Question

Two blocks of masses $3kg$ and $2kg$ respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in $5$ seconds is : $(g=10m/s^2)$

Answer 1

Let the acceleration of the two bodies is $a$.

Then tension in the string is given by:

$3\left(g-a\right)=2\left(g+a\right)=T$

$3g-3a=2g+2a$

$5a=g$

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

$a=\frac{10}{5}$

$a=2m/s^2$

Distance travelled by each mass is,

$s=0\times t+\frac{1}{2}\times 2\times 5^2$

$s=25m$

Now the distance of new centre of mass is,

$s^{\prime }=\frac{3\times 25+2\times \left(-25\right)}{3+2}$

$s^{\prime }=5m$

Thus, new centre of mass will shift $5m$ in downward direction.