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haidy
Answers (1)
Δy = v0t + ½ at2 - 0.8 = (0)t + ½ (-9.8)t2 0.404 s = t Δx = v0t + ½ at2 1.2 = v0 (0.404) + ½ (0)(0.404)2 2.97 m/s = v0 (after the collision as the block leaves the table) p0 = p1 (0. 012 kg )v + (3 kg)(0 m/s) = (0.012 kg)(2.97 m/s) + (3 kg)(2.97 m/s)
