A ball moving with a velocity of 6 m/s strikes an identical stationary ball. After collision each ball moves at angle of 30^@ with the original line of motion. Assuming that the collision is elastic, what are the speeds of the balls after the collision?

A ball moving with a velocity of 6 m/s strikes an identical stationary

1.	A ball moving with a velocity of 6 m/s strikes an identical stationary ball. After collision each ball moves at angle of 30^@ with the original line of motion. Assuming that the collision is elastic, what are the speeds of the balls after the collision?
Answer» `(SQRT3)/2 m//s, 2sqrt(3) m//s` `3 m//s, 2sqrt(3) m//s` `2sqrt(3) m//s, 2sqrt(3) m//s` `SQRT(3) m/s, sqrt(3) m//s` Solution : According to the law of conservation of momentum along vertical and horizontal directions, we get `0 = mv_1 sin 30^@ - m v_2 sin 30^@ ` or `v_1 = v_2 "" …..(i)` `m xx u + m xx 0 = mv_1 cos 30^@ + mv_2 cos 30^@ "" .......(ii)` Putting EQN. (i) in eqn. (ii), we get , `m u = 2 mv_1 cos 30^@` or `m xx 6 = 2 xx m xx v_1 xx (sqrt3)/(2) "or " v_1= (6)/(sqrt3) = 2sqrt(3) m//s` From (i), we get, `v_1 = 2sqrt(3) m//s`.