JEE Main 2017 · Physics · Work, Energy and Power · hard

A bullet of mass 20 g moving horizontally at 400 m s^-1 passes through a wooden block of mass 1 kg hanging at rest from a light string of length 2 m…

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A bullet of mass

20\ \mathrm{g}

moving horizontally at

400\ \mathrm{m\,s^{-1}}

passes through a wooden block of mass

1\ \mathrm{kg}

hanging at rest from a light string of length

2\ \mathrm{m}

, and emerges with speed

100\ \mathrm{m\,s^{-1}}

. Taking

g = 10\ \mathrm{m\,s^{-2}}

, the maximum height risen by the block above its lowest position is

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Answer: (B) $1.8\ \mathrm{m}$

Momentum is conserved during the short passage:

0.02(400) = 0.02(100) + 1\cdot V

, so

V = 8 - 2 = 6\ \mathrm{m\,s^{-1}}

. The block then swings with mechanical energy conserved:

h = V^2/2g = 36/20 = 1.8\ \mathrm{m}

. Since

h < L = 2\ \mathrm{m}

the block stays below the level of the suspension point, so the string remains taut and the energy method is valid. The value

3.07\ \mathrm{m}

results from wrongly assuming the bullet embeds in the block, and

0.45\ \mathrm{m}

from using

V = 3\ \mathrm{m\,s^{-1}}

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake