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 data: `m_1 = 20 g = 20 xx 10^(-3) kg, m_2 = 5 kg, s = 10 xx 10 ^(-2)` m.
Let the speed of the bullet be v. 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))=( 20 xx 10^(-3) v )/( 5 +20 xx 10^(-3)) = ( 0.02)/( 5.02 ) v= 0.004 v `
The bob with bullet go up with a deceleration of `g=9.8.ms^(-2)`. Bob and bullet come to me at a height of `10 xx 10^(-2) m.`
from III rd equation of motion .
` V^2 = u^2 + 2as` ,
here ` v=u + at `
` v^2 - 2gs =0 " " s=((u+v) /(2))t `
` ( 0.004 v)^2 = 2 xx 9.8 xx 10 xx 10^(-2) " " v^2 = u^2 + 2as `
` v^2 = ( 2xx 9.8 xx 10xx 10^(-2))/(( 0.004 )^2)" " s = ut + 1/2 at^2 `
` v= 350 ms^(-1)`