Question

Two identical balls come into a collision with a coefficient of restitution e=0.8. Before the impact, VA = 7 m/s and V3 = 3 m/s. Their directions are as shown. Determine the magnitude of velocity of each ball after the impact. Include the following steps in your VB solution: a) Define a coordinate system appropriately. (3 pts) b) Calculate the two components of each velocity before the impact. (6 pts) c) Apply the right principle(s)/equation(s) to determine the component velocities of each ball after impact. (12 pts) B d) Determine the magnitude of the velocity of each ball after the impact. (4 pts) A 2001

Answer 1

• Collisoon of Two Identical Balls Two balls H & B are collided with each other as shown en figure below. y VB 70 x VA B 2001 (6) The velocities of (a) balls A & B before impact YB = 3 m/sec respectively. are VA 7 m/sec Coefficient of restitution = e = 0.8 Consider plane of of collision a directi- on and co.ordinate afstem (x-4) looks like figure (6) as shown above. Let (VA) is (VB) ia are initial velocities respectively A & B of balls impact. along X direction before (Yaliy & (Ve) in are the velocities of balls A & B along & direction respectively before impact :

= - 7 sino = – 2.3941 m/sec (VB) fx are velocities of A х similarly (VA) fa balls along direction after Also (VA) fy (VB) balls along Y respectively impact. are velocities of IfL A & B direction respectively after impact Applying impulse momentum principal along axis, we can write impulse momentum of balls before impact impulse momentum of balls after impact along along X axis Х axis. MA (VA) ix + MB (VB) ia MA (VA) fx + MB (VB)ga as balls are edentical, mA = mB (both balls having identical mass) (VA) ex + (VB) ia (VA) fx +(VB) fa from figure, the velocity components before impact are, Wix - VACO 520 = 7 cos200 = 6.5778 m/sec (VA) is V sinao A (Ve) in - VB Cos 70 = -3 costo = -1.02606 m/sec -V B sinto = -3 sin 700 – 2.81907 m/sec

(VA)ga + (VD) se = ( Vix + (vo) ja fa 6.5778 -1.02606 + (VA) = 5.55 174 (0 fa Applying impulse momentum principle along Y axis we get 1 (VA) iz + (VB), (Vx) +(VB) sy iL - 2.3941 -2.81 907 (VA) + (va) Fiat - fi (Vx) fx + (VB) t& 10 - 5.2132 definition of Applying a coefficient restitution along i e. along Х axis can plane of impact write e = (VB) sx - (YA) fx (VB) (VA).se (Ve) - (VA) fx 0.8 - -1.02606 - 6.5778

... (Ve)ca - (a) fa = -6.08308 fa (3) Adding equation (1) & (3), we get 2 (VB) = -0.531348 fa :.(VB) fx -0.26 56 m/sec From equation (1) (VA) fa - 5.55174 - (VB) fa = 5.55174 - (-0.2656) (VA) fa = 5.8174 m/sec The tangential force acting normal to impact plane remains constant during entire impact action. Therefore tangential force for mass A along Y axis written i's as mA (VA) it = (m) (y) (Va) (Va) ft iL NA) = - 2.3941 m/sec fL

From equation (2) (VB) ft - 5.2132 (Va) ft = -5.2132-(-2.3944 (VB) Fy -2.819 1 m/sec u :: Components of velocities after impact are (VA) fx = 5.8174 m/sec (Valty -2.3941 m/sec (VB) fx - 0.2656 m/sec (VB) Fo -2.8191 m/sec 11 velocities of balls A & B after impact are 2 Vo = S(W) - (5.8174) + (-2.3341 fy VA = 6.29 m/sec ((VB) at (VR) is V1.0.26563+ (-2.8151) VB = = 2.83158 m/sec