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

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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?
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Looks good to me.