What Determines the Velocity in a Perfectly Inelastic Collision?

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Homework Help Overview

The discussion revolves around a perfectly inelastic collision involving two identical automobiles, each traveling at 12 m/s from perpendicular directions. Participants are exploring the calculation of the final velocity of the combined mass post-collision, referencing a specific answer provided by the original poster's professor.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the conservation of momentum to determine the final velocity. Some are questioning the reasoning behind the final speed being less than the initial speeds, while others are discussing the vector components of momentum.

Discussion Status

There is an ongoing exploration of the momentum calculations and the vector components involved in the collision. Some participants have identified algebraic errors in their reasoning, while others are providing insights into the momentum properties relevant to the problem.

Contextual Notes

Participants are working within the constraints of a homework problem, indicating a need for clarity on the underlying principles of momentum and collision mechanics. There is a repeated emphasis on understanding rather than simply arriving at the answer.

JinSu
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This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.
 
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You have this

[tex]m \vec{v}_{1} + m \vec{v}_{2} = (m + m) \vec{v}_{both}[/tex]

[tex]m \vec{v}_{1} + m \vec{v}_{2} = 2m \vec{v}_{both}[/tex]

[tex]\frac{\vec{v}_{1}}{2} + \frac{\vec{v}_{2}}{2} = \vec{v}_{both}[/tex]

then you have

[tex]\frac{-12 \vec{i}}{2} + \frac{12 \vec{j}}{2} = \vec{v}_{both}[/tex]

Thus

[tex]\sqrt{6^2 + 6^2} = | \vec{v}_{both} |[/tex]
 
Last edited:
JinSu said:
This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.
The momentum of the two cars is 12 m/s North and 12 m/s West. So the momentum is ((12m)^2 + (12m)^2)^1/2 = 2mv which can be written as ((6^2 x (2m)^2 + 6^2 x (2m)^2)^1/2 = 2mv

AM
 
JinSu said:
This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.

You can use the properties of momentum to calculate this, so you can use vectors for the individual components. So just look up :biggrin:
 
Ah thanks, I see what I did wrong. Gosh, such a silly algebra mistake.
 

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