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(C) Case $I$: Simultaneous jump.Let $V_1$ be the velocity of the buggy. The velocity of each man relative to the ground is $(10 - V_1)$.Using conserva

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$7 : 6$

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(C) Case $I$: Simultaneous jump.Let $V_1$ be the velocity of the buggy. The velocity of each man relative to the ground is $(10 - V_1)$.Using conservation of linear momentum: $0 = 100 V_1 - 2 	imes 50(10 - V_1)$.$100 V_1 = 100(10 - V_1) \implies V_1 = 10 - V_1 \implies 2 V_1 = 10 \implies V_1 = 5\, m/s$.Case $II$: Sequential jump.First man jumps: $0 = (100 + 50) v_a - 50(10 - v_a) \implies 150 v_a = 500 - 50 v_a \implies 200 v_a = 500 \implies v_a = 2.5\, m/s$.Second man jumps from the buggy (now mass $150\, kg$): $150 v_a = 100 V_2 - 50(10 - V_2)$.$375 = 100 V_2 - 500 + 50 V_2 \implies 150 V_2 = 875 \implies V_2 = 875 / 150 = 35 / 6 \approx 5.83\, m/s$.Ratio $V_1 : V_2 = 5 : (35/6) = 30 : 35 = 6 : 7$. The ratio of recoil velocities (sequential to simultaneous) is $7 : 6$.

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