Question

A ball of mass $1\text{kg}$ is attached to an inextensible string. The ball is released from the position shown in figure. Find the magnitude of impulse imparted by the string to the ball immediately after the string becomes taut. Take $g=10{\text{m/s}}^2$.

• A. $4\sqrt{40}\text{N-s}$
• B. $4\sqrt{10}\text{N-s}$
• C. $4\sqrt{20}\text{N-s}$
• D. $4\sqrt{5}\text{N-s}$

Answer 1

The correct option is B $4\sqrt{10}\text{N-s}$

The string will become taut when the particle falls through a distance of $8\text{m}$ downwards.

From the initial position to the point where the string becomes taut, it's a case of free fall because the only external force is gravity (since $T=0$ in the string until it gets taut)

Considering vertically downwards as $+vey-$ axis:
$v^2=u^2+2gh$
$\Rightarrow v^2=0+2\times \left(10\right)\times 8$
$\Rightarrow v=\sqrt{160}\text{m/s}$

The impulsive tension will bring the ball at rest immmediately i.e final velocity $v^{\prime }=0$


Impulse on ball is given by:
$J=m\Delta v=m(v_f-v_i)$
$=m(v^{\prime }-v)$
$=0-\sqrt{160}=-\sqrt{160}=-4\sqrt{10}\text{N-s}$

$\therefore$Magnitude of impulse $=4\sqrt{10}\text{N-s}$

The negative sign of impulse represents that impulsive force has retarded the motion of the ball.