In a collinear collision, a particle with an initial speed v0 strikes 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))/2 `
B. `(v_(0))/(sqrt(2))`
C. `(v_(0))/4`
D. `sqrt(2v_(0))`

Answer

D

Explanation

Before collision

According to conservation of linear momentum we can write the following :
`mv_(1) +mv_(2) =mv_(0) rArr v_(1) +v_(2) =v_(0) " " …(i)`
It isgiven that final kinetic energy of the system is 50 % greater than the intial kinetic energy ( may be due to release of internal energy ) , hence we cna write the following :
`1/2mv_(1)^(2) +1/2mv_(2)^(2)=3/2(1/2mv_(0)^(2))`
`rArr " "v_(1)^(2) +v_(2)^(2) = 3/2 v_(0)^(2) " " ...(ii)`
Squaring equation (i)
`v_(1)^(2) +v_(2)^(2)=3/2v_(0)^(2) " " ...(ii)`
Squaring equation (i)
`v_(1)^(2)+v_(2)^(2)+2v_(1)v_(2) =v_(0)^(2)`
Substituting from equation (ii) we get the following :
`3/2v_(0)^(2)+2v_(1)v_(2) =v_(0)^(2) rArr 4v_(1)v_(2) = - v " " ....(iii)`
Substituting from equation (ii) we get teh following :
`3/2v_(0)^(2) +2v_(1)v_(2)=v_(0)^(2) rArr 4v_(1)v_(2) = - v " " ...(iii)`
Now we can write the following equation :
`(v_(2) -v_(1))^(2) = (v_(2)+v_(1))^(2) -4v_(1)v_(2)`
Substituting from equation (ii) we get the following :
`(v_(2)-v_(1))^(2) =(v_(2)+v_(1))^(2)-4v_(1)v_(2)`
Substituting from equation (i) and (iii) ,
`(v_(2)-v_(1))^(2) = v_(0)^(2) -(-v_(0)^(2)) = 2v_(0)^(2)`
`rArr v_(2) -v_(1) = v_(0)sqrt(2)`
We can see that relative velocity of the particles after collision is `v_(0)sqrt(2)`