Question

A meteorite with mass 1.127E+3 kg has a speed of 123. m/s when 854. km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 6.33 m. What is the average force (in N) exerted by the sand on the meteorite?

Answer 1

given values:

Vi = 123 m/s
d = 854,000 m (SI units)
g = 9.81 m/s^2
m = 1127 kg
y = 6.33 m

This is simple uniform acceleration.

Vf = SQRT { Vi^2 + [ 2 * g * d ] }
Vf = SQRT { (123 m/s)^2 + [ 2 * (9.81 m/s^2) * (854,000 m) ] }
Vf = 4,095 m/s

To find acceleration,

a = [ Vf^2 - Vi^2 ] / [2d]
a = [ (0.0 m/s)^2 - (4095 m/s)^2 ] / [ 2 * (6.33m) ]

a = 1324567 m/s^2

Average force actually comes from momentum, as Favg = Final Momentum-Initial Momentum divided by time

Favg = [ P2 - P1 ] / t

Momentum is mass * velocity

P1 = (1.127*10^3 kg) * (4095 m/s)
P1 = 4,615,065 kg-m/s

P2 = 0 kg-m/s, because it's not moving

Now we need to find the time that it took to slow the meteorite

t = [ Vf - Vi ] / a
t = [ (0.0 m/s) - (4095 m/s) ] / (1,324,567 m/s^2)

t = 3.09*E-3 s

So solve for Favg

Favg = [ P2 - P1 ] / t
Favg = [ (0.0 kg-m/s) - (4,615,065 kg-m/s) ] / (3.09*E-3 s)
Favg = [ 4,615,065 kg-m/s ] / (3.09E-3 s)
Favg = 1,493,548,544 N