A 15 gm bullet is fired horizontally into a 3 kg block of wood suspended by a string. The bullet sticks in the block, and the impact causes the block to swing 10 cm above the initial level. The velocity of the bullet nearly is

281 m s^{- 1}

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326 m s^{- 1}

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184 m s^{- 1}

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58 m s^{- 1}

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Solution

The correct option is A $281 m s^{- 1}$
After collision initial K.E of combined mass $= 3 \times 10 \times 0.1$
$= 3 J$

$\Rightarrow \frac{1}{2} \times 3 \times v^{2} = 3$

$v = \sqrt{2}$ m/s
conserving momentum
$15 \times v = 3000 \times \sqrt{2}$
$v = 281 m / s$

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A 15 gm bullet is fired horizontally into a 3 kg block of wood suspended by a string. The bullet sticks in the block, and the impact causes the block to swing 10 cm above the initial level. The velocity of the bullet nearly is

The correct option is A 281ms−1After collision initial K.E of combined mass =3×10×0.1=3J ⇒12×3×v2=3v=√2 m/sconserving momentum15×v=3000×√2v=281m/s

The correct option is A $281 m s^{- 1}$
After collision initial K.E of combined mass $= 3 \times 10 \times 0.1$
$= 3 J$

$\Rightarrow \frac{1}{2} \times 3 \times v^{2} = 3$

$v = \sqrt{2}$ m/s
conserving momentum
$15 \times v = 3000 \times \sqrt{2}$
$v = 281 m / s$