The correct option is C m1m2=3−2√2
Given,
Two particles of mass m1 & m2 with velocities u1 & αu1 respectively.
If the initial kinetic energies of the two particles are equal, and after the collision, m1 comes to rest.
12m1u21=12m2(αu1)2
⇒m1m2=α2 ... (1)
By the conservation of linear momentum,
m1u1+m2(αu1)=0+m2v2
Here, v2= velocity of m2 particle after collision
Using eq. (1)
m2α2u1+m2αu1=m2v2
⇒v2=u1α(α+1) ... (2)
Collision is elastic, hence the kinetic energy of the particles before and after the collision will be equal.
12m1u21+12m2(αu1)2=12m2v22
On putting v2 from equation (2)
m1u21+m2α2u21=m2u21α2(α+1)2
⇒m1m2=α4+2α3
From equation (1),
α4+2α3=α2
α2+2α−1=0
⇒α=−1+√2 (taking +ve root)
⇒m1m2=α2=(−1+√2)2
Hence, m1m2=3−2√2