Why do objects bounce at a 30 degree angle?

  • Thread starter Thread starter Physicsissuef
  • Start date Start date
  • Tags Tags
    Force Gravity
AI Thread Summary
Objects bounce off surfaces at approximately the same angle they strike due to the conservation of momentum and energy, with the angle of incidence equaling the angle of reflection. The coefficient of restitution, which measures the elasticity of the collision, varies between materials and determines how much energy is conserved during the impact. Different materials exhibit different bouncing behaviors based on their microstructural properties, such as rigidity and the interactions of surface electrons. Inelastic collisions, where energy is lost as heat or sound, are common in real-world scenarios, leading to reduced bounce heights. Understanding these principles is crucial for analyzing collisions and the behavior of various materials upon impact.
Physicsissuef
Messages
908
Reaction score
0
Hi! I want ask you something. Why when I throw (lets say rock) in some wall from angle of 30, the rock will bounce off, also for approximately 30? Why some object bounce off more than the other ones? Why the angle of bouncing is like that? What happens with the micro structure of the rock and the wall?
 
Physics news on Phys.org
Physicsissuef said:
Hi! I want ask you something. Why when I throw (lets say rock) in some wall from angle of 30, the rock will bounce off, also for approximately 30? Why some object bounce off more than the other ones? Why the angle of bouncing is like that? What happens with the micro structure of the rock and the wall?

The rock bounces at approximately 30 degrees because the force acting on the rock during the collision is perpendicular to the wall and if energy is conserved the velocity perpendicular to the wall will have the same magnitude before and after the collision.

As for why some collisions are more elastic then others that depends on the structure of the rock and wall and I'll leave that for someone else to answer.
 
John Creighto said:
The rock bounces at approximately 30 degrees because the force acting on the rock during the collision is perpendicular to the wall and if energy is conserved the velocity perpendicular to the wall will have the same magnitude before and after the collision.

As for why some collisions are more elastic then others that depends on the structure of the rock and wall and I'll leave that for someone else to answer.

But if the force is perpendicular to the wall, it will bounce of for 90 degrees? Or the force is not big enough to bounce it for 90 degrees?
 
No, the force si perpendicular so the rck's motion perpendicular to the wall should remain the same. Let's say you throw the rock at 60kph at an angle of 45o; then the rock is only moving 30kph relative to the wall's vertical plane (and the other 30khp is vertical or horizontal motion, or some combination of these). If the bounce off the wall is perfectly efficient, and no energy gets absorbed into the structures or lost as heat and sound, then the rock's motion away from the wall after the impact should still be 30kph, and the total velocity of the rock should still be 60. therefore, you end up with an angle of 45o, just like before the impact.
 
why some materials bounce off more than the others?

Why some materials bounce (jump) off more than the others? What happens in the microstructure?
 
interesting thought, i would like to find out as well :D
 
Physicsissuef said:
Why some materials bounce (jump) off more than the others? What happens in the microstructure?

It depends on something called the coefficient of restitution (or essentially a measure of elasticity between two surfaces). If the coefficient of restitution between two surfaces is one, then energy is completely conserved.

This is why if you drop a tennis ball it will bounce up but only to a portion of the height from where it was originally dropped. In real-world situations, such as throwing a rock against a wall, it is not a perfect elastic collision, because the coefficient of restitution is not 1 energy will not be completely conserved. The coefficient of restitution is different for all objects so as a result objects that hit each other will all react differently.
 
Feldoh said:
It depends on something called the coefficient of restitution (or essentially a measure of elasticity between two surfaces). If the coefficient of restitution between two surfaces is one, then energy is completely conserved.

This is why if you drop a tennis ball it will bounce up but only to a portion of the height from where it was originally dropped. In real-world situations, such as throwing a rock against a wall, it is not a perfect elastic collision, because the coefficient of restitution is not 1 energy will not be completely conserved. The coefficient of restitution is different for all objects so as a result objects that hit each other will all react differently.
But what happens with the microstructure?
 
Feldoh said:
It depends on something called the coefficient of restitution (or essentially a measure of elasticity between two surfaces). If the coefficient of restitution between two surfaces is one, then energy is completely conserved.

This is why if you drop a tennis ball it will bounce up but only to a portion of the height from where it was originally dropped. In real-world situations, such as throwing a rock against a wall, it is not a perfect elastic collision, because the coefficient of restitution is not 1 energy will not be completely conserved. The coefficient of restitution is different for all objects so as a result objects that hit each other will all react differently.
That's a mathematical tool and it doesn't really explain what's going on.

Imaginably, it would be the same thing as standard elastic deformation - chains of molecules flex around to distribute the force/energy, and then flex back due to electrostatic repulsion and other intermolecular forces between chains/parts of the structure. I mean, graphite has a very rigid structure as does diamond and most sort of things.

In addition, some things (i.e. flesh) compress in other ways.As for rocks, they have a very rigid structure and so they obviously don't flex much. It's all to do with how electrons on the surface interact; the electrons within the surface of the rock will repel the electrons within the surface of the wall due to their charges, so you will get it bouncing back. And conservation of momentum is the "reason" why it will bounce at a similar angle. There are all sorts of things to consider.
 
