Work Energy And Power Question 134

Question: A mass ’m’ moves with a velocity ‘v’ and collides inelastically with another identical mass. After collision the Ist mass moves with velocity $ \frac{v}{\sqrt{3}} $ in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision [AIEEE 2005]

Options:

A) $ \frac{2}{\sqrt{3}}v $

B) $ \frac{v}{\sqrt{3}} $

C) v

D) $ \sqrt{3}v $

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Answer:

Correct Answer: A

Solution:

Let mass A moves with velocity v and collides in elastically with mass B, which is at rest.

According to problem mass A moves in a perpendicular direction and let the mass B moves at angle q with the horizontal with velocity v.

Initial horizontal momentum of system (before collision) = mv ….(i)

Final horizontal momentum of system (after collision) $ ~~=mVcos\theta $ ….(ii)

From the conservation of horizontal linear momentum mv = mV cosq therefore v = V cosq …(iii)

Initial vertical momentum of system (before collision) is zero.

Final vertical momentum of system $ \frac{mv}{\sqrt{3}}-mV\sin \theta $

From the conservation of vertical linear momentum $ \frac{mv}{\sqrt{3}}-mV\sin \theta =0 $ therefore $ \frac{v}{\sqrt{3}}=V\sin \theta $ …(iv) By solving (iii) and (iv) $ v^{2}+\frac{v^{2}}{3}=V^{2}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta ) $

therefore $ \frac{4v^{2}}{3}=V^{2} $

therefore $ V=\frac{2}{\sqrt{3}}v $ .



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