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# Solve the following :

Question:

A heavy particle is suspended by a $1.5 \mathrm{~m}$ long string. It is given a horizontal velocity of $\sqrt{57} \mathrm{~m} / \mathrm{s}$. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

(a) $m g \cos \theta=\left(m v^{2}\right) / l$

$v=(\sqrt{g l} \cos \theta) \ldots \ldots 1$

Change in $\mathrm{K} . \mathrm{E}$. is qiven as :

$\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mgh}$

$\frac{\frac{1}{2}}{v^{2}-\frac{1}{2}}(\sqrt{57})^{2}=q(1.5[1+\cos \theta])$

${ }^{\frac{1}{2}} v^{2}-\frac{1}{2}(\sqrt{57})^{2}=g(1.5[1+\cos \theta])$

$v=(\sqrt{5} 7-3 g(1+\cos \theta) \ldots \ldots 2$

from equation 1 and 2

$1.5 \times g \cos \theta=57-3 g(1+\cos \theta)$

$\Theta=53^{0}$

(b) $v=\sqrt{57}-3 g(1+\cos \theta) \ominus=53$

$\mathrm{v}=3 \mathrm{~m} / \mathrm{s}$

(c) After, point B projectile motion will take place

$H=O E+D C$

$\mathrm{H}=1.5 \cos \theta+\left(u^{2} \sin ^{2} \theta\right) / 2 \times g$

$H=1.5 \times \frac{\frac{3}{5}}{H}+\left(9 \times 0.8^{2}\right) / 2 \times 10$

$\mathrm{H}=1.2 \mathrm{~m}$