A ball of mass m moving with a speed v makes a head on collision with an identical ball at rest. The kinetic energy after collision of the balls is three fourth of the original kinetic energy. The coefficient of restitution is 

Options

A. `1/2`
B. `1/3`
C. `1/(sqrt2)`
D. `1/(sqrt3)`

Answer

C

Explanation

Applying the principle of conservation of linear momentum , we get
`mv = mv_1 + mv_2 " or " v = v_1 + v_2 " " ….(i)`
By defination of coefficient of restitution
`e = (v_2 - v_1)/(u_1 - u_2) = (v_2 - v_1)/(v - 0) " or " v_2 - v_1 = ev " " .....(ii)`
As KE after collision = `3/4 KE` before collision
`:. 1/2 m (v_1^2 + v_2^2) = 3/4 xx 1/2 mv^2 " or " v_1^2 + v_2^2 = 3/4 v^2 " ".....(iii)`
Squaring eq. (i), we get `v_1^2 + v_2^2 + 2v_1 v_2 = v^2 " ".........(iv)`
Subtracting (iiii), from equation (iv), we get
`2v_1 v_2 = 1/4 v^2" ".....(v)`
Squaring (ii), we get
`v_2^2 + v_1^2 - 2v_1 v_2 = e^2 v^2 " " ....(vi)`
Using (iii) and (v) in (vi), we get
`3/4 v^2 - 1/4 v^2 = e^2 v^2 , 1/2 v^2 = e^2 v^2 or e = 1/(sqrt2)`.