Per Newton's third law the force the surface exerts on the object is equal and opposite to the force the object exerts on the surface. Shouldn't there be an equal force in magnitude and opposite in direction when the object hits the surface in both cases? therefore in both case shouldn't the object bounce back to the same height? All other things being equal the COR will likely be lower in a collision with a softer surface depending on how much it deforms and the degree to which the deformation is permanent. That is the ratio of the final to initial relative velocity between two two objects after they collide, and is a number less than 1. The degree of inelasticity of a collision is reflected in the coefficient of restitution (COR) of the colliding objects. Since most softer surfaces will undergo more deformation, it stands to reason that the collision with such surfaces will be more inelastic than harder surfaces as they potentially undergo more deformation. Kinetic energy will be lost due to friction associated with the inelastic deformation of the colliding objects. However, at the macroscopic level all real collisions are inelastic. (Assumes the object itself is perfectly rigid).
If the two springs are ideal (no friction losses) an object bouncing off each of them will reach the same height because the collisions would be considered perfectly elastic. It is rather the degree of inelasticity of the collision that determines how high the object bounces.įor example, you can have a spring with a low spring constant that you might consider "softer" than a spring with a higher spring constant which is "stiffer". You need to be careful as to exactly what you mean by the terms "softer" and "harder".
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