Question

1.) A heavy particle hanging from a string of length L
is projected horizontally with speed root gL. Find the
speed of the particle at the point where the tension
in the string equals weight of the particle.
(1) √3gl
(2) root gl/3

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Answer 1

Answer:

root gl/3

Explanation:

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Answer 2

Given that,

Length = L

Speed = √gl

Let the tension at angle θ

$T=mg$

The height is

$h=l(1-\cos\theta)$....(I)

We need to calculate the speed of the particle

Using conservation of energy

$\dfrac{1}{2}m(u^2-v^2)=mgh$

$v^2=u^2-2gh$.....(II)

Now, Using equation of the tension in the string

$T-mg\cos\theta=\dfrac{mv^2}{l}$

Put the value of T

$mg-mg\cos\theta=\dfrac{mv^2}{l}$

$v^2=gl(1-\cos\theta)$....(III)

We need to calculate the angle

Put the value of v, u and h in equation (II)

$gl(1-\cos\theta)=\sqrt{gl}-2gl(1-\cos\theta)$

$\cos\theta=\dfrac{2}{3}$

Put the value of θ in equation (III)

$v^2=gl(1-\dfrac{2}{3})$

$v=\sqrt{\dfrac{gl}{3}}$

Hence, The  speed of the particle is $\sqrt{\dfrac{gl}{3}}$

(2) is correct option.