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+3 votes
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in Physics by (48.5k points)
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A particle of mass m with an initial velocity \(u\hat{i}\) collides perfectly elastically with a mass 3 m at rest. It moves with a velocity v\(\hat{j}\) after collision, then, v is given by

(1)  v = \(\frac{1}{\sqrt{6}}u\)

(2)  v = \(\frac{u}{\sqrt{3}}\)

(3)  v = \(\sqrt\frac{2}{3}u\)

(4)  v = \(\frac{u}{\sqrt{2}}\)

2 Answers

+1 vote
by (15.0k points)
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Best answer

Correct option is (4) \(v = \frac u{\sqrt 2}\)

The mass of particle 1 is m, and its initial velocity is \(u\hat i\). The mass of particle 2 is 3m, and its initial velocity is 0; the velocity of particle 1 after collision is \(v\hat j\).

Let velocity of particle -2 after collision be v'

By law of conservation of linear momentum: 

\(mu\hat i + 0 = mv\hat j + 3mv'\)

\(v' = \frac u3 \hat i - \frac v3 \hat j\)

By law of conservation of kinetic energy:

\(\frac 12 mu^2 + 0 = \frac 12mv^2 + \frac 12 (3m)v'\)

\(u^2 = v^2 + 3 \left[(\frac u3)^2 + (\frac v3)^2\right]\)

\(u^2 - \frac{u^2}3 = v^2 + \frac{v^2}3\)

\(\frac 23 u^2 = \frac 43 v^2\)

\(u^2 = 2v^2\)

\(v = \frac u{\sqrt 2}\)

+3 votes
by (43.7k points)

Answer is (4)  v = \(\frac{u}{\sqrt{2}}\)

Conservation of linear momentum

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