Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 5a


Velocity of the father $=2.4$ $m/s$

Work Step by Step

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 : $\frac{1}{2}m(v_{f})^{2} = \frac{1}{2} \times [ \frac{1}{2}\frac{m}{2}(v_{s})^{2} ]$ $mv_{f}^{2} = \frac{m}{4}v_{s}^{2}$ $4mv_{f}^{2}=v_{s}^{2}$ $2v_{f} = v_{s}$ . . . . . . . . . . . . . .(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 $v_{s}$ as $2v_{f}$ (from equation 1 ) in the above equation and solving gives: $\frac{1}{2}m (v_{f}+1)^{2} = \frac{1}{2}(\frac{m}{2}) 2v_{f}^{2}$ Solving this quadratic equation gives: $v_{f}= \frac {2+ \sqrt 8}{2} = 1+\sqrt 2 = 1+1.4$ $v_{f}= 2.4$ $m/s$ So, velocity of father is $2.4$ $m/s$.
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