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 centre 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)

Center of Mass PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q50 with Solution

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Correct answer: A

This is a ballistic pendulum type problem.

After the bullet embeds in the block, the combined mass swings upward. During this upward swing, mechanical energy is conserved:

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

So,

$\frac{1}{2}V^2 = gh$

$V = \sqrt{2gh}$

Substitute values:

$V = \sqrt{2 \times 9.8 \times 0.098}$

$V = \sqrt{1.9208}$

$V \approx 1.386\,\text{m/s}$

The collision is perfectly inelastic, so momentum is conserved during impact:

$mu = (M+m)V$

where $u$ is the speed of the bullet before collision.

Thus,

$u = \frac{(M+m)V}{m}$

Now,

$M+m = 5.99 + 0.01 = 6.00\,\text{kg}$

Hence,

$u = \frac{6.00 \times 1.386}{0.01}$

$u = 600 \times 1.386$

$u \approx 831.6\,\text{m/s}$

With rounding to the nearest listed option,

$u \approx 831.4\,\text{m/s}$

So the correct option is A.

Stored correct answer: A

My derived answer: A

They agree.