Question

In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass $B$?

• A. $\frac{g}{3}$
• B. $\frac{5g}{2}$
• C. $\frac{2g}{2}$
• D. $\frac{2g}{5}$

Answer 1

The correct option is D

$\frac{2g}{5}$

Step 1: Write the Equations of Motion

Let the acceleration of mass $A$ be $a_A$ and acceleration of mass $B$ be $a_B$.

Therefore, the accelerations can be expressed as,

$a_A=\frac{a_B+0}{2}$

$\Rightarrow 2a_A=a_B$ $\left(1\right)$

Let, $T_A$ be the tension on the string for mass $A$, and $g$ be the acceleration due to gravity.

Then the equation of motion of mass $A$ is

$mg-T_A=m{a}_A$

$\Rightarrow mg=m{a}_A+T_A$ $\left(2\right)$

Similarly, the equation of motion of mass $B$ is

$T_B-mg=m{a}_B$

$\Rightarrow mg=T_B-m{a}_B$

From equation $\left(1\right)$ we get,

$mg=T_B-2m{a}_A$

$2mg=2T_B-4m{a}_A$

Since the mass of the pulleys can be neglected we get

$T_A=2T_B$

$\therefore 2mg=T_A-4m{a}_A$ $\left(3\right)$

Step 2: Solve the Equations

Upon Subtracting $\left(2\right)$ from $\left(3\right)$ we get

$2mg-mg=T_A-T_A-4m{a}_A-m{a}_A$

$\Rightarrow mg=-5m{a}_A$

Upon neglecting the negative sign we get

$a_A=\frac{g}{5}$

From equation $\left(1\right)$ we get

$a_B=\frac{2g}{5}$

Hence, the correct answer is option (D).