Question

a simple pendulum of length 1m has a wooden bob of mass 1kg. it is stuck by a bullet of mass 10-² kg. moving with a speed of 2×10²m/s. the bullet gets embedded within the Bob.obtain the height to which the Bob rises before swinging back.​

Answer 1

Refer to attachment mates.

Thanks

Answer 2

Answer-

$0.2m$

Explanation-

applying principal of conservation of linear momentum,

$mu = (m + m)v$

$= = > {10}^{ - 2} \times (2 \times {10}^{2}$

$= = > (1 + .01)v$

$= = > v = \frac{2}{1.01} \frac{m}{s}$

initial KE of the block with bullet in it, is fully converted into PE as it rises through a height h, given by,

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

$= = > {v}^{2} = 2gh$

$= = > h = \frac{ {v}^{2} }{2g} = ( \frac{2}{1.01} ) {}^{2} \times \frac{1}{2 \times 9.8}$

$= = > 0.2m$

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