sa asked in Science & MathematicsPhysics · 9 months ago

# kinetic energy?

A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.3 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?

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• NCS
Lv 7
9 months ago

son's mass is M and his velocity is V and his kinetic energy is ke.

initially,

father KE = ½ke

½*3M*v² = ½*½*M*V²

6v² = V²

later

father KE = ke

½*3M*(v+1.3m/s)² = ½*M*V²

3(v+1.3m/s)² = V²

and 6v² = V²

so

3(v+1.3m/s)² = 6v²

solving for v gives

v = 3.14 m/s ◄ father

V = v*√6 = 7.69 m/s ◄ son

• 9 months ago

Kf = 1/2 M vf^2    Ks = 1/2 m vs^2

Initially Kf = 1/2 Ks  and we are told m = M/3  so

1/2 (3ms) vf^2 = 1/2 ms vs^2  -->  3vf^2 = vs^2

Now vf --> vf + 1.3 m/s and Kf' = Ks

Kf' = 1/2 (3m)(vf + 1.3)^2 = 1/2 m vs^2

vf^2 + 2.6vf + (1.3)^2 = vs^2   use vf^2 = vs^2/s  and vf = vs/sqrt(3)

vs^2 + 2.6 vs/sqrt(3) + (1.3)^2 = vs^2 --->  2.6 vs/sqrt(3) + (1.3)^2 = 0

vs = -(1.3)^2*sqrt(3)/(2.6) = 1.125 m/s  (drop the "-")

since vf = vs/sqrt(3) = 0.65 m/s