Maximally Inelastic Collision

Maximally Inelastic Collisions between objects. claimed by dwinegarden3 (spring 2017)

The Main Idea

A collision is a brief interaction between large forces. Collisions can include such phenomena as a ball bouncing off a wall, two people running into each other on their way to class, or atoms bouncing around within a star. There are two primary types of collisions: elastic and inelastic. Elastic collisions occur when the total kinetic energy of the system is conserved while inelastic collisions occur when the objects' kinetic energies are not conserved. Maximally inelastic collisions are a subset of inelastic collisions in which the objects in the system collide and stick to form one object. The new object has a new velocity and the mass of the object equals the total mass of all the objects that have become one.

A Mathematical Model

Maximally inelastic collisions are first and foremost momentum problems, so modeling such a collision usually begins with the fundamental momentum principle, modeled in the equation:

[math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}{Δt} }[/math]

where p is the momentum of the system, F is the net force from the surroundings, Δt is the change in time for the process.

Because the momentum principle states that the change in momentum, also called impulse, within a system in which the objects stick is not changed, the time of interaction is negligible, so [math]\displaystyle{ {Δt} ≈ {0} }[/math]. By substituting Δt with zero and rewriting the equation yields:

[math]\displaystyle{ {ΔP_{system}} = {0} }[/math] where we can break the change in momentum of the system to it's initial and final components to get

[math]\displaystyle{ {P_{final}} - {P_{initial}} = {0} }[/math] which becomes

[math]\displaystyle{ {P_{final}} = {P_{initial}} }[/math].

If we substitute the mass and velocity of multiple objects into the above equation, we see that:

[math]\displaystyle{ m_1 v_1 + m_2 v_2 + ... m_n v_n = M V }[/math]

where M is sum of the masses of all collided objects and V is the final velocity of the amalgamated object

[math]\displaystyle{ V=\frac{\sum {m v}}{M} }[/math]

In maximally inelastic collisions, kinetic energy is not conserved between the initial and final states of the objects. However, total energy is conserved. This is possible because some of the energy is lost to heat and changes in shape, accounted for in the change of internal energy called ΔU. Thus, we can us the energy principle:

[math]\displaystyle{ {E} = {Q} + {W} }[/math] where E is the total energy of the system, Q is the change in thermal energy, and W is the work done.

If we input the various types of energy in for the total energy such as kinetic, potential, and internal we get

[math]\displaystyle{ {ΔE_k}+{ΔE_p}+{ΔU}= {Q} + {W} }[/math] where ΔE_k is the change in kinetic energy of the system,ΔE_p is the change in the potential energy of the system, and ΔU is the change in internal energy.

Since we assume that the objects all end up as one combined object with no significant loss of mass, we can also assume that the work done is negligible and the process is adiabatic, or a process that produces negligible heat, and the change in overall potential energy of the system is negligible. With these assumptions in place, the equation above simplifies to

[math]\displaystyle{ {ΔE_k} + {ΔU} = 0 }[/math] which we can break into the initial and final components to get: [math]\displaystyle{ {ΔU} = -({K_{final}} - ({K_{initial,1}} + {K_{initial,2}})) }[/math]

Kinetic Energy for objects that have a velocity smaller than the speed of light is defined as [math]\displaystyle{ \frac{1}{2}{mv^2} }[/math], so putting in the values for mass and velocity we get that:

[math]\displaystyle{ {ΔU} = \frac{1}{2}{(m_1 v_1 +m_2 v_2)^2}{(m_1+m_2)} - (\frac{1}{2}{m_1 (v_1)^2} + \frac{1}{2}{m_2 (v_2)^2}) }[/math]

A Computational Model

Maximally Inelastic Collisions using Glowscript

Examples

Simple

Problem: Greco wants to eat a very large flan, but Fenton and Gumbart both have small flans, so they decide to combine the flans. Fenton throws a 9 kg mass flan at a velocity of 14 m/s which strikes Gumbart's flan that weighs 5 kg mass with a velocity of -5 m/s head-on, and the two flans stick together to make an ultra mega flan for Greco to eat. At what speed will the giant flan have? Greco can only catch objects that are flying at 5 m/s, will he catch it or will it go past him? Assume negligible air resistance.

Solution: Use conservation of momentum to solve for the final velocity : [math]\displaystyle{ m_1 v_1 + m_2 v_2 = \left( m_1 + m_2 \right) v \, }[/math]. Solving for the final velocity we get the equation: [math]\displaystyle{ v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} }[/math]. Plugging in the numbers yields us a final velocity of 7.214 m/s, therefore Greco will be unable to catch the flan.

Middling

Problem: A .1 kg bullet is launched at 100 m/s a stationary block that weighs 5kg. The bullet embeds itself into the block. When someone goes to retrieve the bullet and remove it from the block, they notice the block is slightly warmer than it was before. By how much did the block's thermal energy change? The specific heat of the block and bullet is 6 J/(g*C) so by how many degrees did the block warm up by?

