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+//to find speed of truck immidiately after collision and the maximum deflection of spring during impact. Moreover if k=0.5 then determine hoe the final speedsw will be affected and amount of dissipated energy 
+clc
+//given
+m1=15//tons
+u1=12//m/h
+m2=5//tons
+u2=8//m/h
+k=2//ton/in
+e1=0.5//coefficient of restitution
+printf("\n")
+//conservation of linear momentum
+v=(m1*u1+m2*u2)/(m1+m2)
+printf("velocity at the instant of  collision is %.2f mph",v)
+e=(m1*m2*(88/60)^2*(u1-u2)^2)/(2*32.2*(u1+u2))
+printf("\n")
+printf("The difference between the kinetic energy before and during the impact is %.2f ft tons\n",e)
+//energy stored in spring equals energy dissipated
+//s=(1/2)*k*x^2
+//s=e
+//since there are 4 buffer springs ,4x^2=24 inches (2 ft=24 inches)
+x=((e*12)/4)^.5
+printf("Maximum deflection of the spring is %.2f in\n",x)
+// maximum force acting between pair of buffer = stiffness of spring*deflection
+f=k*x
+printf("Maximum force acting between each buffer is %.2f tons\n",f)
+//assuming perfectly elastic collision
+//for loaded truck 
+v1=2*11-12
+//for unloaded truck 
+v2=2*11-8
+printf("Speed of loaded truck after impact %.2f mph\n",v1)
+printf("speed of unloaded truck after impact %.2f mph\n",v2)
+//if coefficient of restitution =o.5
+//for loaded truck 
+ve1=(1+.5)*11-.5*12
+//for unloaded truck 
+ve2=(1+.5)*11-.5*8
+printf("Speed of loaded truck after impact when e=0.5 %.2f mph\n",ve1)
+printf("Speed of unloaded truck after impact when e=0.5 %.2f mph\n",ve2)
+//net loss of kinetic energy=(1-e^2)*energy stored in spring
+l=(1-(e1^2))*2//ft tons
+printf("Net loss of kinetic energy is %.2f ft tons\n",l)