Question

A father racing his son has three-fifths the kinetic energy of the son, who has one-third the mass of the father. The father speeds up by 2.5 m/s and then has the same kinetic energy as the son. (a) What is the original speed of the father? (b) What is the original speed of the son?

Answer 1

Let mass of dad=M and mass of son = m
let speed of dad = Vi (initially) & Vf (final speed after he sppeds up by 2.5) and speed of son = v

lets make this question into two parts: before dad speeds up and after he speeds up.

1: Before he speeds up:
k.en of dad = (k.e of son)/3
ie; M(Vi^2)/2 = m(v^2)/6
ie; M(Vi^2) = m(v^2)/3 ..............equation 1

we already know that m=(2M)/5

substitute this value for the ''m'' in equation 1:
then you will get a relation :
(Vi^2)/(v^2) = 2/15 ...................equation 2

2 : after dad speeds up :
Vf = Vi plus 2.5 ........ (P.s. : my plus sign is not working so im using the word ''plus'')
now : k.en of dad = k.en Of son
ie; M(Vf^2)/2 = m(v^2)/2
now in this equation substitue the value of ''m'' as you did earlier.

You will get an equation like this :

(Vf^2)/(v^2) = 2/5 ...............equation 3

divide equation 2 by eqaution 3:

(equation 2)/(equation 3) will give :
(Vi^2)/(Vf^2) =1/3
ie; Vf^2 =3Vi^2
ei; (Vi plus 2.5)^2 = 3Vi^2
expand the above equation to get a quadratic equation.

Solve the quadratic equation.

( how to solve a quadratioc equation of the type : ax^2 plus bx plus c =0 :
here the value of
x = {(-b) plus or- (root of b^2 - 4ac)}/2a )

you will get 2 values : one positive and another
-ve value.
The speed is obviously ve.
It will be nearly 3m/s
this is speed of dad : Vi

to find speed of son, put value of Vi in equation 2 and solve it.

Speed of son will be nearly 8m/s

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