Solution: Using the conservation of momentum principle, we can find the final velocity of the object: [math]\displaystyle{ m_1 v_1 = \left( m_1 + m_2 \right) v \, }[/math] which is 1.961 m/s. Now we can use the conservation of Energy principle to solve for the thermal energy: [math]\displaystyle{ {ΔE_k} + {ΔU} = 0. }[/math] Then we break the change of kinetic energy into it's initial and final conditions to get : [math]\displaystyle{ {ΔU} = -({K_{final}} - ({K_{initial,bullet}})) }[/math] Plugging in the values we get : [math]\displaystyle{ {ΔU} = \frac{1}{2}{(v_{final})^2}{(m_{final})} - (\frac{1}{2}{m_{bullet} (v_{bullet})^2}) }[/math]. Entering the values from the word problem, our final answer is 490J.

To find the change in temperature, we need to use the heat equation : [math]\displaystyle{ Q = mCΔT }[/math] Plugging in the given information of specific heat, thermal energy that we solved for, and the mass of the final object, we get an increase of 16 degrees.

Difficult

Problem: Sally (5 years old) is at the playground and decides that she wants to go on the ride that spins really fast in circles. The playground ride consists of a currently stationary disk of mass M = 45 kg and radius R = 2.4 m mounted on a low-friction axle. Sally weighs about 18 kg. She is super excited to play and runs at speed of 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. She initially decided to just run around and spin the wheel, but later she decides that she wants to jump on and let the wheel keep going. What is the change of energy of the system?

Solution: This problem is difficult because it requires the use of the three big equations of classical physics 1. First we need to use the conservation of angular momentum and calculate the initial angular momentum which is [math]\displaystyle{ L = RPsinθ }[/math]. In this case, the momentum is 60 [math]\displaystyle{ m^2 kg/s }[/math]. Next we need to find the angular velocity of the ride with Sally. Using the equation: [math]\displaystyle{ L_i = Iω + RPsinθ }[/math] where we can use the relationship that [math]\displaystyle{ v = ωr }[/math] to find the more useful equation : [math]\displaystyle{ L_i = Iω + RmwR }[/math]. Since this is a wheel, the moment of inertia will be [math]\displaystyle{ .5MR^2 }[/math]. Solving for ω gives us .639 radians.

Now we can solve for the change in energies using the formula : [math]\displaystyle{ {ΔE_k} + {ΔU} = 0 }[/math]. However in this case we need to break the kinetic energies into rotational and translational and then their respective final and initial parts. So we get the rather large equation : [math]\displaystyle{ K_{rot,f} + K_{trans,f} + ΔU = K_{rot,i} + K_{trans,i} }[/math] where we can convert [math]\displaystyle{ K_{rot} }[/math] to [math]\displaystyle{ .5Iω^2 }[/math]. Using the conservation of linear momentum theorem, we can find Sally's final velocity with the equation : [math]\displaystyle{ v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} }[/math]. Plugging in values, we find that her final velocity is 1.02m/s. Now we can plug all our numbers into the previous equation and solve for the change in energy which is 373J.

Connectedness

This topic is very important in my major, Chemical Engineering. Chemical processes deal with with heat flow and transporting chemicals. A hypothetical situation would be creating polyethylene glycol. This is a compound made by combining many ethylene oxides. This process is essentially an maximally inelastic collision, because the ethylene oxides will have their own flow rate from a a previous process that creates the molecule. This ethylene oxide will enter a new chamber and combine with the previous polyethylene glycol chain that is moving around to form a new molecule that now is a combination of the two and has a new velocity. So there will be a change of kinetic energy and in the form of heat since there are new bonds being formed. To keep the process going the heat must either be cooled or heated depending the chemical engineer's desire to stop or continue formation of polyethylene glycol. The change in thermal energy is key to this process and is created by this maximally inelastic collision that occurs.

History

The idea of maximally inelastic collisions is part of the conservation of linear momentum which is implied by Newton's Laws. Most objects tend to bounce off each other creating the idea of collision. Most objects that are visibly colliding tend to lose energy through a variety of ways but theoretically objects could collide and not lose kinetic energy. This created the two types of collision inelastic and elastic. Elastic defines collisions that have no change in kinetic energy, these are usually microscopic such as Rutherford Scattering. All other objects tend to be inelastic since some energy in lost between the objects and so their respective kinetic energy changes. However, some objects can stick together and therefore combine their masses. This is a special case of inelastic collisions and tend to be a small portion of collisions that actually occur in real life, but some examples seen are when car's collide or when a sticky substance stays on the object it is throw at.

Maximally Inelastic Collision

Contents

The Main Idea

A Mathematical Model

A Computational Model

Examples

Simple

Middling

Difficult

Connectedness

History

See also

Further reading

External links

References

Navigation menu