How can momentum be conserved in an inelastic collision?

AI Thread Summary
In an inelastic collision, momentum is conserved when considering the system as a whole, including both the ball and the Earth. While the ball loses kinetic energy and its velocity decreases after rebounding, the total momentum of the ball and Earth remains constant due to Newton's third law, which states that forces between the two are equal and opposite. The discussion clarifies that the terms "elastic" and "inelastic" pertain to energy conservation, not momentum conservation. Thus, despite the ball's energy loss, momentum conservation still holds true in the system. Understanding this distinction resolves the initial confusion regarding momentum conservation in inelastic collisions.
pyman12
Messages
5
Reaction score
0
Say I drop a ball of mass 5kg from a height of 2 meters, assuming no air resistance, it will hit the ground with a velocity of 6.26m/s and collide with the ground, and rebound. From the law of conservation of momentum, 6.26 * 5 = v * 5, where v is the new velocity of the ball. But if the ball has lost kinetic energy, then through Ek = 0.5 * 5 * v^2, doesn't this mean the velocity of the ball has decreased? And as a result, it is impossible for momentum to be conserved?
 
Physics news on Phys.org
pyman12 said:
Say I drop a ball of mass 5kg from a height of 2 meters, assuming no air resistance, it will hit the ground with a velocity of 6.26m/s and collide with the ground, and rebound. From the law of conservation of momentum, 6.26 * 5 = v * 5, where v is the new velocity of the ball. But if the ball has lost kinetic energy, then through Ek = 0.5 * 5 * v^2, doesn't this mean the velocity of the ball has decreased? And as a result, it is impossible for momentum to be conserved?
The total momentum of ball & Earth is conserved. The momentum of the ball alone isn't conserved even if it rebounds with the same speed, because the direction of momentum changes.
 
A.T. said:
The total momentum of ball & Earth is conserved. The momentum of the ball alone isn't conserved even if it rebounds with the same speed, because the direction of momentum changes.

So the Earth has also gained some momentum as a result from the collision, thus balancing the momentum before and after the impact? That makes sense, thanks.
 
pyman12 said:
So the Earth has also gained some momentum as a result from the collision, thus balancing the momentum before and after the impact?
Yes, that is basically what Newtons 3rd Law is about. There are equal but opposite forces on Earth & ball. So the changes in their momentum are also equal but opposite. Hence the total momentum doesn't change.
 
I am puzzled by your question, "How can momentum be conserved in an inelastic collision". "Elastic" or "inelastic" refers to conservation of energy and has nothing to do with conservation of momentum.
 
HallsofIvy said:
I am puzzled by your question, "How can momentum be conserved in an inelastic collision". "Elastic" or "inelastic" refers to conservation of energy and has nothing to do with conservation of momentum.

That was what I was confused about, but I understand now.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top