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In the figure, mass of a ball is $\frac{9}{5}$ times mass of the rod. Length of rod is $1 \,m$. The level of ball is same as rod level. Find out time taken by the ball to reach at upper end of rod. (in $S$)
$1.4$
$2.45$
$3.25$
$5$
Solution
Let $a_{1}$ and $a_{2}$ be accelerations of a ball (upward) and rod (downward), respectively.
Clearly, from the diagram
$2 a_{1}=a_{2} \ldots( i )$
Now, for the ball
$2 T-\frac{9}{5} m g-\frac{9}{5} m a-(i i)$
and for the rod, $m g-T=m a_{2} \ldots$ $(iii)$
On solving equations $(i)$ and $(iii),$ we get
$a_{1}=\frac{g}{29} m / s^{2} \uparrow$ (upward)
$a_{2}=\frac{2 g}{29} m / s ^{2} \downarrow$ (upward)
So, acceleration of ball $w.r.t$ rod $=a_{1}+a_{2}=\frac{3 g}{29}$
Now, displacement of ball $w.r.t.$ rod when it reaches the upper end of rod is $1\, m$.
Using equation of motion,
$s=u t+\frac{1}{2} a t^{2}$
$s=0+\frac{1}{2} \times \frac{3 \times 10}{29} t^{2}$
$t=\sqrt{\frac{58}{30}}=1.4 s$ (approx)
