Question

. A marble of mass m slides on a horizontal smooth surface with velocity Vo until it col lides elastically with a second marble of mass m23m, which was initially at rest. The marble of mass m2 falls from a step of height h 0.5m. Find the value of vo for which the n 2 le touches the ground from the step. Answer |31.3 m/s

Answer 1

From conservation of momentum m_1 v_1 = m_2 v_2 - m_1 v_1 implies v_0 = 3 v_2 - v_1 bar (1) from conservation of energy 1/ 2 m_1 v^2_0 = 1/ 2 m_1 v^2_1 + 1/ 2 m_2 v^2_2 implies v^2_0 = v^2_1 + 3 v^2_2 bar (2) we know time required to land (t) = squareroot 2h/ g d = v_2 t implies d = v_2 squareroot 2h/ g implies v_2 = d squareroot g/ 2h = 5 squareroot 9.8/ 2 times 0.5 m/s = 15.652 m/ s now putting v_2 in (2) v_0 = 3 v_2 - v_1 bar (1) implies v_0 = 46.95 - v_1 and putting (2) v^2_0 = v^2_1 + 735 bar (4) solving (3) and (4) v^2_0 = (46.95 - v_0)^2 + 735 implies v^2_0 = 2204.3 + v^2_0 - 93.9 v_0 + 735 implies 93.9 v_0 = 2939.3 implies v_0 = 31.30 m/s.
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