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

[princessfrost]

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?

 

[G01]

is this correct?
[/b]

Looks good to me.

 

[ogb p]

Yes, your calculation is correct. The new speed of the train after the additional load is dropped will be 10.2 m/s. This is because of the conservation of momentum, where the initial momentum of the train (10,500 kg * 15 m/s) is equal to the final momentum after the load is dropped (15,500 kg * vf). Solving for vf gives a value of 10.2 m/s, which is the new speed of the train.