Question

11. A pendulum bob of mass M is raised to a height
h and then released. At the bottom of its swing, it
picks up a mass m. To what height will the
combined mass rise?
express as[ M<2 h/(m+M)<2]
here (<2 means square) ​

Answer 1

A rough diagram is shown in figure. here a pendulum bob of mass M is raised to a height j and then released, then at bottom of its swing , it picks up a mass m.

so, first of all, find velocity of bob at bottom,

using formula, v² = u² + 2as

or, v² = 0 + 2(-g)(-h) [ as initial velocity of bob is zero because bob is released under gravity. ]

so, v = √(2gh)

now, applying law of conservation of linear momentum.

momentum of bob and mass m at bottom = momentum of bob and mass m at H .

here H is maximum height attained by system of bob and mass after striking.

i.e., $m_1v_1+m_2v_2=(m_1+m_2)v'$

or, Mv = (M + m)v'

or, v' = M√(2gh)/(M + m)

now, height attained by system of bob and mass m, H = v'²/2g

= {M√(2gh)/(M + m)}²/2g

= M²h/(M + m)²

Answer 2

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