Two balls, of masses mA = 28 g and mB = 68 g are suspended. The lighter ball is
ID: 2193562 • Letter: T
Question
Two balls, of masses mA = 28 g and mB = 68 g are suspended. The lighter ball is pulled away to a 60Explanation / Answer
The height of A at 60° is 30cm - x, and x = 30*cos60° = 15--> height of A = 15 cm = 0,15 m Now set potential energy equal to kinetic energy: mgh = 1/2 mv^v^2 v^2 = 2gh = 2*9,81*0,15 vA = 1,7155 m/s = velocity of A before it hits B The velocities of both balls after collision are v1' = (m1v1 + m2(2v2-v1))/(m1+m2) v1' = (0,028*1,7155 + 0,068(- 1,7155))/0,096 v1' = - 0,71479 m/s = velocity of A v2' = (m2*v2 + m1(2v1 - v2))/(m1+m2) v2' = 0,028(2*1,7155)/0,096 v2' = 1 m/s = velocity of B for the height they reach use again mgh = 1/2 mv^2 h = v^2/(2g) height of A = 0,71479^2/(2*9,81) height of A = 0,026 04 m = 2,604 cm height of B = 1/(2*9,81) height of B = 0,05096 m = 5,096 cm