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 directionof the second sphere after the collision .

A sphere of steel of mass 1 kg moving with a velocity of 12m/s ,along

1.	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 directionof the second sphere after the collision .
Answer» Solution :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)` Takingthe 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)`