What is the New Speed of a 15,500kg Train?

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SUMMARY

The discussion focuses on calculating the new speed of a 15,500kg train after a 5,000kg load is dropped onto it. Using the principle of conservation of momentum, the initial momentum of the 10,500kg railroad car traveling at 15.0 m/s is calculated as 157,500 kg·m/s. After the load is added, the final speed (vf) is determined to be 10.2 m/s, confirming the accuracy of the calculations presented.

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A 10,500kg railroad car travels alone on a level, frictionless track with a constant speed of 15.0 m/s. An additional 5,000kg load is dropped from a tower onto the car. What will then be its new speed?



2. Homework Equations

Momentum of conservation: m1i+m2i=(m1+m2)vf





3. The Attempt at a Solution

Plugging in the numbers i get:

10,500kg(15m/s) + 0 = (10,500 kg+5000 kg)vf

157500kgm/s+0=(15,500kg)vf
10.16 m/s=vf
vf=10.2m/s

is this correct?
 
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princessfrost said:
is this correct?
[/b]

Looks good to me.
 

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