A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their center of mass rising a vertical distance h=9.8 cm before the (block+ bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8 ms^–2)

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821.4m/s

841.4m/s

811.4m/s

831.4m/s

answer is D.

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Detailed Solution

After bullet gets embedded till the system comes momentarily at rest,

$(M + m) gh = \frac{1}{2} (M + m) {v_{c}}^{2}$

$v_{c} = \sqrt{2 gh} = \sqrt{2 \times 9.8 \times 9.8 \times 10^{- 2}}$

Applying momentum conservation, after and before collision,

$mv = (M + m) v_{c} v = (\frac{M + m}{m}) v_{c} v = \frac{6}{10 \times 10^{- 3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{- 2}} v \approx 831.4 m / s$

Hence the correct answer is 831.4m/s.

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