## Fundamentals of Physics Extended (10th Edition)

Velocity of son $=4.8$ $m/s$
Let the mass and the velocity of the father be $m$ and $v_{f}$ respectively, and the mass and the velocity of the son be $\frac{m}{2}$ and $v_{s}$ respectively. The expression of kinetic energy is: $Kinetic$ $Energy= \frac{1}{2}mv^{2}$ Therefore "father's Kinetic energy is half of the son's Kinetic Energy implies : $\frac{1}{2}m(v_{f})^{2} = \frac{1}{2} \times [ \frac{1}{2}\frac{m}{2}(v_{s})^{2} ]$ $mv_{f}^{2} = \frac{m}{4}v_{s}^{2}$ $4mv_{f}^{2}=v_{s}^{2}$ $2v_{f} = v_{s}$ Or, $v_{f}=\frac {v_{s}} {2}$ . . . . . . . . . . . . (1) If father's velocity is increased by $1 m/s$, it becomes $v_{f}+1$, and their kinetic energies become equal. Therefore we can write: $\frac{1}{2}m (v_{f}+1)^{2} = \frac{1}{2}(\frac{m}{2})v_{s}^{2}$ Substituting the value of $v_{f}$ as $\frac{v_{s}}{2}$ (from equation 1 ) in the above equation and solving gives: $\frac{1}{2}m (\frac{v_{s}}{2}+1)^{2} = \frac{1}{2}(\frac{m}{2}) v_{s}^{2}$ Solving this quadratic equation gives: $v_{s}= \frac {4+ \sqrt {32}}{2} = 2+2\sqrt 2 = 2+2.8$ $v_{s}= 4.8$ $m/s$ So, velocity of son is $4.8$ $m/s$.