Momentum/conservation of momentum problem

AI Thread Summary
The discussion revolves around a physics problem involving conservation of momentum and energy during an inelastic collision between a falling lump of putty and a suspended frame. The key confusion arises from the application of conservation of momentum, as external forces like weight are present, leading to the belief that momentum cannot be conserved. However, it's clarified that during the very short collision time, the momentum change is negligible, allowing for the assumption of conservation of momentum. Additionally, the problem highlights that kinetic energy is not conserved due to the inelastic nature of the collision, with energy lost primarily as heat. Understanding these principles is crucial for solving the problem accurately.
fab123

Homework Statement


A 0,15kg frame, when suspended from a coil spring, stretches the spring 0,070m. A 0,200kg lump of putty is dropped from rest onto the frame from a height of 30,0cm. Find the Maximum distance the frame moves downward from its initial position.

Homework Equations


I tried to solve the problem several times now but i can't seem to get the right answer, so i looked into the solution guide and there they use Conservation of momentum to the collision of the putty and the frame, however that is what gets me confused. I may be wrong but this is how i understood Conservation of momentum: Momentum of a system of particles is conserved if the net external forces Equal zero such that the total momentum of the system is constant. If that is the case, one can use that the change in momentum over any given period of time is Equal to zero. Now back to the problem:
Since the motion is happening along the y-axis there are external forces due to weight which (at least as I can tell) don't cancel out by other forces. Hence the external net force is not zero such that momentum is not conserved.
Now assuming that the solution in the solution guide is correct i have to misunderstood something which I think might be the part about the external forces not cancelling out each other.

The Attempt at a Solution


I found the spring constant k by using the initial displacement of the spring due to the frame. Since the frame is in equilibrium one can solve kx-mg=0 for the spring konstant k since all the other variables are given. Then one can use, for example, constant Acceleration Equations to find the speed of the putty when it hits the frame. Also, one can later use Conservation of kinetic energy (ignoring air ressistance) since the total work done is the sum of potential gravitational and potential elastic energy.

PS:
I have found a post on this website about the same problem however addressing a different aspect of the problem.
 
Physics news on Phys.org
This is an inelastic collision. The collision time (from the instant the putty touches the frame till they move together with the same velocity) is supposed to be very short so the external forces can not change the momentum appreciably. You have three stages: first: the putty falls, gains velocity and reaches the frame. Second: an inelastic collision in a very short time, the overall momentum conserved. Third: Motion of the frame and putty together , stretching the spring till new equilibrium .
 
  • Like
Likes fab123
Okay, I see. So momentum is changing by a very tiny amount which has such a small impact such that the change in momentum is approximately zero, am I getting that right?
 
fab123 said:
Okay, I see. So momentum is changing by a very tiny amount which has such a small impact such that the change in momentum is approximately zero, am I getting that right?
Yes.
 
fab123 said:
Also, one can later use Conservation of kinetic energy (ignoring air ressistance) since the total work done is the sum of potential gravitational and potential elastic energy.

You likely already figured this out, but the problem with using conservation of energy is that there's an inelastic collision which, by definition, loses kinetic energy. In this case it would be mainly lost in the form of sound from the impact. Something similar to this fooled a lot of people, including myself.

 
person123 said:
mainly lost in the form of sound
No, heat mostly.
 
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