Question

A bullet of mass m passes through a pendulum Bob of mass M with velocity v and comes out of it with a velocity v/2.The minimum value of v so that the bob completes one revolution in the vertical circle is

Answer 1

In vertical circular motion velocity at lowest point is = √5gl

Using conservation of energy,

mv = M√5gl + m(v/2)

or, m(v/2) = M√5gl

or, v = (2M√5gl)/m

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Answer 2

Answer:

The minimum value of v is $\dfrac{2M}{m}\sqrt{5gl}$

Explanation:

Given that,

Mass of bullet = m

Mass of bob = M

Velocity of bob = v

Velocity with which the of bob returns $v'=\dfrac{v}{2}$

In vertical circular motion velocity at lowest point is given by

$v'' = \sqrt{5gl}$

Using conservation of energy

$mv=Mv''+mv'$

$mv=M\sqrt{5gl}+m(\dfrac{v}{2})$

$mv-\dfrac{mv}{2}=M\sqrt{5gl}$

$\dfrac{mv}{2}=M\sqrt{5gl}$

$v = \dfrac{2M}{m}\sqrt{5gl}$

Hence, The minimum value of v is $\dfrac{2M}{m}\sqrt{5gl}$