A small steel ball `A` is suspended by an inextensible thread of length `l= 1.5 m` from `O`. Another identical ball is thrown vertically downwards such that its surface remains just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just completes vertical circle after collision, calculate the velocity (in `cm//s`) of the falling ball just before collision `(g = 10 m s^(-2))`.

Answer

`1250cm/s`

Explanation

`v_(2)=sqrt(5gl)=sqrt(5xx10xx1.5)=5sqrt(3)m//s`
implies `Jxx1/2=5sqrt(3)impliesJ=10sqrt3m`

For `B, J=m(v_(1)+ucostheta)`
`implies10sqrt(3)m=(v_(1)+(usqrt(3))/2)`
`implies20sqrt(3)=2v_(1)+usqrt(3)`……..i
By newton's experiment law:
`e=1=(v_(2)sintheta+v_(1))/(ucostheta)implies(usqrt3)/2=v_(2)/2+v_(1)`
`impliesusqrt(3)=5sqrt(3)+2v_(1)`.........ii
From i and ii `u=12.5m//s=1250cm//s`