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Question
A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it. If the centre of the pendulum rises by a distance of 10 cm, find the speed of the bullet.
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Solution
Given:
Mass of bullet, m = 25 g = 0.025 kg
Mass of ballistic pendulum, M = 5 kg
Vertical displacement, h = 10 cm = 0.1 m
Let the bullet strikes the pendulum with a velocity u.
Let the final velocity be v.
Using the law of conservation of linear momentum, we can write:
\[mu = (M + m)v\]
\[ \Rightarrow v = \frac{m}{(M + m)}u\]
\[ \Rightarrow v = \frac{0 . 25}{5 . 025} \times u = \frac{u}{201}\]
Applying the law of conservation of energy, we get:
\[\left( \frac{1}{2} \right)(M + m) v^2 = (M + m)gh\]
\[ \Rightarrow \frac{u^2}{(201 )^2} = 2 \times 10 \times 0 . 1\]
\[ \Rightarrow u = 201 \times \sqrt{2} = 280 \text{ m/s}\]
The bullet strikes the pendulum with a velocity of 280 m/s.
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