A trolley of length L has a pistol fixed on i | vedclass

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(A) Since there is no external horizontal force acting on the system (trolley + pistol + bullet),the position of the center of mass of the system rema

Vedclass · Accepted Answer

$-\frac{m}{M} L$

Vedclass · Answer

(A) Since there is no external horizontal force acting on the system (trolley + pistol + bullet),the position of the center of mass of the system remains unchanged.Let $\Delta x_1$ be the displacement of the trolley and $\Delta x_2$ be the displacement of the bullet relative to the ground.The displacement of the center of mass is given by $\Delta x_{cm} = \frac{M \Delta x_1 + m \Delta x_2}{M + m} = 0$.Here,the bullet travels a distance $L$ relative to the trolley. If the trolley moves by $\Delta x_1$ (to the left,so $\Delta x_1$ is negative),the bullet's displacement relative to the ground is $\Delta x_2 = L + \Delta x_1$.Substituting this into the equation: $M \Delta x_1 + m(L + \Delta x_1) = 0$.$M \Delta x_1 + mL + m \Delta x_1 = 0$.$(M + m) \Delta x_1 = -mL$.$\Delta x_1 = -\frac{m}{M+m} L$.Note: The provided option $A$ is $-\frac{m}{M} L$,which is an approximation assuming $m \ll M$. However,based on the standard derivation,the correct expression is $-\frac{m}{M+m} L$. Given the options,$A$ is the intended answer under the assumption $M+m \approx M$.

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