A bullet of mass 20 g strikes a pendulum of mass 5kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

Explanation

Given: `m_(1)=20g=20xx10^(-3)kg, m_(2)=5kg,s=10xx10^(-2)m`
Let the speed of the bullet be. The common velocity of bullet and pendulum bob is V. According to law of conservation of linear momentum
`v=(m_(1)v)/((m_(1)+m_(2)))=(20xx10^(-3)v)/(5+20xx10^(-3))=(0.02)/(5.02)=0.004v`
The bob with bullet go up with a deceleration of `g = 9.8ms^(-2)`. Bob and bullet come to rest at a height of `10 x x10^(-2) m`.
from lind equation of motion
`v^(2)=u^(2)+2"as here"`
`v^(2)-2gs=0`
`v^(2)=2gs`
`(0.004v)^(2)=2xx9.8xx10xx10^(-2)`
`v^(2)=(2xx9.8xx10xx10^(-2))/((0.004)^(2))`
`v=350ms^(-1) `