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? https://s1.lite.msu.edu/res/mcgrawhill/BauerWestfall/1stEdition/Chapter08/P036figure.png

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 equations

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