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+//Example 8.4

+m1=0.500;//Mass of object 1 (kg)

+m2=3.50;//Mass of object 2 (kg)

+v1=4.00;//Initial velocity of object 1 (m/s)

+v2=0;//Initial velocity of object 2 (m/s)

+//Using equations of conservation of momentum and conservation of internal kinetic energy, we can derive a quadratic equation with v1_final as the variable

+//(1/2*m1+1/2*m1^2/m2)v1_final^2-(m1^2/m2*v1)v1_final-(1/2*m1*v1^2-1/2*m1^2/m2*v1^2)

+

+p=[(1/2*m1+1/2*m1^2/m2) -(m1^2/m2*v1) -(1/2*m1*v1^2-1/2*m1^2/m2*v1^2)];//Coefficients of above polynomial

+

+r=roots(p);//Finding the roots of the equation

+if r(1,1)==v1 then 

+    v1_final=r(2,1);

+else 

+    v1_final=r(1,1);

+end//Assigning a meaningful value to final velocity of object 1 (m/s)

+v2_final=m1/m2*(v1-v1_final);//Final value of object 2 from momentum equation (m/s)

+

+printf('Final velocity of object 1 = %0.2f m/s',v1_final)

+printf('\nFinal velocity of object 2 = %0.2f m/s',v2_final)

+//Openstax - College Physics

+//Download for free at http://cnx.org/content/col11406/latest