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A and B are two particles having the sam...

A and B are two particles having the same mass m. A is moving along X-axis with a speed of `10ms^(-1)` and B is at rest. After undergoing a perfectly elastic collision with B, particle A gets scattered through an angle of `30^(@)`. What is th edirection of motion of B, and the speeds of A and B, after the collision?

Text Solution

Verified by Experts

Here, `u=1m//s, theta=30^(@), phi=?, upsilon=?`
As `theta+phi=90^(@) :. =90^(@)-theta=90^(@)-30^(@)=60^(@)`
From `u=upsilon_(1)cos theta+upsilon_(2)cosphi`
`1=upsilon_(1)cos 30^(@)+upsilon_(2)cos 60^(@)=(upsilon_(1)sqrt(3))/(2)+(upsilon_(2))/(2) or upsilon_(1)sqrt(3)+upsilon_(2)=2` ...(i)
Again, as `0=upsilon_(1)sintheta-upsilon_(2)sinphi :. 0=upsilon_(1)sin30^(@)-upsilon_(2)sin60^(@)=(upsilon_(1))/(2)-(upsilon_(2)sqrt(3))/(2)` or `upsilon_(2)=(upsilon_(1))/(sqrt(3))`
Put in (i), `upsilon_(1)sqrt(3)+(upsilon_(1))/(sqrt(3))=2`
`3upsilon_(1)+upsilon_(1)=2sqrt(3)`
`upsilon_(1)=(sqrt(3))/(2)m//s`
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