A small block of mass M is released from rest at the top ofcurved frictionless r

Question

A small block of mass M is released from rest at the top ofcurved frictionless ramp. The block is at the left of theramp at a height h above the horizontal. The block slidesdown the ram and is moving with speed of 5.0 v when it collideswith a larger block of mass 1.5 M at rest at the bottom of theincline on the right. The large block moves to the right withspeed of 2.5 v immediately after collision. Express answersto questions in terms of given quantities and fundamentalconstants. Determine height h of ramp from which small block wasreleased Determine speed of small block after collision Larger block slides a distanceD before coming to rest. Determine value of coefficient of kinetic friction betweenblock and surface on which it slides Indicate whether collision between 2 blocks is elastic orinelastic. Justify answer A small block of mass M is released from rest at the top ofcurved frictionless ramp. The block is at the left of theramp at a height h above the horizontal. The block slidesdown the ram and is moving with speed of 5.0 v when it collideswith a larger block of mass 1.5 M at rest at the bottom of theincline on the right. The large block moves to the right withspeed of 2.5 v immediately after collision. Express answersto questions in terms of given quantities and fundamentalconstants. Determine height h of ramp from which small block wasreleased Determine speed of small block after collision Larger block slides a distanceD before coming to rest. Determine value of coefficient of kinetic friction betweenblock and surface on which it slides Indicate whether collision between 2 blocks is elastic orinelastic. Justify answer

Explanation / Answer

energy conservation: Mgh = M*(5.0v)2/2 h = 12.5 v2/g initial momentum = M*5.0v final momentum = 1.5M*2.5v + M*V momentum conservation: M*5.0v = 1.5M*2.5v + M*V V = 1.25 v (1.5M)*(2.5v)2/2 = (1.5M)g*D = 3.125 v2/(gD) energy before collision E = M*(5.0v)2/2 =12.5Mv2 energy after collision E' = M*(1.25v)2/2 +(1.5M)*(2.5v)2/2 = 5.47Mv2 E'

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