Conservation of momentum and SHM of a spring

[boredaxel]

Homework Statement

A spring of negligible mass
and force constant k = 400 N/m is hung vertically, and a 0.200-kg
pan is suspended from its lower end. A butcher drops a 2.2-kg
steak onto the pan from a height of 0.40 m. The steak makes a
totally inelastic collision with the pan and sets the system into ver-
tical SlIM. What are (a) the speed of the pan and steak immedi-
ately after the collision;

Homework Equations

P = Mv

The Attempt at a Solution

I can't really see how linear momentum can be conserved in this case. Doesnt the weight of the steak constitue a net external force on the system? How then can momentum be conserved? I can do the rest of the problem if i clarify this doubt. Thanks

 

[Hootenanny]

At the *instant* the steak collides with the pan, there is no net external force on the pan-steak system. Therefore, one may apply conservation of momentum at the *instant* of the collision to determine the initial velocity of the system. However, as you correctly point out, both immediately before and after the collision there is a net external force acting on the pan-steak system and therefore momentum is not conserved.

Does that make sense?

 

[boredaxel]

But doesn't the weight of the steak constitue a net external force even at the instant of collision?

 

[D H]

But doesn't the weight of the steak constitue a net external force even at the instant of collision?

Yes, but it is finite. The collision occurs over a very short period of time [itex]\Delta t[/itex]. If this is sufficiently short, the collision forces [itex]\Delta \boldsymbol{p}/\Delta t[/itex] will be much, much greater in magnitude finite external forces. In the limit of an instantaneous collision, you can ignore those external forces during the collision process because the internal collision forces become infinitely large (they are impulses). The meat+pan system is effectively a closed system during the infinitesimally short collision process. While conservation of kinetic energy does not apply here (the collision is inelastic), conservation of momentum does.

 

[boredaxel]

Thanks for the clarification. I get it now.