A small block of mass m1 = 0.580 kg is released from rest at the top of a fricti

Question

A small block of mass m1 = 0.580 kg is released from rest at the top of a frictionless, curve-shaped wedge of mass m2 = 3.00 kg, which sits on a frictionless, horizontal surface as shown in Figure a. When the block leaves the wedge, its velocity is measured to be vf = 3.40 m/s to the right as shown in Figure b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? m/s to the left (b) What is the height h of the wedge? m

Explanation / Answer

momentum conservation gives that mv = m1v1 => v1 is 0.657333 using conservation of energy we get mgh = 0.5(m1v1^@ + m2v2^2) =>> height h = 6.89m

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