Maximally Inelastic Collision: Difference between revisions
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===Middling=== | ===Middling=== | ||
'''Problem:''' | '''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 increase? The specific heat of the block and bullet is | 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 increase? The specific heat of the block and bullet is 6 J/(g*C) so by how many degrees did the block warm up by? | ||
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<math> {ΔU} = -({K_{final}} - ({K_{initial,bullet}})) </math> | <math> {ΔU} = -({K_{final}} - ({K_{initial,bullet}})) </math> | ||
Plugging in the values we get : | Plugging in the values we get : | ||
<math> {ΔU} = \frac{1}{2}{(v_{final})^2}{(m_{final})} - (\frac{1}{2}{m_{bullet} (v_{bullet})^2}) </math> | <math> {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> 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=== | ===Difficult=== | ||
Revision as of 16:39, 29 November 2015
This topic covers Maximally Inelastic Collisions. claimed by apatel404
The Main Idea
A collision is a brief interaction between large forces. This could include two objects or several depending on the situation and how they collide is important. Collisions can be either inelastic,elastic, or maximally inelastic which is a subset of inelastic. Inelastic collisions occur when the object's kinetic energies are not conserved in the final and initial state. In maximally inelastic collisions, the objects in the system collide and stick together to form one object which has a new velocity and the mass of the object is the total mass of all the objects that have now combined into one.
A Mathematical Model
Maximally Inelastic Collisions can be based off the fundamental principle of momentum:
- [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.
Using the principle of momentum, one can derive the final velocity of the object where the two initial objects have combined to become one since the interaction between the objects is brief so, [math]\displaystyle{ {Δt} ≈ {0} }[/math]. So rewriting the equation with the concept that the change in time is 0 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}} }[/math].
Now if we plug in the mass and velocity of object 1 and the mass and velocity of object 2 we see that: [math]\displaystyle{ m_1 v_1 + m_2 v_2 = \left( m_1 + m_2 \right) v \, }[/math] where v is the final velocity, which becomes
- [math]\displaystyle{ v=\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} }[/math]
Now if we apply the concept of conservation of energy:
- [math]\displaystyle{ {E} = {Q} + {W} }[/math] where E is the total energy, Q is the heat given off, 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,ΔE_p is the change in potential energy, and ΔU is the change in internal energy.
Based on the concept that the two objects have initial velocities and are going to combine into one, we can assume that the work done is negligible, the process is adiabatic,and the change in potential energy is negligible. The equation is simplified 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 increase? 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
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