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Question

A block of mass m moving with velocity v0 on a smooth horizontal surface hits the spring of constant K as shown. The maximum compression in spring is

A
2mKv0
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B
mKv0
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C
m2Kv0
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D
m4Kv0
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Solution

The correct option is B mKv0
According to law of conservation of energy
Ei=Ef
Ui+Ki=Uf+Kf
[Sum of potential energy stored in spring and kinetic energy of block will be constant]
[Ui=0, as spring is unstretched]
0+12mv20=12Kx2
[for maximum compression Kf=0]
x=mKv0
Maximum compression in spring, x=mKv0

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