A) 1.4 J
B) 3.5 J
C) 5.5 J
D) 1.1 J
Correct Answer: A
Solution :
Momentum (p) is defined as product of mass (m) and velocity (u) i.e., \[\therefore \] Given \[=\frac{H}{H}=\frac{2R}{R/2}=\frac{4}{1}\] \[=k{{a}^{2}}\] From Newtons second law \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{9}{1}\] where a is acceleration. \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{1}\] From equation of motion, we have \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[\phi \] Change in kinetic energy is \[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}=k{{({{a}_{1}}+{{a}_{2}})}^{2}}\] \[\Rightarrow \] \[{{I}_{\max }}=k{{({{a}_{2}}+3{{a}_{2}})}^{2}}=16a_{2}^{2}k\]
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