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Questions using constraints

Question

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. [AIIMS 2018]

Difficult

A

1.4 s
B

2.45 s
C

3.25 s
D

5 s

Solution

Let a₁ and a₂ be accelerations of a ball (upward) and rod (downward), respectively
Clearly, from the diagram,
$2 a_{1} = a_{2} . . . . . . (i)$
Now, for the ball,
$2 T - \frac{9}{5} m g = \frac{9}{5} m a_{1} . . . . . . (ii)$
and for the rod, mg - T = ma₂ …………..(iii)
On solving Eqs. (i), (ii) and (iii), we get
$a_{1} = \frac{g}{29} m / s^{2} ↑ (upward)$
$a_{2} = \frac{2 g}{29} m / s^{2} ↓ (downward)$
So, acceleration of ball w.r.t. rod = $= a_{1} + a_{2} = (3 g / 29) m / s^{2}$
Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is 1m.
Using second equation of motion, $s = u t + \frac{1}{2} a t^{2}$
$1 = 0 + \frac{1}{2} \times \frac{3 \times 10}{29} t^{2}$
$\Rightarrow t = \sqrt{58 / 30} = 1.4 s (approx)$

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