Question

A father racing his son has half the kinetic energy of the son, who has two-fifths the mass of the father. The father speeds up by 3.0 m/s and then has the same kinetic energy as the son.

(a) What is the original speed of the father?
m/s

(b) What is the original speed of the son?
m/s

Answer 1

First you need the equation of KE which is:

KE = 1/2 * m * v^2

Call the mass of the father M, and assume he's running at speed V. The son has speed v. Just writing down what you are told at the beginning gives you:

2*1/2 M V^2 = (2/5)*(1/2)M v^2

Also you're told in the second sentence:

1/2 M (V+3)^2 = (2/5)*(1/2)*M*v^2

These are two equations and at first sight have three unknowns (M, V and v), but the M cancels in both equations so there are only two unknowns. They are thus simultaneous equations. The way you solve these is to find an expression for one of the unknowns in terms of the other: take the first equation:

2*1/2 M V^2 = (2/5)*(1/2)M v^2

V^2 = (2/5)*(1/2) v^2

V^2 = (1/5) v^2

V = 0.447 v

Now put this into equation 2

1/2 M (0.447v+3)^2 = (2/5)*(1/2)*M*v^2

1/2 (0.447v+3)^2 = (2/5)*(1/2)*v^2

(0.447v+3)^2 = (4/5)*(1/2)*v^2

(0.447v+3)^2 = (2/5)*v^2

Multipying the LHS out

0.2v^2 + 2.682v + 9 = 0.4v^2

-0.2v^2 + 2.682 v + 9 = 0

Using the usual formula for the roots of a quadratic gives:

taking the positive root (you can't have a negative speed)

16.189 m/s (son's speed)

Father is 0.447 times this or

7.236 m/s

You can verify these are the right answers by plugging them back into the first two equations above.