# Momentum change of a ball during rebound

Consider a ball of mass 'm' moving with velocity 'v' and striking a wall, and it rebounds after striking. Assume, the collision is elastic. Now, if i have to find the force on the ball, that would be equal to rate of change of momentum of the ball. This is where, i have trouble, since momentum is a vector quantity therefore the change in momentum should be the vector sum of initial and final momentum. The initial momentum would be $mv$ along positive x-axis, and after collision it will be $mv$ along the negative x-axis. Since the two vectors are at 180° angle, therefore momentum change should be zero.

But if i calculate the momentum change as,

Final momentum - Initial momentum $= m (V_2-V_1) = m[-v - (v)] = -2mv$.

The second answer is right, but what is the flaw in the first approach?

• the flaw is that you have used positive and negative x-axis, rather than keep the same vector direction for before and after – Rory Alsop Sep 13 '17 at 13:47
• @RoryAlsop: Kindly elaborate on your statement. The ball will change its direction after striking, so how can i keep the same direction, for before and after. – Mohammad Nayef Sep 13 '17 at 14:02
• before: mv. after: -mv. change is mv- (-mv) = 2mv but this only works when you keep the same frame of reference. You can't just reverse it and expect the answer to work :-) – Rory Alsop Sep 13 '17 at 14:03
• Hmm, now i get it, i was merely adding the initial and final momentum vectorally, without regard to the change part i.e final-initial. Qualitatively speaking, i think the first approach is wrong, because the zero change in momentum suggests that no force is acting on the ball. But since the ball is undergoing a change in direction, then some force must have acted on the ball, which is reflected by the second result, change of momentum = 2mv. – Mohammad Nayef Sep 13 '17 at 14:12

That becomes clear when considering $\vec{p}=\text{const.}$, say, $\vec{p}=3\hat{\imath}$ in your units; then, according to the quoted text above, the change would be $\Delta \vec{p} = \vec{p}_i +\vec{p}_f = 3\hat{\imath}+3\hat{\imath} = 6\hat{\imath}$, which doesn't make sense, since the change of a constant quantity is zero (since constant means that it doesn't change), which is what you get from your second way of calculating: $$\Delta \vec{p} = \vec{p}_f -\vec{p}_i.$$
Then, for the constant $\vec{p}$ example, you get $\Delta \vec{p} = 3\hat{\imath}-3\hat{\imath} = \vec{0}$.