# Solve the following :

Question:

A bullet of mass $25 \mathrm{~g}$ is fired horizontally into a ballistic pendulum of mass $5.0 \mathrm{~kg}$ and gets embedded in it. If the center of the pendulum rises by a distance of $10 \mathrm{~cm}$, find the speed of the bullet.

Solution:

Use C.O.L.M

$0.025 \times u=(0.025+5) V$

$V=\frac{u \times 0.025}{5.025}=\frac{u}{201}$

From C.O.E.L (conservation of energy law)

$\frac{-1}{2}(0.025+5) \cdot V^{2}=-(0.025+5) \times g \times h$

$\Rightarrow \frac{u^{2}}{(201)^{2}}=2$

$\Rightarrow u=201 \sqrt{2}=280 \mathrm{~m} / \mathrm{s}$