Two masses \(M_1\) and \(M_2\) are arranged as shown in the figure. Let \(a\) be the magnitude of the acceleration of the mass \(M_1\). If the mass of \(M_1\) is doubled and that of \(M_2\) is halved, then the acceleration of the system is
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For smooth connected systems, acceleration is found by dividing the net driving force by the total mass of the system.
Step 1: Set up the original system. Two masses $M_1$ and $M_2$ are connected over a smooth pulley, with $M_2$ providing the driving pull along an incline of angle $\theta$. The driving force is $M_2 g\sin\theta$ and the total mass being accelerated is $M_1 + M_2$. Step 2: Write the initial acceleration. \[ a = \frac{M_2 g\sin\theta}{M_1 + M_2} \] Step 3: Apply the changes to the masses. Now $M_1$ is doubled to $2M_1$ and $M_2$ is halved to $\dfrac{M_2}{2}$. The driving force becomes $\dfrac{M_2}{2} g\sin\theta$ and the total mass becomes $2M_1 + \dfrac{M_2}{2}$. Step 4: Write the new acceleration. \[ a' = \frac{\frac{M_2}{2} g\sin\theta}{2M_1 + \frac{M_2}{2}} \] Step 5: Clear the fractions by multiplying top and bottom by 2. \[ a' = \frac{M_2 g\sin\theta}{4M_1 + M_2} \] Step 6: Express $a'$ in terms of $a$. Divide $a'$ by $a$: the common factor $M_2 g\sin\theta$ cancels, leaving \[ \frac{a'}{a} = \frac{M_1 + M_2}{4M_1 + M_2} \] So \[ a' = \left(\frac{M_1 + M_2}{4M_1 + M_2}\right)a \] which is option (1). \[ \boxed{\left(\dfrac{M_1 + M_2}{4M_1 + M_2}\right)a} \]
