Question

A `12` `kg` meteorite is traveling at `550` `ms^-1`, and then collides with the ground and comes to rest in `0.22` `m`. What is the average force acting in the object during its deceleration?

Answer 1

This is a question about impulse. The impulse is the change in momentum in a collision. The average force in this case was `8.3xx10^6` `N`.

Explanation:

The object was stationary after the collision, so its momentum at that point was `0` `kgms^-1`. Before the collision, its momentum was `p=mv=12*550=6600` `kgms^-1`.

The impulse is given by `I=Ft=6600` `kgms^-1`

We need to calculate the time taken for the object to decelerate, a two-step process:

`a=(v^2-u^2)/(2s)=(0^2-550^2)/(2*0.22)=687,500` `ms^-2`

`t=(v-u)/a=(0-550)/(687,500)=-8xx10^-4` `s`

(we can ignore the minus sign, it just reflects how long the collision began before it ended, i.e. the time from collision until the metorite stopped)

Now we can rearrange `I=Ft` to find the average force:

`F = I/t = 6600/(8xx10^-4)=8.25xx10^6` `N`

You were asked for an answer to two significant figures, which would be `8.3xx10^6` `N`.

This seems like a very large force, but it stopped a 12 kg mass traveling at `550` `ms^-1` (almost 2000 km/h) in `0.22` `m`, so that makes sense.