) cm//sec`.