An impulse `I` is applied at the end of a uniform rod if mass `m`. Then:

Options

A. `KE` of translation of the rod is `(I^(2))/(2m)`
B. `KE` of rotation of the rod is `(I^(2))/(6m)`
C. `KE` of rotation of the rod is `(3I^(2))/(2m)`
D. `KE` of the rod is `(2I^(2))/m`

Answer

A, C, D

Explanation

`KE_(T)=1/2mv^(2).....(1) & I=mv....(2)`
so `KE_(T)=(I^(2))/(2m),` again `KE_(R)=1/2Iomega^(2)...(3)`
we have `Ixxl/2=1/12 ml^(2)xxomegarArromega=(6I)/(ml)`
Thus `KE_(R)=1/2xx1/12 ml^(2)xx(36I^(2))/(m^(2)l^(2))=(3I^(2))/(2m)`
and `KE=KE_(T)+KE_(R)=(2I^(2))/m`