## What can be the maximum velocity

1. The problem statement, all variables and given/known data
A shell flying with velocity v = 500 m/s bursts into three identical fragments so that the kinetic energy of the system increases η = 1.5 times. What maximum velocity can one of the fragments obtain?

2. Relevant equations
Conservation of Momentum
Conservation of Energy

3. The attempt at a solution
$K.E._{initial}=\frac{1}{2} mv^{2}$

Let final velocities be $v_{1}, v_{2} and v_{3}$

$K.E._{final}=\frac{1}{2} \left[ \frac{m}{3} v_{1}^{2}+ \frac{m}{3} v_{2}^{2}+ \frac{m}{3} v_{3}^{2}\right]$

From question
$K.E._{final}= η K.E._{initial}$

Putting the values of K.E. I get a relation
$=3v^{2}η = v_{1}^{2}+ v_{2}^{2}+ v_{3}^{2}$

Also using conservation of momentum

$mv = \frac{m}{3}\left(v_{1}+v_{2}+v_{3}\right)$
which simplifies to
$3v = (v_{1}+v_{2}+v_{3})$

But now I'm stuck here. What to do next to find the maximum velocity?

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 The maximum velocity will occur if two pieces fly off in a direction opposite the third piece. When something explodes in flight, the center of mass follows the same path as before the explosion. So assuming the projo is going to the right, the maximum velocity will occur if two pieces are sent to the left relative to the center of mass (one upward and to the left while the other is downward and to the left). The horizontal component of velocity of the two pieces going to the left are equal. The vertical components are equal and opposite. Conservation of momentum then imparts the maximum velocity to the remaining piece that moves horizontally to the right. Set this up and you'll arrive at two equations and two unknowns. The energy equation will be quadratic while the momentum equation will be linear.
 Thanks. I got my answer correct.