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?

2 Answers

Relevance
  • NCS
    Lv 7
    9 months ago
    Favorite Answer

    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

    If you find this helpful, please award Best Answer. You get points too!

  • 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

Still have questions? Get your answers by asking now.