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A mass of 0.98 kg attached to a spring of constant K = 100 N $m^{- 1}$ is hit by a bullet of 20 gm moving with a velocity 30 ${ms}^{- 1}$ horizontally. The bullet gets embedded and the system oscillates with the mass on horizontal friction less surface. The amplitude of oscillations will be

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0.6 cm

6 cm

1.2 cm

12 cm

answer is B.

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Detailed Solution

Conservation of momentum

(0.98+0.02)v = 0.02 $\times 30 \Rightarrow v = 0.6 {ms}^{- 1}$

Conserving energy in oscillation we get

$\frac{1}{2} (0.98 + 0.02) v^{2} = \frac{1}{2} {KA}^{2}$

$∴ A^{2} = \frac{v^{2}}{K} = \frac{{(0.6)}^{2}}{100} \Rightarrow A = \frac{0.6}{10} = 0.06 m = 6 cm$

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