Two carts with masses of 8.0 kg and 1.8 kg, respectively, move on a friction

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SUMMARY

The problem involves two carts with masses of 8.0 kg and 1.8 kg moving on a frictionless track at velocities of 8.0 m/s and 3.5 m/s, respectively. After colliding head-on, the final speed is calculated using the equation m1v1 - m2v2 = v_final(m1 + m2). The correct final speed is determined to be 5.89 m/s, accounting for the direction of motion by defining one cart's velocity as negative. This adjustment is crucial for accurate momentum calculations.

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[SOLVED] Two carts with masses of 8.0 kg and 1.8 kg, respectively, move on a friction

Homework Statement



Two carts with masses of 8.0 kg and 1.8 kg, respectively, move on a frictionless track with velocities of 8.0 m/s and 3.5 m/s. The carts stick together after colliding head-on. Find the final speed.


Homework Equations



m1v1+m2v2 = v (m1 + m2)

v_final = (m1*v1 + m2*v2)/(m1+m2)

The Attempt at a Solution



(8*8)+(1.8*3.5)=70.3
so
70.3=(8+1.8)V
which is then 70.3=9.8V
V=7.17


but it says I'm doing it wrong. (webassign problem)
please show me how i did it wrong?? or what's wrong?
 
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You're adding the initial momenta, this is your problem. Define a coordinate system and determine which cart is moving in the positive direction, and which cart is moving in the negative direction (that is, which cart has a positive momentum, which cart has a negative momentum?).
 
Everything is perfect...except you forgot one little thing.

Since both Cars are going in opposite directions when they hit each other "head on" (one going left and one is going right), then one of them must be going in the "negative" direction. Let's define "left" as the positive direction:

m1v1-m2v2 = v (m1 + m2)

8(8) - 1.8(3.5) = Vfinal (8+1.8)
57.7=Vfinal (9.8)
Vfinal = 5.89 m/s
 

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