Speed of two bodies that sticks together after collision

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Homework Help Overview

The problem involves two railway trucks with different masses and speeds colliding and sticking together. The subject area pertains to the conservation of momentum in a collision scenario.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the conservation of momentum equation and question the correctness of the calculations leading to the final speed after the collision. There are attempts to clarify the initial momentum calculations and the interpretation of the results.

Discussion Status

The discussion is ongoing with participants examining different interpretations of the problem and the calculations involved. Some participants express doubt about the correctness of the provided answer, while others suggest revisiting the setup and calculations.

Contextual Notes

There is a noted discrepancy between the calculated speed and the answer provided in the reference material, leading to questions about the accuracy of the problem statement and assumptions made during the calculations.

coconut62
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Homework Statement



Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together after the collision. What is the speed of the trucks after the collision?

Homework Equations



Conservation of momentum: m1v1 + m2v2 = (m1+m2)Vtotal


The Attempt at a Solution



m(2v) + 3m(-v) = (m+3m)V

V= -4/v
∴ speed = 4/v

Answer given: 5v/4

Apparently they added the 2mv and 3mv together for the initial momentum. Is the book wrong or am I wrong?
 
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coconut62 said:

Homework Statement



Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together after the collision. What is the speed of the trucks after the collision?

Homework Equations



Conservation of momentum: m1v1 + m2v2 = (m1+m2)Vtotal


The Attempt at a Solution



m(2v) + 3m(-v) = (m+3m)V

Correct so far.

coconut62 said:
V= -4/v
∴ speed = 4/v

Wrong, do it again.

coconut62 said:
Answer given: 5v/4

Apparently they added the 2mv and 3mv together for the initial momentum. Is the book wrong or am I wrong?

The book is wrong.


ehild
 
From " m(2v) + 3m(-v) = (m+3m)V "

2mv -3mv = 4mV

-mv = 4mV

V= -v/4

Speed does not have a direction, so answer is v/4. Why wrong?
 
You wrote speed=4/v in the original post:biggrin:

ehild
 
coconut62 said:
Answer given: 5v/4.
That would be the right answer if m had speed v and 3m had speed 2v. Are you sure you quoted the question correctly?
 
Yes. I quoted it correctly.
 

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