A block A having a mass ‘mA ’ is released from rest at the position P shown and slides freely down the smooth inclined ramp. When it reaches the bottom of the ramp it slides horizontally onto the surface of a cart of mass mc for which the coefficient of friction between the cart and the box is ‘µ’. If ‘h’ be the initial height of A, determine the final velocity of the cart once the block comes to rest in it. Also determine the position ‘x’ of the box on the cart after it comes to rest relative to cart. (The cart moves on smooth horizontal surface.)

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Applying energy conservation for the blockmgh=12mv2v=2gh...1Applying momentum conservation for the block and the carmA×2gh+0=(mA+mC)v1v1=mA×2gh(mA+mC) (when relative motion will cease they will aquire common velocity v1)Force of friction on the block in the backward direction =μN=μ×mAgmAaA=μ×mAgaA=μg ...1(acceleration of a is in backward direction that is it is retarded)Force of friction on the car in the forward  direction =μN=μ×mAgmCaC=μ×mAgaC=mAμgmC ...2Relative acceleration of block with respect to the car in the forward direction isablock,car=-μg-mAμgmC=-μg(1+mAmC)Applying the formula for stopping distancesblock,car=u2block,car2ablock,car=v22μg(1+mAmC)=2gh2μg(1+mAmC)=hμ(1+mAmC)Regards

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