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+// SAMPLE PROBLEM 3/30
+clc;funcprot(0);
+// Given data
+v_1=6;// m/s
+v_2=0;// m/s
+e=0.6;// The coefficient-of-restitution
+theta=30;// degree
+
+// Calculation
+// Assume a for '
+v_1n=v_1*cosd(theta);// m/s
+v_1t=v_1*sind(theta);// m/s
+v_2n=0;// m/s
+v_2t=v_2n;// m/s
+function[X]=velocity(y)
+    X(1)=(v_1n+v_2n)-(y(1)+y(2));
+    X(2)=(e*(v_1n+v_2n))-(y(2)-y(1));
+endfunction
+y=[1,1];
+z=fsolve(y,velocity);
+v_1an=z(1);// m/s
+v_2an=z(2);// m/s
+v_1at=v_1t;// m/s
+v_2at=v_2t;// m/s
+v_1a=sqrt((v_1an)^2+(v_1at)^2);// m/s
+v_2a=sqrt((v_2an)^2+(v_2at)^2);// m/s
+thetaa=atand(v_1an/v_1at);// m/s
+// The kinetic energies just before and just after impact, with m=m1=m2,are
+T=18;// m
+T_a=13.68;// m
+E_l=((T-T_a)/T)*100;// The percentage energy loss(%)
+printf("\nThe final speeds of the particles v_1a=%1.2f m/s ,v_2a=%1.2f m/s \nThe angle which v_1a makes with the t-direction,theta=%2.2f degree \nThe percentage energy loss is %2.0f percentage.",v_1a,v_2a,thetaa,E_l);