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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 $v_{2}>v_{1}$



Answers

(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, $$ \begin{gathered} M \overrightarrow{v_{1}}+2 m\left(\vec{u}+\overrightarrow{v_{1}}\right)=0 \\ \overrightarrow{v_{1}}=\frac{-2 m \vec{u}}{M+2 m} \end{gathered} $$ (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): $$ (M+m) \vec{v}=M \overrightarrow{v_{2}}+m\left(\vec{u}+\vec{v}_{2}\right) $$ 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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