A shell flying with velocity `v=500m//s` burts into three identical fragements so that the kinetic energy of the system increases `eta=1.5` times. What maximum velocity can one of the fragments obtain?

Explanation

From the conservation of linear momentum of the shell just before and after its fragementation
`3vecv=vecv_1+vecv_2+vecv_3` (1)
where `vecv_1`, `vecv_2` and `vecv_3` are the velocities of its fragments.
From the energy conservation `3etav^2=v_1^2+v_2^2+v_3^2` (2)
Now `overset~vecv_i` or `vecv_(iC)=vecv_i-vecv_C=vecv_i-vecv` (3)
where `vecv_C=vecv`=velocity of the C.M. of the fragements the velocity of the shell. Obviously in the C.M. frame the linear momentum of a system is equal to zero, so
`overset~vecv_1+overset~vecv_2+overset~vecv_3=0` (4)
Using (3) and (4) in (2), we get
`3etav^2=(vecv+overset~vecv_1)^2+(vecv+overset~vecv_2)^2+(vecv-overset~vecv_1-overset~vecv_2)^2=3v^2+2overset~v_1^2+2overset~v_2^2+2overset~vecv_1*overset~vecv_2`
or, `2overset~v_1^2+2overset~v_1overset~v_2costheta+2overset~v_2^2+3(1-eta)v^2=0` (5)
If we have had used `overset~vecv_2=-overset~vecv_1-overset~vecv_3`, then Eq. 5 were contain `overset~v_3` instead of `overset~v_2` and so on.
The problem being symmetrical we can look for the maximum of any one. Obviously it will be the same for each.
For `overset~v_1` to be real in Eq. (5)
`4overset~v_2^2cos^2thetage8(2oversetv_2^2+3(1-eta)v^2)` or `6(eta-1)v^2ge(4-cos^2theta)overset~v_2^2`
So, `overset~v_2levsqrt((6(eta-1))/(4-cos^2theta))` or `overset~v_(2(max))=sqrt(2(eta-1))v`
Hence `v_(2(max))=|vecv+overset~vecv_2|_(max)=v+sqrt(2(eta-1))v=v(1+sqrt(2(eta-1)))=1km//s`
Thus owing to the symmetry
`v_(1(max))=v_(2(max))=v_(3(max))=v(1+sqrt(2(eta-1)))=1km//s`