#### superwillis

##### Senior Member
10+ Year Member
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I was doing a problem in the TPR book today, and now I'm kind of confused about the conservation of momentum law.

So, momentum is conserved in both elastic and inelastic collisions. But, the TPR book says dropping an object to the ground is NOT a case where momentum is conserved.

But, I feel like momentum IS conserved for falling objects. If the object is falling towards the earth, isnt it true that the earth is also technically "falling" towards the object, due to the mutual gravitational attraction? which means that when the object is falling towards the earth with some speed v at some point in time, the earth is also moving toward the object with a certain speed as well (although its infinitesmally small). Considering the gigantic mass of the earth and its tiny velocity, isnt it possible that the momentum of the object is equal and opposite to the momentum of the earth (m1v1 = m2v2), so that the collision is still inelastic and momentum is conserved?

Maybe I'm just not seeing something. Could someone explain please?

#### UofT_475

##### Member
10+ Year Member
Hey, I didn't fully read your post, but I recall the same thing...

I believe TPR is suggesting that the
---TOTAL momentum of the SYSTEM is conserved
But

The momentum of that ball that fell is not conserved...it does not make up the system per se.

momentum is a vector...bouncing back up alone suggets a change in momentum

#### doc G

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Sorry for double post

#### doc G

##### RESPECT MY AUTHORITA!!
10+ Year Member
superwillis said:
I was doing a problem in the TPR book today, and now I'm kind of confused about the conservation of momentum law.

So, momentum is conserved in both elastic and inelastic collisions. But, the TPR book says dropping an object to the ground is NOT a case where momentum is conserved.

But, I feel like momentum IS conserved for falling objects. If the object is falling towards the earth, isnt it true that the earth is also technically "falling" towards the object, due to the mutual gravitational attraction? which means that when the object is falling towards the earth with some speed v at some point in time, the earth is also moving toward the object with a certain speed as well (although its infinitesmally small). Considering the gigantic mass of the earth and its tiny velocity, isnt it possible that the momentum of the object is equal and opposite to the momentum of the earth (m1v1 = m2v2), so that the collision is still inelastic and momentum is conserve
Maybe I'm just not seeing something. Could someone explain please?

This would be a reach for the MCAT...REMEMBER MOMEMTUM OF A SYSTEM IS ALWAYS CONSERVED...and move on.

##### Member
10+ Year Member
TPR is right

momentum is not always conserved in all collisions...........inelastic and elastic collisions are on the ground where g is ignored and velocity is supposedly constant......that is when momentum is conserved.......and the equation mv=mv works out perfect............when an object is dropped.......it accelerates towards the earth so the mv=mv won't work anymore............does this help? Am I right?

#### doyourealize1

##### Member
10+ Year Member
I asked this question in physics last quarter.

Your explanation is technically correct but it's not practical in real life terms. You can define Earth and the ball as a system in which the ball has an elastic collision with the Earth, but the problem with defining the system as such is that Earth's mass is so large in comparison. The tiny momentum that the ball imparts on Earth's surface hardly causes Earth to move. Also, take the case of someone driving a car into a very large brick wall. In this case, the brick wall won't be accelerating much in the other direction, and even if it did, the car and the wall can't be treated like a closed system because the wall is actually implanted into the Earth. The Earth probably absorbed a good deal of the energy absorbed by the car.

People usually define this case as an open system in which the Earth imparts an impulse (an instananeous force) upwards twice in magnitude upwards as that that the ball imparted upon the Earth downwards. That way, the ball comes back up into your hand. If you just drop a flat basketball on the ground Earth still imparts an upwards impulse on it, but it's not necessarily great enough to have the baseketball bounce back into your hand.

hope this helps. (i hope it's right)