In a collinear collision a particle with an intial speed `v_(0)` strike a stationary particle of the same mass. If the final total kinetic energy is 50 % greater than the original kinetic energy the magnitude of the relative velocity between the two particles after collision is .

Options

A. `(v_(0))/4`
B. `sqrt(2) v_(0)`
C. `(v_(0))/2`
D. `(v_(0))/(sqrt(2))`

Answer

B

Explanation

Initial Kinetic energy = `1/2 mv_(0)^(2)`
Final Kinetic energy = 50 % of `1/2 mv_(0)^(2) +1/2 mv_(0)^(2)`
` = 1/2 mv_(0)^(2)+1/2 mv_(0)^(2) xx1/2 `
`= 3/4mv_(0)^(2)` . ` :. 1/2 mv_(1)^(2) +1/2 mv_(2)^(2) = 3/4 mv_(0)^(2)`
` :. v_(1)^(2) +v_(2)^(2)=v_(0)`
` :. v_(1)^(2) +2v_(1)v_(2) +v_(2)^(2) =v_(0)^(2) " "....(2)`
` rArr` From ewu . (1) and (2)
`2v_(1)v_(2) =v_(0)^(2) -3/2v_(0)^(2) [ :. v_(1)^(2) +v_(2)^(2) =3/2v_(0)^(2)]`
` :. 2v_(1)v_(2) = - (v_(0)^(2))/2 `
and `(v_(1)-v_(2))^(2) = v_(1)^(2)+v_(2)^(2) -2v_(1)V_(2)`
`= 3/2 v_(0)^(2) - (-(v_(0)^(2))/2)`
` =3/2 v_(0)^(2) +(v_(0)^(2))/2`
` = 2v_(0)^(2)`
` :. v_(1) - v_(2) = sqrt(2) v_(0)` Relative velocity .