Inelastic collision (not homework)

In summary, a head-on collision between two balls with different properties (radius, mass, elasticity modulus, and Poisson ratio) and a relative velocity (v) will result in a transfer of momentum and deformation/compression of the balls. The amount of internal energy generated as a result of this collision can be estimated by considering the principles of conservation of energy and momentum, as well as the properties of the balls. Further research may be needed for a more accurate calculation.
  • #1
Count Iblis
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Note: This is not a homework problem (it is far too advanced for a suitable homework problem). So don't move it to the homework category as that would not lead to interesting replies/attempts to solve this problem. This is simply a discussion problem for everyone here to think about.

Consider two balls. Ball i has radius Ri, mass Mi, elasticity modulus Ei and Poisson ratio [tex]\nu_i[/tex]. Ball 1 moves at velocity v relative to ball 2 and has an impact parameter of zero. Both balls have an angular momentum of zero.

Compute (or estimate) the amount of internal energy of both balls that will be generated as a result of the collsion (this will eventually be dissipated, of course).
 
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  • #2


I would approach this problem by first gathering all the necessary information and defining the parameters involved. In this case, we have two balls (Ball 1 and Ball 2) with different properties (radius, mass, elasticity modulus, and Poisson ratio) and a relative velocity (v) between them. The impact parameter is also given as zero, indicating a head-on collision.

Next, I would consider the conservation of energy and momentum principles. The total kinetic energy before the collision must equal the total kinetic energy after the collision, as there are no external forces acting on the system. However, since the impact parameter is zero, there will be a transfer of momentum between the balls, resulting in a change in their velocities.

To calculate the amount of internal energy generated, we need to consider the deformation and compression of the balls upon impact. This will depend on their elasticity modulus and Poisson ratio. The higher the elasticity modulus, the more energy will be stored in the balls as they deform. The Poisson ratio also plays a role, as it determines the ratio of transverse to axial strain. A higher Poisson ratio means more energy will be dissipated as heat.

To accurately calculate the internal energy generated, we would need to know the exact properties of the balls and their material composition. However, we can make some estimates based on the given information. Assuming the balls are made of a uniform material, we can use the Young's modulus (E) to estimate the elasticity modulus (Ei = E/2(1+ \nu_i)). We can also assume a value for the Poisson ratio, such as 0.3 for most materials.

Using these estimates, we can calculate the deformation and compression of the balls upon impact and the resulting internal energy. This would require some mathematical calculations and possibly simulations, depending on the complexity of the problem.

In conclusion, as a scientist, I would approach this problem by gathering all the necessary information, defining the parameters, and using the principles of conservation of energy and momentum to calculate or estimate the amount of internal energy generated in the collision between the two balls. Further research and experimentation may be needed to accurately determine the exact amount of energy dissipated.
 
  • #3


Inelastic collisions occur when the kinetic energy of the system is not conserved, meaning that some energy is lost in the form of heat or deformation. In this scenario, the impact parameter is zero, meaning that the two balls will collide head-on. This type of collision is highly inelastic, as the two balls will deform upon impact and the kinetic energy will be dissipated in the form of heat.

To compute the amount of internal energy generated, we can use the equation for elastic potential energy:

U = (1/2)kx^2

Where U is the potential energy, k is the elastic modulus, and x is the deformation distance. Since the balls are colliding head-on, the deformation distance will be equal to the sum of their radii, Ri + R2. We can also assume that the two balls will deform in opposite directions, canceling out their angular momentum, and resulting in a net deformation distance of 2(Ri + R2).

Substituting these values into the equation, we get:

U = (1/2)Ei(2(Ri + R2))^2

To estimate the amount of internal energy generated, we can use a typical value for the elastic modulus of a rubber ball, which is around 10^6 Pa. We can also make assumptions about the masses and radii of the two balls, for example, let's say Ball 1 has a mass of 1 kg and a radius of 0.1 m, while Ball 2 has a mass of 2 kg and a radius of 0.2 m.

Substituting these values into the equation, we get:

U = (1/2)(10^6 Pa)(2(0.1 m + 0.2 m))^2 = 800 J

This means that approximately 800 joules of internal energy will be generated as a result of the collision. This energy will eventually be dissipated in the form of heat, as the two balls continue to deform and come to a stop.

It is important to note that this is just an estimate and the actual amount of internal energy generated will depend on various factors such as the materials and properties of the balls, as well as the exact conditions of the collision. But this calculation gives us an idea of the amount of energy that can be generated in an inelastic collision and the importance of considering these factors in order to accurately predict the outcome of such collisions.
 

FAQ: Inelastic collision (not homework)

What is an inelastic collision?

An inelastic collision is a type of collision in which the kinetic energy of the objects involved is not conserved. This means that after the collision, the objects will have less total kinetic energy than before the collision.

How is an inelastic collision different from an elastic collision?

In an elastic collision, the kinetic energy of the objects is conserved, meaning that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an inelastic collision, some of the kinetic energy is lost as heat or sound, resulting in a decrease in total kinetic energy.

What factors affect the amount of energy lost in an inelastic collision?

The amount of energy lost in an inelastic collision depends on the materials of the objects involved, the shape and size of the objects, and the speed and angle of the collision. Generally, the more dissimilar the objects are and the faster and more oblique the collision, the more energy will be lost.

Can an inelastic collision be perfectly inelastic?

Yes, a perfectly inelastic collision is a special case in which the objects stick together after the collision and move with a common velocity. This means that all of the kinetic energy is lost and the objects become one combined mass.

What are some real-life examples of inelastic collisions?

Some common examples of inelastic collisions include a car crash, where the kinetic energy of the cars is converted into heat and sound upon impact, and a baseball being caught by a catcher's mitt, where the kinetic energy of the ball is dissipated as the glove compresses and the ball comes to a stop. Other examples include a bowling ball hitting the pins, a tennis ball hitting a racket, and a bullet hitting a target.

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