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