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A mass m is at rest on top of the frictionless incline of mass M. It is initially at a height h above the surface. After m slides down the incline, the incline moves to the left while m moves to the right with a velocity vo. The surface m and M slide along are frictionless. a) Find an expression for the velocity, v, of the incline after the mass m slides down. Your expression should only contain m, M, and vo. b) Find an expression for the height h the mass m was originally released from. Your expression should only contain m, M, vo, and various constants.

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Answer:

Explanation:

a)

Apply the law of conservation of momentum

Since there is no external force(friction) oon the system, Momentum in horizontal direction must remain constant.

Initial Momentum = Final Momentum

[tex]M(0) + m(0) = M(-V) + mv_0\\\\0=-MV+mv_0\\\\V=\frac{mv_0}{M}[/tex]

b)

Apply law of conservation of energy

Loss in Potential Energy = Gain in Kinetic Energy

[tex]mgh=\frac{1}{2}mv_0^2+\fac{1}{2}MV^2\\\\mgh=\frac{1}{2}mv_0^2+\fac{1}{2}M(\frac{mv_0}{M})^2\\\\mgh=\frac{1}{2}mv_0^2+\fac{1}{2}\frac{m^2v_0^2}{M}\\\\gh=\frac{v^2_0}{2}+\frac{v^2_0}{2}\frac{m}{M}\\\\h=\frac{v_0^2}{2g}[1+\frac{m}{M}][/tex]

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