VITEEE 2018 Paper

Kinetic energy is defined as the half of the product of mass and square times the velocity. $K.E. = \frac{1}{2} M v^{2}, K.E.$ where is the kinetic energy, $M$ is the mass and $v$ is the velocity of the object whose kinetic energy We are calculating. First Case: The kinetic energy of the man is half the kinetic energy of the boy with half the mass. The man has a mass $M$ and velocity $v_{1}$ whereas the boy has mass $\frac{M}{2}$ and velocity $v_{2}$ . $\frac{1}{2} M v_{1}^{2} = \frac{1}{2} \left( \frac{1}{2} \frac{M}{2} v_{2}^{2} \right)$ $v_{1}^{2} = \frac{1}{4} (v_{2}^{2})$ Second Case: The kinetic energy of the man is equal to the kinetic energy of the boy with half the mass. The man has a mass $M$ and velocity $v_{1}+1$ whereas the boy has mass $\frac{M}{2}$ and velocity $v_{2}$. $\frac{1}{2} M (v_{1}+1)^{2} = \frac{1}{2} \frac{M}{2} v_{2}^{2}$ $(v_{1}+1)^{2} = \frac{1}{2} v_{2}^{2}$ By taking the ratio both these relations we get $\frac{v_{1}^{2}}{(v_{1}+1)^{2}} = \left( \frac{v_{1}}{v_{1}+1} \right)^{2} = \frac{1}{2}$ Taking the reciprocal, $\left( \frac{v_{1}+1}{v_{1}} \right)^{2} = 2$ $\frac{v_{1}+1}{v_{1}} = \sqrt{2}$ $1 + \frac{1}{v_{1}} = \sqrt{2}$ $v_{1} = \frac{1}{\sqrt{2}-1} \,\text{m/s}$