Question

As shown in the figure, two masses $6\text{kg}$ and $4\text{kg}$ respectively are tied to the ends of a string which passes over a light frictionless pulley. Initially, the masses are at rest and are at the same horizontal level. Now, they are released. Find the distance travelled by their center of mass in $10\text{s}$.

• A. $10\text{m}$
• B. $20\text{m}$
• C. $30\text{m}$
• D. $25\text{m}$

Answer 1

The correct option is B $20\text{m}$


Assume acceleration of the system is $a$, as shown.

$F.B.D$ of $6\text{kg}$ block:


$6g-T=6a$
$T=6(g-a)$ ...(1)

$F.B.D$ of $4\text{kg}$ block:


$T-4g=4a$
$T=4(g+a)$ ...(II)

From eq. $\left(I\right)$ and $\left(II\right)$
$4(g+a)=6(g-a)$
$6g-4g=4a+6a$
$10a=2g$
or $a=2{\text{m/s}}^2$

Distance travelled by $6\text{kg}$ mass in $10\text{s}$ is (taking downward +ve)
$S=ut+\frac{1}{2}a{t}^2$
Here $u=0$ (initially at rest)
$\Rightarrow S=\frac{1}{2}\times 2\times {10}^2=100\text{m.}$ {Downwards}

Similarly, distance travelled by $4\text{kg}$ mass block is $100\text{m}$ {upwards}

So, the distance travelled by $COM$ is
$S^{\prime }=\frac{6\times \left(100\right)+4(-100)}{6+4}$
$S^{\prime }=\frac{200}{10}=20\text{m}$ {Downwards}

Hence, the distance travelled by the $COM$ of the blocks in $10\text{s}$ is $20\text{m}$. {Downwards}