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A particle of mass m with an initial velocity $$u\hat{\text{i}}$$ collides perfectly elastically with a mass 3 m at rest. It moves with a velocity $$v\hat{\text j}$$after collision, then, vv is given by

a)  $$v=\sqrt{2\over3}u$$

b)  $$v= {u\over\sqrt3}$$

c)  $$v={u\over\sqrt6 }$$

d)  $$v={u\over\sqrt2}$$

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The correct answer is option d) $$v={u\over \sqrt 2 }$$

Explaination ::

$$mu\hat{i}+0=mv\hat{j}+3m\hat{v'}$$

$$\hat{v'}={u\over3}\hat{i}-{v\over3}\hat{j}$$

$${1\over2}mu^2={1\over2}mv^2+{1\over2}(3m)\Big(\big({u\over3}\big)^2+\big({v\over3}\big)^2\Big)$$   --------{ K.E.}

$$mu^2=m(v^2+({u^2\over 3}+{v^2\over3})$$

$$u^2=(v^2+{u^2\over 3}+{v^2\over3})$$

$$3u^2-3v^2=u^2+v^2$$

$$2u^2=4v^2$$

$$u^2=2v^2$$

$$v={u\over\sqrt2}$$

by Expert (10.9k points)