Question 7: Laws of Motion

Question

In the figure, mass of a ball is $$\frac{9}{5}$$ times mass of the rod. Length of rod is $$1 \mathrm{~m}$$. The level of ball is same as rod level. Find out time taken by the ball to reach at upper end of rod.

Accepted Answer

Let $$a_1$$ and $$a_2$$ be accelerations of a ball (upward) and rod (downward), respectively.
Clearly, from the diagram
$$2 a_1=a_2$$ ..... (i)
Now, for the ball
$$2 T-\frac{9}{5} m g=\frac{9}{5} m a$$ ..... (ii)
and for the rod,
$$m g-T=m a_2$$ ...... (iii)
On solving Eqs. (i), (ii) and (iii), we get
$$\begin{aligned}
& a_1=\frac{g}{29} \mathrm{~m} / \mathrm{s}^2 \uparrow 	ext { (upward) } \\
& a_2=\frac{2 g}{29} \mathrm{~m} / \mathrm{s}^2 \downarrow(	ext { downward ) }
\end{aligned}$$
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 $$1 \mathrm{~m}$$.
Using equation of motion,
$$\begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& s=0+\frac{1}{2} 	imes \frac{3 	imes 10}{29} t^2 \\
& t=\sqrt{\frac{58}{30}}=1.4 \mathrm{~s} 	ext { (approx) }
\end{aligned}$$

Answer

1.4 s

Answer

2.45 s

Answer

3.25 s

Answer

5 s

Laws of Motion

In the figure, mass of a ball is $\frac{9}{5} times mass of the rod. Length of rod is 1 \mathrm{~m}$. The level of ball is same as rod level. Find out time taken by the ball to reach at upper end of rod.

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