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+//Two identical balls
+//refer fig. 18.5 and 18.6 (a) and (b)
+//Before impact
+uAY=4.5 //m/sec
+uAX=7.794 //m/sec
+uBY=10.392 //m/sec
+uBX=-6 //m/sec
+vAY=uAY //m/sec
+vBY=uBY //m/sec
+//Applying principle of conservation of momentum
+//vAX+vBX=1.794
+//From defination of coefficient of restitution
+//vBX-vAX=12.415
+//Solving
+vBX=(12.415+1.794)/(2) //m/sec
+vAX=1.794-7.104 //m/sec
+vA=sqrt((5.31^2)+(4.5^2)) //m/sec
+thetaA=atand(4.5/5.31) //degree
+vB=sqrt((7.104^2)+(10.392^2)) //m/sec
+thetaB=atand(10.392/7.104) //degree
+printf("\nvA=%.3f m/sec\nthetaA=%.3f degree\nvB=%.3f m/sec\nthetaB=%.3f degree",vA,thetaA,vB,thetaB)