Calculating Post-Collision Velocities Using Momentum and Energy Formulas

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Homework Help Overview

The discussion revolves around calculating post-collision velocities of a car and a truck using momentum and energy conservation principles. The original poster presents a scenario involving a 450 kg car and a 550 kg truck, both traveling east, and seeks assistance in determining their velocities after an elastic collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and energy equations, noting that there are two equations with two unknowns. Some express confusion about how to solve for the final velocities, while others suggest using the relationship between relative speeds in elastic collisions.

Discussion Status

Several participants have provided insights into the equations needed to approach the problem, with some suggesting methods to solve the equations together. There is acknowledgment of the challenge posed by the unknowns, but no consensus on a specific solution method has been reached.

Contextual Notes

The original poster expresses frustration about their understanding and mentions a lack of clarity from external resources. There is an indication of prior instruction on the topic that may not have been fully grasped.

beanryu
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Please help me!
I have been thinking on this problem

A car of mass 450kg travels east at 4.5m/s, and collides with a 550kg truck infront of it traveling east at 3.7m/s. What are their velocity after the elastic collision.

I know the energy formula and the momentum formula, and I know that when the energy formula is divided by the momentum formula, it came out that:

V1-V2=V'2-V'1
the ' sign means prime/final

But I can't figure it out anyway... due to my stupidity... because V'2 and V'1 is unknown.
Would anyone help me? If so, please explain to me how you would do it using the energy and momentum formula.
 
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Conservation of momentum

[tex]m_1v_1 + m_2v_2 = m_1v'_1 +m_2v'_2[/tex]

Conservation of energy
[tex]m_1{v_1}^2 + m_2{v_2}^2 = m_1{v'_1}^2 +m_2{v'_2}^2[/tex]

2 equations, 2 unkowns.
 
But there must be answer since its my textbook problem...
 
beanryu said:
But there must be answer since its my textbook problem...
Of course there's an answer. Just solve those two equations that Astronuc provided.
 
HOW you solve it?
V'2 and V'1 is unknown...

I found a link http://physics.bu.edu/~duffy/py105.html where it has a similar problem situation under momentum and collision section in the Energy and Momentum section, but it didn't clear explain how to do it... can somobody help me!...
 
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Astronuc said:
Conservation of momentum

[tex]m_1v_1 + m_2v_2 = m_1v'_1 +m_2v'_2[/tex]

Conservation of energy
[tex]m_1{v_1}^2 + m_2{v_2}^2 = m_1{v'_1}^2 +m_2{v'_2}^2[/tex]

2 equations, 2 unkowns.

[tex]1/2{m_1}{v_1}^2 + 1/2{m_2}{v_2}^2 = 1/2{m_1}{v'_1}^2 +1/2{m_2}{v'_2}^2[/tex]
Conservation of KE has a coefficient of 1/2.

Also just a hint, In an elastic collision the relative speeds of the two objects will be the same before and after the collision. Just opposite directions. :p And mass does not matter, unless your dealing with objects of considerable mass difference.
 
beanryu said:
HOW you solve it?
V'2 and V'1 is unknown...
Sure [itex]v'_1[/itex] and [itex]v'_2[/itex] are unknown--they are what you are trying to find. Luckily everything else is known and you have two equations.

One way to solve the problem is to solve the two conservation equations together (as Astronuc was suggesting). In the momentum equation, solve for [itex]v'_2[/itex] in terms of [itex]v'_1[/itex]. Then plug that into the second equation and solve the quadratic for [itex]v'_1[/itex]. It's a bit tedious, but you can do it.

An easier way is to make use of that equation for relative speeds that you posted: [itex]v_1 - v_2 = v'_2 - v'_1[/itex]. (This is what Tjl was suggesting, but you already knew it.) Combine that equation with the conservation of momentum equation. Two equations and two unknowns again, but no quadratic this time.
 
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thanx a lot! Now I m enlightened by you people thanks a lot!

THANK YOU!
THANK YOU!
actually my teacher went over this... but i wasn't listening... and... i don't wana get a lecture... DANX a lot!
 
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