question_answer

What happens to kinetic energy of a body when 1/4 mass is removed and the velocity is doubled?

Answer:

                               

Case-I	Case-II
\[{{m}_{1}}={{m}_{1}}\]	\[{{m}_{2}}={{m}_{1}}-\frac{1}{4}{{m}_{1}}=\frac{3}{4}{{m}_{1}}\]

(1/4 th mass is removed from the body) \[{{v}_{1}}={{v}_{1}}{{v}_{2}}=2{{v}_{1}}\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] Applying the formula for both cases, we get \[K.E{{.}_{1}}=\frac{1}{2}{{m}_{1}}v_{1}^{2}\]                                    ?????.(1) \[K.E{{.}_{2}}=\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[=\frac{1}{2}\left( \frac{3}{4}{{m}_{1}} \right){{(2{{V}_{1}})}^{2}}\]                         ?????.(2) Dividing equation (1) by (2), we get, \[\Rightarrow \frac{K.E{{.}_{1}}}{K.E{{.}_{2}}}=\frac{\frac{1}{2}{{m}_{1}}v_{1}^{2}}{\frac{1}{2}\times \frac{3}{4}{{m}_{1}}\times 4{{v}_{1}}}=\frac{1}{3}\] \[\Rightarrow K.E{{.}_{2}}=3K.E{{.}_{1}}\] \[\therefore \]K.E. is tripled.