Question

A small ball of $m$ is connected by an inextensible massless string of length $\ell =4$ meter with another ball of mass $M=4m$. They are released with zero tension in the string from a height $h=0.2$ meter as shown. How long ( in seconds ) after the release, the string becomes taut for the first time after the mass $M$ collides with the ground? (all collisions are elastic, assume only one collision takes place between the mass $M$ and the ground)

• A. $2$
• B. $12$
• C. $13$
• D. $3$

Answer 1

The correct option is A $2$

The speeds of both the balls just after the bigger ball hits the ground will be $\sqrt{2gh}=2㎧$. Then both the balls move towards each other and cover 4 meters to collide again. Since accelerations are same for both the balls, the time taken for collision is: $\frac{4}{2+2}=1$ second.

At this instant, since the impulsive forces will be much larger than the weights, we can use impulse momentum equations on both balls and add the results to get: $4v_1-v_2=5\sqrt{2gh}$. And a kinetic energy conservation will give $4v_1^2+v_2^2=10gh$. Solving the above two equations, we get, $v_1=-v_2=\sqrt{2gh}=2㎧$. Now the balls move apart until the string is tight and since the accelerations are same for both the balls, we get the time taken from this point for the string to become tight again as $1$ second.

Thus total time taken is $2$ seconds.