A block of mass mb = 1.23 kg slides to the right at a speed of 2.89 m/s on a fri

Question

A block of mass mb = 1.23 kg slides to the right at a speed of 2.89 m/s on a frictionless horizontal surface, as shown in the figure. It "collides" with a wedge of mass mw, which moves to the left at a speed of 1.11 m/s. The wedge is shaped so that the block slides seamlessly up the Teflon (frictionless!) surface, as the two come together. Relative to the horizontal surface, block and wedge are moving with a common velocity vb+w at the instant the block stops sliding up the wedge. a) If the block's center of mass rises by a distance h = 0.383 m, what is the mass of the wedge? b) What is vb+w? Figure: http://i.imgur.com/5lbk5.png Hint Given: To solve this problem, write two equations. One for the conservation of momentum, and another for the conservation of energy (kinetic and potential will be in this equation). You should have a V_w+b in both equations. Solve each equation for V_w+b and then set each equation equal to eachother to solve for the mass. This is something you should all be familiar with. If you undress the v's m's and other variables from the equations, and replace them with the familiar x's, y's and z's, you will quickly see that this is a simple system of equations that can be solved by 10th grade methods. Don't let the variables confuse you. You will probably never be given a system of equations that has infinitely many solutions (one or more free variables) in this class. With this in mind, there will always be some trick that will enable you to solve for variables. In this case, the trick is to simply use substitution. Most of the time, that's the hardest it will be.

Explanation / Answer

initial momentum =1.23 X 2.89 - mw X 1.11 final momentum = (1.23+mw) Vf they are equal. now, theinitial energy = 1/2 mbU1^2 + 1/2 mw U2^2 = 0.5( 1.23 X 2.89^2 + mw X 1.11^2) final energy = 1/2( mw+ mb )Vf^2 + mb g 0.383 we have two equations , two variables. mw and Vf Vf is the velocity of block and wedge wrt ground frame. ie horizontal surface, which is what is necessary solve the

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