A particle of mass 5m at rest explodes into three fragments. Two fragments of mass m each travel along mutually perpendicular directions with equal speed v. The energy released is:
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A particle of mass 5m at rest explodes into three fragments. Two fragments of mass m each travel along mutually perpendicular directions with equal speed v. The energy released is:

G1 Physics · Accepted Answer

Step 1 — Principle: During a very brief explosion, external impulse is negligible; total momentum is conserved. Step 2 — Working: Momentum conservation makes the third fragment (mass 3m) have momentum magnitude √2mv and speed √2v/3. Step 3 — Working: Total KE=2(½mv²)+½(3m)(2v²/9)=mv²+mv²/3=4mv²/3. Correct answer: D — 4mv²/3.