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
[tex]E_2 = 2E_1[/tex]
[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]
[tex]V^2 + 5V + 6.25 = 2V^2[/tex]
[tex]V^2 - 5V - 6.25 = 0[/tex]
[tex]V \approx 6.04m/s[/tex]
b) The son kinetic energy initially is:
[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]
We can solve for the son speed by the following formula
[tex]E_s = \frac{mv^2}{2}[/tex]
[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]
[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]
