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

Solution : Real lousy one and you can find even lousier solutions. We also made a lousy assumption that the fastest fragment should move along the CM of the system, nothing else. This is not possible to get the answer given in the book without such an assumption. Though other cases are possible and I can provide you many of them but we are mostly relying on the word "maximum:. This is better if we work out in a frame the CM of the system is at rest. I will be using ' notation for physical quantities appearing in the CM frame.

No doubt about \(\vec{p'_{1}}+\vec{p'_{2}}+\vec{p'_{3}} =0\)

Let \(\vec{p'_{2}}+\vec{p'_{3}} = \vec{P}\)

i.e. \(\vec{p'_{1}}\) is equal and opposite to \(\vec{P}\).

Thus their corresponding kinetic energies can be given as \(\frac{p_{1}'^{2}}{2m}\) and \(\frac{P^{2}}{4m}\), where \(p_{1}' = P = mv'_{1}\).

Now you know that the change in KE in Lab frame and CM frame are same for a closed system obeying momentum conservation.

We have the change in KE in Lab frame = \((3/2 -1)\frac{1}{2} 3mv^{2}\) and the change in KE in CM frame = \(\frac{p_{1}'^{2}}{2m}\) + \(\frac{P^{2}}{4m}\),where \(p_{1}' = P = mv'_{1}\). After equating, we get \(v_{1}' = v\). Thus the velocity of the fastest part in Lab frame is 2v.