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A solid ball and a hollow ball, each with a mass of 2 kg and radius of 0.1 m, start from rest and roll down a ramp of length 2 m at an incline of 34°. An ice cube of the same mass slides without friction down the same ramp.

What is the speed of the solid ball at the bottom of the incline?

The answer is 3.96 m/s. How do I get this answer? At first I thought I could use E=1/2mv^2; however, I do not have the total amount of energy, so I cannot use this equation. Any help is greatly appreciated.

  • Physics -

    For each object
    PE → KE
    h =s•sinα = 2•sin34º =1.12 m

    Solid ball I = 2•m•R²/5, ω=v/R
    PE =KE = KE (translational) + KE (rotational)
    m•g•h = m•v²/2 + I•ω²/2 = m•v²/2 + 2•m•R²•v²/2•5R² =
    = m•v²/2 + m• v²/ 5 =0.7•m• v².
    v = sqrt(g•h/0.7) =sqrt(9.8•1.12/0.7) =3.96 m/s.

    Hollow ball I = 2•m•R²/3 , ω=v/R
    m•g•h = m•v²/2 + I•ω²/2 = m•v²/2 + 2•m•R²•v²/2•3R² =
    = m•v²/2 + m• v²/ 3 =5•m• v²/6.

    Ice cube
    m•g•h = m•v²/2

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