A ball moving on a horizontal frictionless plane hits an identical ball at rest with a velocity of `0.5m//s`. If the collision is elastic, calculate the speed imparted to the target ball, if the speed of projectile after the collision is `30cm//s`. Show that the two balls will move at right angles to eachother, after the collision.

Explanation

Let `m` be the mass of each ball. The situation is shown in figure,
As the collision is elastic, KE is conserved.
`:. (1)/(2)mu_(1)^(2)+(1)/(2)mu_(2)^(2)=(1)/(2)m upsilon_(1)^(2)+(1)/(2)m upsilon_(2)^(2)`
or `u_(1)^(2)+u_(2)^(2)=upsilon_(1)^(2)+upsilon_(2)^(2)`
`(0.5)^(2)+0=(0.3)^(2)+upsilon_(2)^(2)`
`upsilon_(2)=sqrt((0.5)^(2)-(0.3)^(2))=0.4m//s`
According to law of conservation of linear momentum along X `-` axis,
`0.5m+0=0.3m cos theta_(1)+0.4m cos theta_(2)`
or `5=3cos theta_(1)+4 cos theta_(2)`
or `3 cos theta_(1)=5 -4 cos theta_(2)` ...(i)
Again, applying law of conservation of linear momentum along `Y-` axis,
`0.3m sin theta_(1)=0.4m sin theta_(2)`
or `3 sin theta_(1)=4 sin theta_(2)` ...(ii)
Squaring (i) and (ii) and adding, we get
`9(cos ^(2)theta_(1)+sin^(2)theta_(1))=(5-4 costheta_(2))^(2)+(4 sin theta_(2))^(2)`
`9xx1=25+16 cos^(2)theta_(2)-40 cos theta_(2)+16 sin^(2)theta_(2)`
`=16(cos^(2)theta_(2)+sin^(2)theta_(2))-40cos theta_(2)+25`
`40 cos theta_(2)=16+25-9=32`
`cos theta_(2)=(32)/(49)=(4)/(5)`
From (i), `cos theta_(1)=(5)/(3)-(4)/(3)xx(4)/(5)=(3)/(5)`
`:. sin theta_(1)=sqrt(1-cos^(2)theta_(1))=4//5`
`:. sin theta_(2)=sqrt(1-cos^(2)theta_(1))=3//5`
`sin(theta_(1)+theta_(2))=sin theta_(1)cos theta_(2)+cos theta_(1)sin theta_(2)`
`=(4)/(5)xx(4)/(5)+(3)/(5)xx(3)/(5)=1`
`:. theta_(1)+theta_(2)=90^(@)`