A bullet of mass 10gmoving horizontally with a velocity of 400 ms-1 strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5m. As a result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

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100 ms-1

80 ms-1

120 ms-1

160 ms-1

answer is C.

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Detailed Solution

We can apply work energy theorem for the bullet and block before strike and after the bullet is ejected out of the blockFrom conservation of linear momentum 101000×400 + 0 = 2v1 + 101000v2 From work energy theorem Wgr = ∆K.E⇒2g×yC.G=12×2×v12 solving we get v1=2m/sec and v2=120m/sec

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