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Almost in every book on physics, there's an example of conservation of momentum when the ball that is moving horizontally in the air, hits some massive wall. They claim that the return speed of the ball when it bounces off is the same as it was before the hit. If there were no external forces acting on the system (or their net force was zero) that would be fine. But in this case, there is a gravitational force acting on the ball, and because there is no surface underneath it, there's no normal force and therefore it doesn't "cancel out" the gravitational force. So my question is, why they say that the momentum conserved? Do they neglect the gravitational force or what? I'm quite confused. Thanks!

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The assumption in these problems is that the collision takes place instantaneously so that gravity has no time to change the momentum of the ball during the collision.

To see why this is makes sense, let $y$ denote the vertical direction, and notice that if the collision took some small amount of time $\delta t>0$ then the change in vertical momentum of the ball would be (by integrating both sides of Newton's second law) $$ \delta p_y = \int_{t_0}^{t_0+\delta t}dt \,F(t) = F(t_0)\delta t + \mathcal O(\delta t^2) $$ so we see that as the collision time goes to zero, so does the change in momentum in the vertical direction.

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Thank you very much sir. So basically it would never be the same velocity in real life. We just assume that everything happens very quick. I got it. –  brmch8 Mar 31 '13 at 16:07
    
@brmch8 Sure thing. I have a follow-up question (that I think will also strengthen your understanding ultimately): "Doesn't the vertical wall itself exert a force on the ball in the horizontal direction? In this case, why is the momentum in the $x$ direction conserved? There is an external force from the wall on the ball in that direction." –  joshphysics Mar 31 '13 at 16:20
    
there is a change in the momentum of the ball ($mv_f-mv_i=-2mv_i$), therefore there is an impulse acting on it. Impulse, by definition, is force that is acting on object during the time $\delta t$. It means that the wall itself got that amount of momentum ($2mv_i$), but because the mass of the wall is huge compared to the ball, the velocity it gets can be neglected. Am I correct? –  brmch8 Mar 31 '13 at 16:37
    
@brmch8 Yep! For this reason, I'd say it's actually misleading to say that momentum is conserved in the collision. It's true that its magnitude remains the same, but its direction reverses as you note. –  joshphysics Mar 31 '13 at 17:05
    
you said that the magnitude is the same but the direction is opposite - why then you say that it is misleading to say that the momentum is conserved (if we neglect the gravitation in that very short period of time of the collision)? As I know, conservation of momentum implies that the net momentum is conserved. If the wall got a momentum of $2mv_i$ after the collision (where $v_i$ is the initial velocity of the ball) then the momentum conservation equation would be right: $mv_i+0=-mv_i+2mv_i$. Isn't it? Wow, I feel I got confused again, sorry. –  brmch8 Mar 31 '13 at 17:15
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The momentum conservation principle will be valid for horizontal direction. However, for the vertical direction gravity is the external force acting so it would not be valid

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