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Question

A spring-block pendulum is shown in the figure. The system is hanging in equilibrium. A bullet of mass m/2 moving with a speed u hits the block from down as shown in the figure. Find the amplitude of oscillation now.
1605679_9cc19bb6f1e64d219ec57cc988dfc59b.png

A
mu28k+(mg2k)2
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B
mu2k+(mgk)2
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C
mu26k+(mg2k)2
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D
mu24k(mg2k)2
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Solution

The correct option is C mu26k+(mg2k)2
Let h be the elongation in spring
mg=kh

If a bullet of mass m/2 gets embedded in block

New mass, m=m+m2=3m2

New mean position of block will be let say at a depth of h1 from old mean position ; then
32mg=k(h+h1)

h1=mg2k (mg=kh)

Just after impact due to inelastic collision if the velocity of block becomes v1,

According to momentum conservation

m2u=3m2v

v=u3

Now, the block executes SHM and at t=0 block is at a distance h1=mg2k above its mean position and having a velocity u3

If A is amplitude u3=ωA3(mg2k)2

or u29=2k3m[A2(mg2k)2[ [ω=k(3m/2)]

A=mv26k+(mg2k)2 (Option C)

2063244_1605679_ans_a7ae4de814af4c46ae1caf5d44bba453.png

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