Question

Consider the collision depicted in Figure, to be between two billiard balls with equal masses `m_(1)=m_(2)`. The first ball is called the cue and the second ball is called the target. The billiard player wants to sink the target ball in a corner pocket, which is at an angle `theta_(2)=phi=37^(@)`. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain `theta_(1)=theta` .

Answer 1

If `vecu_(1)` is initial velocity of cue, and `vecu_(2)=0` is initial velocity of tartget
`vecupsilon_(1)=` final velocity of cue, and `vecupsilon_(2)=` final velocity of target, then from momentum conservation,
`mvecu_(1)=mvecupsilon_(1)+m vecupsilon_(2) or vecu_(1)=vecupsilon_(1)+vecupsilon_(2)`
`u_(1)^(2)=vecu_(1).vecu_(1)=(vecupsilon_(1)+vecupsilon_(2)).(vecupsilon_(1)+vecupsilon_(2))`
`u_(1)^(2)=upsilon_(1)^(2)+upsilon_(2)^(2)+2vecupsilon_(1).vecupsilon_(2)` ...(i)
Accordin to conservation of K.E.
`(1)/(2)m u_(1)^(2)=(1)/(2)m(upsilon_(1)^(2)+upsilon_(2)^(2)) :. u_(1)^(2)=upsilon_(1)^(2)+upsilon_(2)^(2)`
Put in (i), `upsilon_(1)^(2)+upsilon_(2)^(2)=upsilon_(1)^(2)+upsilon_(2)^(2)+2vecupsilon_(1).vecupsilon_(2)`
`:. 2vecupsilon_(1).vecupsilon_(2)=0`
`upsilon_(1)upsilon_(2)cos(theta+phi)=0`
`cos(theta_(1)+37^(@))=0=cos90^(@)`
`theta_(1)+37^(@)=90^(@)`
`theta_(1)+90^(@)-37^(@)=53^(@)`