In a collinear collision , a particle with an initial speed `v_0` strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greather 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(2)v_0`

Answer

D

Explanation

From the law of conservation of linear momentum
`mv_0 =mv_1+mv_2`
[ m is mass of the particle ,`v_1 and v_2` are the final velocity of the first and second particle respectively ]
`or, v_1+v_2=v_0` …….(1)
According to the question ,`1/2 m(v_1^2+v_2^2)=3/2(1/2 mv_0^2)`
`or, v_1^2+v_2^2=3/2v_0^2 or, (v_1+v_2)^2=3/2 v_0^2+2v_1v_2`
`or, v_0^2-3/2 v_0^2=2v_1v_2` [from (1) ]
or, `2v_1v_2=-(v_0^2)/(2)`
Now, `(v_1-v_2)^2=v_1^2+v_2^2-2v_1v_2=3/2v_0^2+(v_0^2)/2=2v_0^2`
So, `v_1-v_2=sqrt(2)v_0`