A block of mass mb = 1.23 kg slides to the right at a speed of 2.99 m/s on a fri

Question

A block of mass mb = 1.23 kg slides to the right at a speed of 2.99 m/s on a frictionless horizontal surface, as shown in the figure. It "collides" with a wedge of mass m w, 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 v b+w at the instant the block stops sliding up the wedge. If the block's center of mass rises by a distance h = 0.355 m, what is the mass of the wedge? Tries 3/99 Previous Tries What is y b+w? Tries 1/99 Previous Tries

Explanation / Answer

conservation of linear momentum M *1.11-1.23*2.99 =(M+1.23)*V energy conservation 0.5*1.23*2.99*2.99 + 0.5M*1.11*1.11 = 0.5(M+1.23)V^2 + 1.23*9.82*0.355 M=0.87kg V=- 1.28m/s

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