Question:
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
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.
Updated On: Jun 22, 2026
- \(\left(\dfrac{M_1+M_2}{4M_1+M_2}\right)a\)
- \(\left(\dfrac{2M_1+M_2}{4M_1+M_2}\right)a\)
- \(\left(\dfrac{M_1+2M_2}{4M_1+2M_2}\right)a\)
- \(\left(\dfrac{M_1+2M_2}{M_1+M_2}\right)a\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the initial acceleration.
Since all surfaces are smooth and the pulley and rope are massless, the driving force is the component of weight of \(M_2\) along the inclined plane.
Therefore, driving force is \[ M_2g\sin\theta \] Total mass of the system is \[ M_1+M_2 \] Hence, initial acceleration is \[ a=\frac{M_2g\sin\theta}{M_1+M_2} \]
Step 2: Find the new acceleration after changing masses.
Now, mass \(M_1\) is doubled, so new mass is \[ 2M_1 \] Mass \(M_2\) is halved, so new mass is \[ \frac{M_2}{2} \] New driving force is \[ \frac{M_2}{2}g\sin\theta \] New total mass of the system is \[ 2M_1+\frac{M_2}{2} \] Therefore, new acceleration is \[ a'=\frac{\frac{M_2}{2}g\sin\theta}{2M_1+\frac{M_2}{2}} \] Multiplying numerator and denominator by \(2\), \[ a'=\frac{M_2g\sin\theta}{4M_1+M_2} \]
Step 3: Express new acceleration in terms of \(a\).
From the initial acceleration, \[ a=\frac{M_2g\sin\theta}{M_1+M_2} \] So, \[ M_2g\sin\theta=a(M_1+M_2) \] Substituting this in \(a'\), \[ a'=\frac{a(M_1+M_2)}{4M_1+M_2} \] \[ a'=\left(\frac{M_1+M_2}{4M_1+M_2}\right)a \]
Step 4: Final conclusion.
Hence, the acceleration of the system is \[ \boxed{\left(\frac{M_1+M_2}{4M_1+M_2}\right)a} \]
Since all surfaces are smooth and the pulley and rope are massless, the driving force is the component of weight of \(M_2\) along the inclined plane.
Therefore, driving force is \[ M_2g\sin\theta \] Total mass of the system is \[ M_1+M_2 \] Hence, initial acceleration is \[ a=\frac{M_2g\sin\theta}{M_1+M_2} \]
Step 2: Find the new acceleration after changing masses.
Now, mass \(M_1\) is doubled, so new mass is \[ 2M_1 \] Mass \(M_2\) is halved, so new mass is \[ \frac{M_2}{2} \] New driving force is \[ \frac{M_2}{2}g\sin\theta \] New total mass of the system is \[ 2M_1+\frac{M_2}{2} \] Therefore, new acceleration is \[ a'=\frac{\frac{M_2}{2}g\sin\theta}{2M_1+\frac{M_2}{2}} \] Multiplying numerator and denominator by \(2\), \[ a'=\frac{M_2g\sin\theta}{4M_1+M_2} \]
Step 3: Express new acceleration in terms of \(a\).
From the initial acceleration, \[ a=\frac{M_2g\sin\theta}{M_1+M_2} \] So, \[ M_2g\sin\theta=a(M_1+M_2) \] Substituting this in \(a'\), \[ a'=\frac{a(M_1+M_2)}{4M_1+M_2} \] \[ a'=\left(\frac{M_1+M_2}{4M_1+M_2}\right)a \]
Step 4: Final conclusion.
Hence, the acceleration of the system is \[ \boxed{\left(\frac{M_1+M_2}{4M_1+M_2}\right)a} \]
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