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.

A bullet of mass 20 g strikes a pendulum of mass 5kg. The centre of ma

1.	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.
Answer» Solution : Given DATA: `m_1= 20 g = 20xx10^(-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= 0.004v ` 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 xx10^(-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 9.8xx 10 xx 10^(-2) ""v^2= u^2+ 2as ` `v^2 = ( 2xx 9.8xx 10xx 10^(-2))/(( 0.004 )^2)"" s = ut+ 1/2at^2 ` ` v= 350ms^(-1)`