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` .

Explanation

Let `m_(1) = m_(2) = m `
Using the law of conservation of momentum
` m vec(v)_(1i) +0 = m vec(v)_("If") + m vec(v) _(2f) " " ….(i)`
Using law of conservation of energy
` 1/2mv_(li)^(2) = 1/2 mv_("If")^(2) +1/2 m v_("If)^(2)`
` rArr v_(li)^(2) = v_(1f)^(2) +v_(2f)^(2) " "...(ii)`
From (i)
`vec(v)_(li).vec(v)_(li) = (vec(v)_(1f)+vec(v)_(2f)) . (vec(v)_(1f) +vec(v)_(2f))`
`rArr = v_(li)^(2) = vec(v)_(1f).vec(v)_(1f)+vec(v)_(1f).vec(v)_(2f)+vec(v)_(1f) +vec(v)_(2f).vec(v)_(2f)`
`=v_(1f)^(2)+2vec(v)_(1f).vec(v)_(2f)+vec(v)_(2f)^(2)`
Using (ii) ,
`v_(li)^(2)==v_(li)^(2)+2vec(v_(1f)).vec(v_(2f))`
` rArr vec(v_(1f)).vec(v_(2f)) = 0 `
or `v_(1f).v_(2f)cos theta = 0 `
`rArr " " theta = 90^(@)`
i.e angle between their final velocities is `90^(@)`
` :. " " theta_(1)+theta_(2) = 90^(@)`
`theta_(1) = 90^(@) - theta _(2) = 90 - 37^(@) = 53^(@)`