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A block A of mass $$m$$ moving with a constant velocity $$v$$ along a smooth horizontal floor collides with another block B of mass $$2m$$ and rebounds with a velocity $$\frac{2v}{5}$$, the velocity of block B after collision is
By conservation of momentum, $$mv = m\left(-\frac{2v}{5}\right) + 2m v_B$$, which gives $$v_B = \frac{7v}{10}$$; this is not among the options and would also increase the kinetic energy, so the data as printed contains a misprint. If A rebounds with $$\frac{v}{5}$$, then $$mv = -\frac{mv}{5} + 2m v_B$$ gives $$v_B = \frac{3v}{5}$$, which is the intended option.
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