Question

A father racing his son has half the kinetic energy of the son, whohas three-fifths the mass of the father. The father speeds up by2.5 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

Answer:

a) 6.04 m/s

b) 11.02 m/s

Explanation:

a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s

$E_2 = 2E_1$

$\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}$

$V^2 + 5V + 6.25 = 2V^2$

$V^2 - 5V - 6.25 = 0$

$V \approx 6.04m/s$

b) The son kinetic energy initially is:

$E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J$

We can solve for the son speed by the following formula

$E_s = \frac{mv^2}{2}$

$v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s$

$v = \sqrt{121.4} = 11.02 m/s$