A block of mass mb = 1.0 kg slides to the right at a speed of 3.0 m/s on a frict

Question

A block of mass mb = 1.0 kg slides to the right at a speed of 3.0 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 1.2 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.
V1f=((m1-m2)/(m1+m2))v1 V2f=((2m1)/(m1+m2)) dy=.33 completly inelastic collision? Do I need to find the center of mass for both objects?

Explanation / Answer

By conservation of momentum of the system 1.v(b)-m(w).v(w)=-momentum of the whole system i.e m(b+w).v(b+w) similarly by conservation of energy K E of block + K. E of wedge = KE of block and wedge + PE of block these two equations are unknown in two variables M(w) and v(b+w) and can be solved to get answer

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