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Two blocks of masses of 3 kg and 6 kg rest on a horizontal smooth surface. The 3 kg block is attached to a spring with a force constant k=900Nm1 which is compressed 2 m beyond the equilibrium position. The 6 kg mass is at rest at 1 m from mean position, 3 kg mass strikes the 6 kg mass and the two stick together
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A
velocity of the combined masses immediately after the collision is 10ms1
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B
velocity of the combined masses immediately after the collision is 5ms1
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C
amplitude of the resulting oscillation is 2m
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D
amplitude of the resulting oscillation is 5/2m
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Solution

The correct option is A velocity of the combined masses immediately after the collision is 10ms1
Using work-energy principle,
Just before collision
12k(x22x12)=12m(v2u2)
12k(2212)=12×3×(v2)
v=30m/s
Now, Using momentum conservation principle,
m1u1+m2u2=(m1+m2)v
v=10m/s

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