Conservation of momentum and SHM of a spring

AI Thread Summary
In this discussion, a scenario is presented involving a spring and a pan with a steak dropped onto it, leading to an inelastic collision. The key point is that while the weight of the steak acts as an external force, during the very brief moment of collision, the system can be treated as closed, allowing for the conservation of momentum to be applied. The forces during the collision are significantly larger than the external forces, making it valid to ignore them for that instant. Although kinetic energy is not conserved due to the inelastic nature of the collision, momentum conservation holds true. This clarification helps resolve the initial confusion regarding the conservation of momentum in the context of the problem.
boredaxel
Messages
19
Reaction score
0

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
 
Physics news on Phys.org
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?
 
But doesn't the weight of the steak constitue a net external force even at the instant of collision?
 
boredaxel said:
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 \Delta t. If this is sufficiently short, the collision forces \Delta \boldsymbol{p}/\Delta t 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.
 
Thanks for the clarification. I get it now.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top