Calculating Speed After Collision in Railroad Freight Yard

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In a railroad freight yard scenario, an empty freight car with mass m collides with a stationary loaded boxcar of mass 3.8m, coupling together post-collision. The initial speed of the empty car is 1.0 m/s, and the goal is to determine the final speed of both cars after the collision. The correct momentum conservation equation is (m1 + m2) vf = m1v1, where m1 is the mass of the empty car and m2 is the mass of the loaded boxcar. The error in the initial calculation lies in not simplifying the mass terms correctly, as "m" should not appear in the final expression. The discussion emphasizes the importance of correctly applying the conservation of momentum principle.
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In the railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded boxcar of mass 3.8m. The two cars couple together upon collision.

(a) What is the speed of the two cars after the collision?



This is the equation I have devised so far:
m1= M
m2= 3.8m
V1= 1.0 m/s
p2= 0
pf=p1 or (m1 + m2 )vf = (m1v1)

Vf = (m1)(1.0) / (m1 + 3.8)

I don't think this is correct. With the first mass being indicate as only m, I'm not sure how to compensate. Could someone please let me know what I'm doing incorrectly?
 
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evan,

You're very close! Your equation:

(m1 + m2) vf = m1v1

is exactly right. Do you know what law of physics this equation comes from?

You're mistake is just in plugging in the given values for m1, m2. Try it again.
 
That "m" should not appear in the final expression.It should be simplified through.

Daniel.
 

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