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# (i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation ...

## Question

###### (i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation of linear momentum, or, $$egin{gathered} M overrightarrow{v_{1}}+2 mleft(vec{u}+overrightarrow{v_{1}} ight)=0 \ overrightarrow{v_{1}}=frac{-2 m vec{u}}{M+2 m} end{gathered}$$ (ii) Let $vec{v}^{prime}$ be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man $+$ other man) from th

(i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation of linear momentum, or, $$egin{gathered} M overrightarrow{v_{1}}+2 mleft(vec{u}+overrightarrow{v_{1}} ight)=0 \ overrightarrow{v_{1}}=frac{-2 m vec{u}}{M+2 m} end{gathered}$$ (ii) Let $vec{v}^{prime}$ be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man $+$ other man) from the conservation of linear momentum : $$0=(M+m) vec{v}^{prime}+m(vec{u}+vec{v})$$ Let $overrightarrow{v_{2}}$ be the sought velocity of the buggy when the second man jump off the buggy; then from conservation of linear momentum of the system (buggy + one man): $$(M+m) vec{v}=M overrightarrow{v_{2}}+mleft(vec{u}+vec{v}_{2} ight)$$ Solving equations (2) and (3) we get $$overrightarrow{v_{2}}=frac{m(2 M+3 m) vec{u}}{(M+m)(M+2 m)}$$ From (1) and (4) $$frac{v_{2}}{v_{1}}=1+frac{m}{2(M+m)}>1$$ Hence $v_{2}>v_{1}$

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