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

Step-by-step 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=(1)/(3)K_s`

And
`m_s=(1)/(4)m_f`

Kinetic energy of father is given by :

`K_f=(1)/(2)m_fv_f^2`.............(1)

Kinetic energy of son is given by :

`K_s=(1)/(2)m_sv_s^2`...........(2)

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

`(v_f^2)/(v_s^2)=(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 :

`(K_f)/(K_s)=(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=(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.