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Asked in NEET 2023 Manipur · Energy transfer in elastic collisions
Idea: in a head-on elastic collision with a target at rest, the fraction of the projectile's kinetic energy handed over is (4mM)/((m+M)²).
Write M=km: the fraction becomes (4k)/((1+k)²).
That expression is 1 at k=1 and falls away on either side, since (1+k)²-4k=(1-k)²≥0.
So the transfer is complete only when k=1.
M=m: the bullet stops dead and the block moves off with the whole of the energy.
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