A sphere of steel of mass 1 kg moving with a velocity of 12m/s ,along X-axis collides elastically with a stationary sphere after the collision is 8 m/s and is moving at an angle of `45^(@)` with X -axis , find the magnitude and direction of the second sphere after the collision .

Explanation

Momentum is conserved in collision ,
` :. M_(1)vec(v_(1)) +m_(2)vec(v_(2)) =m_(1)vec(v_(1))+m_(2)vec(v_(2))`
Taking X - components and `vec(v_(2))=0`
`m_(1)v_(1) =m_(1)v_(1)costheta_(1)+m_(2)v_(2)cos theta_(2) `
` :. 180 = 120 xx1/(sqrt(2)) +20 v_(2)cos theta_(2)`
` :. 180 = 120 xx1/(sqrt(2))+20v_(2)cos theta_(2)`
` :. 9- 3sqrt(2)=v_(2) cos theta_(2)" "......(1)`
Taking Y - components ,
` 0 =m_(1)v_(1) sin theta_(1)- m_(2) v_(2) sin theta_(2)`
` :. 0 = 15 xx 8 xx sin 45^(@) -20 v_(2) sin theta_(2)`
` :. 0 = 15 xx 8 xx sin 45^(@) - 20 v_(2) sin theta_(2)`
` :. 0 = 120 xx 1/(sqrt(2)) -20 v_(2) sin theta_(2)`
` :. 0 = 3sqrt(2) -v_(2) sin theta_(2)`
` :. 4.2425 =v_(2) sin theta_(2) " "........(2)`
Taking the ratio of equation (2) and (1) ,
` :. (v_(2)sintheta_(2))/(v_(2)costheta_(2)) =(4.2425)/(4.758)`
` :. tan theta_(2) =0.8916`
` :. tan theta_(2) =0.8916`
` :. theta_(2) =tan^(-1) (0.8916)`
` :. theta_(2) =41^(@)43`
` :. v_(2)sin (41^(@)43)=4.2425`
` :. v_(2) =(4.2425)/(sin (41^(@)43)) =(4.2425)/(0.6654)`
` :. v_(2) = 6.37 ms^(-1)`