Question

A bullet of mass $25g$ is fired horizontally into a ballistic pendulum of mass $5\cdot 0kg$ and gets embedded in it . If the centre of the pendulum rises by a distance of $10cm$, find the speed of the bullet.

Answer 1

Given that,

Mass of bullet $m=25g=0.025kg$

Mass of pendulum $M=5kg$

Vertical displacement $h=10cm=0.1m$

Now,

Let it strike pendulum with velocity u.

Let final velocity be v

$mu=(m+M)v$

$v=\frac{mu}{(M+m)}$

$v=\frac{0.025\times u}{5.025}$

$v=\frac{u}{201}$

Now, using conservation of energy

$0-\frac{1}{2}(M+m)v^2=-(M+m)g\times h$

$\frac{1}{2}(M+m)\times \frac{u^2}{{\left(201\right)}^2}=(M+m)\times 10\times 0.1$

$\frac{1}{2}(5+0.025)\times \frac{u^2}{{\left(201\right)}^2}=(5+0.025)$

$u^2={\left(201\right)}^2\times 2$

$u=284.2$

$u=284m/s$

Hence, the speed of the bullet is $284m/s$