# Momentum doubles at wall ?

1. Nov 30, 2012

### johns123

1. The problem statement, all variables and given/known data

A ball hits a wall with momentum p1. The book says the momentum "returned" by the wall equals -2p1 .. i.e. it doubles. What does that mean? Here's why I ask:

If I throw a ball against a wall and it comes back and I catch it, I KNOW the momentum has not doubled just from the feel of the catch in my glove.

If I were to hit a golf ball against a wall, I KNOW that the ball does not come back at twice the speed. Where p = mv going to the wall, it sure doesn't come back at twice the velocity or m(2v).

The closest I can come to understanding this is ... at the wall, when the ball hits, the wall moves in the direction the ball is coming .. that is one of the mv which the ball puts into the wall. THEN, the wall returns in the - direction another mv ( elastic ), and that is the 2nd mv that makes up the -2mv. But EACH mv is separate, and only jargon connects them into 2mv ... not reality. And it is the 2nd mv returned to the ball by the wall that gives the ball an Impulse Ft = mv ........... NOT 2mv. The first mv was just stored in the wall .. and returned in an elastic response. Is that right ? And if that is right, then should I view the wall as an elastic spring which generates the inpulse AFTER it is compressed? The first mv compresses the spring. The second mv returns the "impulse" to the ball?

2. Nov 30, 2012

### superdave

The ball has mv, and then has -mv when it returns. So the wall changed the momentum by -2mv

3. Nov 30, 2012

### MrWarlock616

when you look at JUST the ball, the CHANGE in its momentum is -2mv which means that the wall gave -2mv to the ball, and the ball finally has a momentum of mv+(-2mv)=-mv

4. Nov 30, 2012

### johns123

I understand what both of you are saying, but there is no way in reality that anything has
2mv. Each of those gives mv in and mv out, but at no time are they added together, or do they occur concurrently. So, in my humble and naive opinion, the entire concept becomes misleading. Would you say the same thing about a bouncing ball moving down the road? For each bounce the Ptotal = bounce1 + bounce2 + bounce3 .... + bounce 217 + ... ????? That's the same concept .. we are saying that the entire earth is slowly filling up with bounces ?????? I guess I'm looking for the "concept" of this problem rather than the mathematics of it. Heh! I think the professor should add in the "spring" action of the ball too. That would give -4mv wouldn't it :-)

5. Nov 30, 2012

### aralbrec

There is a change in direction of the velocity; there is no doubling of the velocity.

Changes in momentum are caused by forces acting on a body for some duration.

F = m dv/dt
F dt = m dv
Impulse = ∫ F dt = m ∫ dv = m Δv = change in momentum

The momentum of the ball did change by 2mv and that was brought about by the wall exerting a force during the period of contact with the ball. The details of that force are not known and depend on how the ball and wall deformed while in contact but the impulse is known as is the total change in momentum.

When the ball initially impacted on the wall, the wall had to exert some force to bring it to 0 velocity. Then it had to exert some force on the ball to bring it back to the same velocity but this time directed to the left. The integral of the force during that time (the impulse) equalled 2mv, the change in momentum. You should note the wall experienced an equal and opposite force during that time that was absorbed by the wall and the structure supporting the wall.

In other words, saying a body experienced a change in momentum is equivalent to saying the body had some force applied to it over some period of time.

Last edited: Dec 1, 2012
6. Dec 1, 2012

### johns123

In other words, saying a body experienced a change in momentum is equivalent to saying the body had some force applied to it over some period of time.[/QUOTE]

I'm aware of all of that. I get the answers in the back of the book. I think I finally found an explanation that I can believe. I read into the next chapter, and Ohanian says at the bottom of the page that the KE is converted to PE as the ball strikes the wall. And at zero velocity when the ball stops, all the KE of the ball has been converted to PE stored in the wall .. which is then returned to the ball ( 100% if elastic ) in the form of Impulse on the ball from the wall. He sure didn't explain that at the start of the chapter, and then gives 15 problems on it before he goes into the conservation of Kinetic Energy. He still doesn't use calcs for the PE stored in the wall. Instead, he goes off into "spring problems". I guess I'm suppose to "get it", and I sort of do now. I sure won't this time next week unless I can picture the whole process of conversion of energy along with conservation of momentum. You should see this pile of equations where he inserts the momentum equations into the energy equations and gives a long list of the final calcs. In them I see the division of velocities and forces between each body involved. At the start, he did not distinguish between which object he was talking about .. or I just didn't "get it" then.