Last edited:
  • #10
I don't think determining the coefficient of restitution is easy. Here is one paper for some types of materials:

Abstract

This paper examines the theoretical regimes underlying the collision and recoil of elasto-plastic particles in low-velocity normal impacts. The coefficient of restitution is shown to be an approximate function of the ratio of the relative impact velocity to the system compressional wave speed, and the ratio of yield pressure to Young's modulus. The system compressional wave speed is further a function of the relative radii of the impacting particles, and it is predicted that the coefficient of restitution for equally sized sphere–sphere impacts is about 19% small than for sphere–plate impacts of otherwise identical particles. This result is confirmed experimentally. Theory is also developed to quantify an increase in restitution coefficient following identical impacts at the same point. Progressive plastic deformation about the impact point leads to a collision that is increasingly more elastic, and the restitution coefficient for typical particles approaches unity after about five to 10 impacts. This is also approximately confirmed experimentally.
http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TFK-4FWK6H7-6&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=8138502b41d03bdd3a7ed8e2f91568fe
 
  • #11
This paper sounds somewhat interesting:

1 Introduction
The coefficient of restitution remains to be the most controversial, and arguably, the most important constant that is used in the solution of collision problems. Following its inception by Newton (1686), and after undergoing several revisions regarding its definition, widely accepted solutions of impact
problems remain to be incomplete without a proper definition of the restitution coefficient. Investigators in the field have proposed alternative methods of solution that do not necessitate the use of this constant. Methods such as the ones based on Hertz contact theory (Maw et al, 1976 and
Marghitu and Hurmuzlu, 1996), or discrete, lumped spring-dashpot based models (Stoianovici and Hurmuzlu, 1996) are few that fall in this classification. The appeal of the restitution coefficient, however, lies in the remarkably simple way of resolving the interaction between the colliding bodies. This simplicity also enables the use of momentum based methods in the solution, which are significantly less complex than other methods of approach.

The concept of the coefficient of restitution has evolved progressively since its introduction by Newton as velocity ratio. Routh (1897) used Poisson’s Hypothesis and introduced an impulse based coefficient followed by Stronge (1990) who introduced the energetic coefficient of restitution. Hurmuzlu and Marghitu (1995) and Marghitu and Hurmuzlu (1996) conducted thorough comparisons of the outcomes using the three definitions and concluded that, pending further experimental evidence, the energetic coefficient yields the most consistent results.Stoianovici and Hurmuzlu (1996) conducted a set of experiments with slender bars falling on a massive surface. This problem was selected because it was benchmark example that was widely used in the related literature. The bars were dropped from various heights, and the pre-impact conditions
were selected such that the three definitions of the coefficient of restitution yield identical results (i.e. no tangential velocity reversals at the contact point). The study confirmed several key aspects of rigid body theory that is widely used in solving collision problems. For example, it was shown that Coulomb’s law of friction remained generally valid during the collision 2
process. In addition, for the velocity range that was considered in the experiments, the coefficient of restitution did not depend on the incidence velocity.

Yet, the most surprising result was the unusual variation (as high as 80% of the 0-1 interval) in the coefficient of restitution as the inclination of the bars were varied from vertical to horizontal. They attributed this variation to the vibrational energy that was trapped in the bar as a result of the impact event. The authors also developed a discrete model to explain and model
this behavior. They demonstrated that the outcomes predicted by the model matched the experimental results. Their study cast further questions on the practical utility of using a constant coefficient of restitution that depended on local contact properties only. The authors concluded that, in its present form, the coefficient of restitution, had a very limited applicability even when the underlying conditions of the theory were met (relatively rigid colliding bodies and low impact speeds).

The goal of the present study is to generalize the concept of coefficient of restitution such that it can be applied to a wider range of impact problems. Our goal is to amend the definition of the coefficient of restitution such that it incorporates the effect of internal vibrations in planar collisions of slender members. We seek to obtain a simple algebraic form that can be used with the standard momentum based methods. We impose the requirement that the new coefficient is consistent with the classical one when the effect of vibrations diminish.

In the ensuing article we will focus on the free collisions of bars with massive external surfaces. We will propose a new method to accurately predict the post impact velocities of these bars. inally, we will verify the practical utility of the method by comparing the computed outcomes with
the experimental results.
http://engr.smu.edu/me/syslab/papers/p5.pdf
 
Last edited by a moderator:
  • #12
Here are some of my thought on how you could lose energy. If you break a piece of it then that should be in-elastic. When you compress the air in a tennis ball it creates heat. If any of that heat escapes to the air then you lose energy as the compression is no longer adiabatic. If there is viscous flow like in the case of plastics you lose energy.
 
  • #13
Some other modes of energy transport are should waves. When the rock collides with the wall it creates compressive waves in the wall which carry energy away from the rock. When some of this energy is reflected back I presume it bounces the rock away from the wall but not all of this vibrational energy is transferred back into the rock.
 
Back
Top