A bullet of mass 50 g is fired rom below into the bob of mass 450 g of a long simple pendulum as hown in figure. The bullet premains inside the bob and the bob rises thrugh a height of 1.8 m. Find the speed of the bullet. Take `g=10 m/s^2`

Explanation

Let the speed of the bullet be `v`. Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, be `v`. By the principle of conservation of linear momentum.
`V=((0.05kg)v)/(0.45g+0.05kg)=v/10`
The string becomes loose and the bob will go up with a deceleration of `g=m//s^(2)`. As it comes rest at a height of `1.8 m`, using the equation `v^(2)=u^(2)+2ax`, we get
`1.8m=((v/10)^(2))/(2xx10m//s^(2))` or `v=60m//s`