The rebound height of a mass on a trampoline

Question

When a mass(metal ball) lands on the end of a trampolne bed, the mass is displaced towards the centre of the trampoline. I have read about the energy changes and forces involved in a trampoline but can't find any information about this aspect. I'm guessing it has something to do with a spring always trying to return to equilibrium but can't work out what forces are involved.

In my Advanced Higher Physics project I am changing the mass landing on the edge of the trampoline and measuring how this affects the amount it is displaced towards the centre. Obviously the displacement increases as mass increases. My other investigation is how the rebound height is affeced when I change the distance from the centre a mass dropped. The rebound height increases as I increase the distance. Why is this?

Any suggestions about the physics behind these movements would be much appreciated and very interesting.

Answer 1

When a heavy enough mass $M$ lands near the circumference of a trampoline, the trampoline is deformed in an asymmetric way. The slope of the trampoline on the outer side of the massive object (even closer to the rim) is steeper – more vertical – than the slope on the inner side (closer to the center). That's guaranteed by the fact that the "height" of the trampoline near the rim is more constrained.

So the trampoline is effectively tilted. The normal vector to the "average cloth" surrounding the object goes up but slightly towards the center. When the massive object gets reflected, it throws the object closer to the center because it's just like a light ray reflecting from a mirror that isn't quite horizontal. A picture would probably help but I hope you will be able to draw yours.

Indeed, you are right that there is no equivalence principle because the angle $\gamma$ by which the trampoline's formerly horizontal surface is distorted in average is de facto proportional to the mass $M$, $\gamma\sim C M$, and the direction by which the reflected objects will deviated from the vertical axis is therefore $2\gamma\sim 2CM$. The constant $C$ may actually be estimated from the radius, the distance of the mass from the rim, and "stiffness" of the trampoline.