From momentum conservation
mui^+0=mvj^+3mv′
v′=3ui^−3vj^
From kinetic energy conservation 21mu2=21mv2+21(3m)((3u)2+(3v)2)
solving v=2u
JEE Main 2020 — Physics Mechanics
A particle of mass m with an initial velocity ui^ collides perfectly elastically with a mass 3m at rest. It moves with a velocity vj^ after collision, then, v is given by
Held on 2 Sept 2020 · Verified 6 Jul 2026.
v=32u
v=3u
v=2u
v=61u
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