Calculate impulse force in inelastic collision

[Point Conception]

Homework Statement

A bag of sand with mass=55 is dropped from h=30 ft
Velocity at impact with Earth is 43.8ft/sec
The impact time ( time for sand bag to come to rest) is 1/10 sec.
What is the impulse force on the sand bag in this inelastic collision ?

Homework Equations

Impulse force = change in momentum/change in time
Change in momentum = mv2-mv1

The Attempt at a Solution

mv1 = (55lbs) (43.8 ft/sec) = 2409 lb-sec
since v2=0 then the impulse force -2409lb-sec/.1sec.=-24090lbs= 12 tons
Is this correct ?

The follow up question: How does the conservation of momentum: m1u1+m2u2 =(m1+m2)v
apply here since the Earth u2 is taken to=0 then does m1u1 = (m1+m2)v ? Or is there a conservation equating momentum being transformed to kinetic energy ?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Filip Larsen]

The method you use in the first part is correct, but if you mean to get metric ton, then the number seems a bit off.

For the second part, I would first say that you seem to use different names for the speeds here than compared to what you used in the first part, so be careful; it is usually best to use the same name for the same thing throughout a problem to reduce the risk of getting yourself confused (talking from experience here ). The question in the text is to make you consider how conservation of momentum "works" if we apply it to the whole Earth as one of the masses. For instance, it would be tempting to model (or think of) the speed of the Earth as zero both before and after collisions like described in the problem, but if the law of conservation of momentum is to be true, would that be strictly correct? How big an error would you make if you assumed the speed of the Earth was zero before and after (that is, assuming the speed of the Earth is zero before the collision calculate the speed of the two masses after collision, and think about the value you get).