| ### Statement | | ### Statement |
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| $2.1.50.$ [Insert the problem statement] | | $2.1.50.$ [Insert the problem statement] |
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| ### Solution | | ### Solution |
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| \section{Kinematics} | | \section{Kinematics} |
| Let us consider the instantaneous kinematics of the system.\\ | | Let us consider the instantaneous kinematics of the system.\\ |
| Let $M$ be the mass of the wedge.\\ | | Let $M$ be the mass of the wedge.\\ |
| Let $V$ be the velocity of the wedge. \\ | | Let $V$ be the velocity of the wedge. \\ |
| Let $v$ be the velocity of the block relative to the wedge. | | Let $v$ be the velocity of the block relative to the wedge. |
| \vspace{6pt} \\ | | \vspace{6pt} \\ |
| The velocity components of the block relative to the ground are: | | The velocity components of the block relative to the ground are: |
| $$ v_x = v\cos{\alpha}-|V|, \qquad v_y= v\sin{\alpha}$$ | | $$ v_x = v\cos{\alpha}-|V|, \qquad v_y= v\sin{\alpha}$$ |
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| From there we obtain \begin{equation} | | From there we obtain \begin{equation} |
| \tan{\beta}=\frac{v\sin{\alpha}}{v\cos{\alpha}-|V|} | | \tan{\beta}=\frac{v\sin{\alpha}}{v\cos{\alpha}-|V|} |
| \end{equation} | | \end{equation} |
| \section{Displacement of CM} | | \section{Displacement of CM} |
| Since no force acts on the system horizontally, the horizontal displacement of the center of mass $\Delta x_{cm_x}$ must equal to 0, therefore we can write: | | Since no force acts on the system horizontally, the horizontal displacement of the center of mass $\Delta x_{cm_x}$ must equal to 0, therefore we can write: |
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| $$ \Delta x_{cm_x} = \frac{m \vec{\Delta x}_{mx}+M \vec{\Delta x}_{Mx}}{m +M} $$ \vspace{1pt} | | $$ \Delta x_{cm_x} = \frac{m \vec{\Delta x}_{mx}+M \vec{\Delta x}_{Mx}}{m +M} $$ \vspace{1pt} |
| $$0 = \frac{mv_x\Delta t - M|V| \Delta t}{m+M}$$ | | $$0 = \frac{mv_x\Delta t - M|V| \Delta t}{m+M}$$ |
| \vspace{1pt} | | \vspace{1pt} |
| \begin{equation} | | \begin{equation} |
| M|V|=m(v \cos{\alpha}-|V|) | | M|V|=m(v \cos{\alpha}-|V|) |
| \end{equation} | | \end{equation} |
| \vspace{1pt} | | \vspace{1pt} |
| $$|V|=\frac{mv\cos{\alpha}}{M+m}$$ | | $$|V|=\frac{mv\cos{\alpha}}{M+m}$$ |
| \section{Substitution} | | \section{Substitution} |
| Now we substitute V into (1): | | Now we substitute V into (1): |
| $$\tan{\beta}=\frac{v}{v-\frac{mv}{M+m}}\tan{\alpha}$$ | | $$\tan{\beta}=\frac{v}{v-\frac{mv}{M+m}}\tan{\alpha}$$ |
| And obtain the following: \boldmath $$M=\frac{m\tan{\alpha}}{\tan{\beta} + \tan{\alpha}}$$ | | And obtain the following: \boldmath $$M=\frac{m\tan{\alpha}}{\tan{\beta} + \tan{\alpha}}$$ |
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