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A ball moving with a velocity of 6 m/s s...

A ball moving with a velocity of 6 m/s strikes an identical stationary ball. After collisions each ball moves at an angle of `30^(@)` with the original line of motion. What are the speeds of the balls after the collision?

Text Solution

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Applying the law of conservation of momentum perpendicular to the initial line of motion:
`0 = mv _(1) sin 30^(@) - mv _(2) sin 30^(@) or v _(1) =v _(2) …(i)`
Hence speed of both will be same after collision. Now along the line of motion:
`mv = mv _(1) cos 30^(@) + mv _(2) cos 30^(@).....(ii)`
Putting equation (i) in equation (ii)
`mv = 2 mv _(1) cos 30^(@) (or) v _(1) = (v)/(sqrt3) = (6)/(sqrt3) = 2 sqrt3 m//s`
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