Question

A father racing his son has half the kinetic energy of the son, who has half the mass of the father.The father speeds up by 1.0 m/s and then has the same kinetic energy as the son.What are the original speeds of (a) the father and (b) the son?

Answer 1

Answer:

(a) $v_f=2.414\ m.s^{-1}$

(b) $v_s=4.828\ m.s^{-1}$

Explanation:

Let:

mass of father be, $m_f$

mass of son be, $m_s=\frac{m_f}{2}$

speed of son be, $v_s$

initial speed of father be, $v_f$

After speeding up, speed of father is  $v_f+1$

We know Kinetic Energy is given as

$KE=\frac{1}{2} m.v^2$ .....................................(1)

where:

m = mass

v = velocity

Hence, according to the initial condition the father is having kinetic energy half the kinetic energy of the son.

$KE_s=2.KE_f$

$\frac{1}{2} m_s.v_s^2=2\times \frac{1}{2} m_f.v_f^2$

$(\frac{m_f}{2})\times v_s^2=2\times m_f\times v_f^2$

$v_s=2v_f$ .................................................(2)

According to the final condition:

$\frac{1}{2} m_s.v_s^2= \frac{1}{2} m_f.(v_f+1)^2$

$(\frac{m_f}{2})\times v_s^2=m_f.(v_f+1)^2$

$v_s^2=2(v_f+1)^2$

$v_s=\sqrt{2}(v_f+1)$.....................................................(3)

(a)

From eq. 2 & 3

$2v_f=\sqrt{2}(v_f+1)$

$\sqrt{2}\ v_f=(v_f+1)$

$v_f(\sqrt{2}-1)=1$

$v_f=\frac{1}{(\sqrt{2}-1)}$

$v_f=2.414\ m.s^{-1}$

(b)

putting the above value in eq. (2)

$v_s=4.828\ m.s^{-1}$