Question

A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 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

Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$. The kinetic energies of father and son are $K_f\ and\ K_s$. The mass of father and son are  $m_f\ and\ m_s$

(a) According to given conditions, $K_f=\dfrac{1}{3}K_s$

And $m_s=\dfrac{1}{4}m_f$

Kinetic energy of father is given by :

$K_f=\dfrac{1}{2}m_fv_f^2$.............(1)

Kinetic energy of son is given by :

$K_s=\dfrac{1}{2}m_sv_s^2$...........(2)

From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

$v_s^2=4(v_f+1.5)^2$

Using equation (3) in above equation, we get :

$v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s$

(b) Put the value of $v_f$ in equation (3) as :

$v_s=7.09\ m/s$

Hence, this is the required solution.