Question

a pendulum Bob of mass M is raised to a height H and then released.At the bottom of it swing ,it picks up a mass m. To what height will the combined mass rise ?
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Answer 1

Answer:

$\frac{M^2H}{(M+m)^2}$

Explanation:

let the velocity of the bob at the bottom of the swing is v

When the bob is released from height H and at the bottom of the swing if its velocity is v then using Newton's second equation of motion

$v^2=0^2+2gH$

or, $v^2=2gH$

At the bottom of the swing, a mass m gets attached to the bob. Since the bob has some momentum, it will still go to some height, let that height be h

Let the velocity just after the mass m is attacged is v'

Then from the conservation of momentum

At the bottom of the swing,

Momentum just before the mass m gets attached = momentum just after the mass is attached

Mv = (M+m)v'

or, $v'= \frac{Mv}{(M+m)}$

again using the second equation of motion

$0^2=v'^2-2gh$

or, $h=\frac{v'^2}{2g}$

or, $h=\frac{1}{2g}\times\frac{M^2v^2}{(M+m)^2}$

or, $h=\frac{1}{2g}\times\frac{M^2\times2gH}{(M+m)^2}$

or, $h=\frac{M^2H}{(M+m)^2}$

Answer 2

Answer:

Explanation:

Here's the solution to your question.

Hope it helps you...!!!

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