Shame on you TH-cam for not letting me post links. I worked all day on this code, got rid of the spring model and got the match between theory and simulation.
@5:40 In the energy balance, it is necessary to introduce the norm of the speed v1 and not its component along x but we do not know a priori the angle bêta of the trajectory of m1 with the horizontal, except for the extreme case m2 >> m1 where this angle corresponds to the inclination of the sliding plane new edit according to my calculations : Nt = m1*m2*mag(g)*cos(theta)/(m2+m1*sin(theta)**2) bêta = atan((m1*mag(g)-Nt*cos(theta))/(Nt*sin(theta))) v1t = (2*mag(g)*h/(1+m1*cos(bêta)**2/m2))**0.5
Shame on you TH-cam for not letting me post links. I worked all day on this code, got rid of the spring model and got the match between theory and simulation.
@5:40 In the energy balance, it is necessary to introduce the norm of the speed v1 and not its component along x but we do not know a priori the angle bêta of the trajectory of m1 with the horizontal, except for the extreme case m2 >> m1 where this angle corresponds to the inclination of the sliding plane
new edit according to my calculations :
Nt = m1*m2*mag(g)*cos(theta)/(m2+m1*sin(theta)**2)
bêta = atan((m1*mag(g)-Nt*cos(theta))/(Nt*sin(theta)))
v1t = (2*mag(g)*h/(1+m1*cos(bêta)**2/m2))**0.5
@~5:22 by the fourth equation when h = 0 we must have v1 = v2 = 0 so the velocities (and momenta) are zero when the block reaches the ground.
So cool