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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?

  • $\begingroup$ the flaw is that you have used positive and negative x-axis, rather than keep the same vector direction for before and after $\endgroup$
    – Rory Alsop
    Sep 13, 2017 at 13:47
  • $\begingroup$ @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. $\endgroup$ Sep 13, 2017 at 14:02
  • $\begingroup$ 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 :-) $\endgroup$
    – Rory Alsop
    Sep 13, 2017 at 14:03
  • $\begingroup$ 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. $\endgroup$ Sep 13, 2017 at 14:12

1 Answer 1


the change in momentum should be the vector sum of initial and final momentum

That's incorrect because you must subtract the quantities in order to obtain the difference between, i.e., by how much it changed.

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}$.

  • $\begingroup$ Indeed, i forgot about the subtraction of the quantities. My folly was to view the situation as two cars having same velocity colliding head on. But the two momentum vectors do not exist simultaneously . Before striking there is only one momentum vector and after striking the same momentum vector undergoes a change in direction. $\endgroup$ Sep 13, 2017 at 14:49
  • $\begingroup$ @MohammadNayef, actually the head-on collision is more subtle than that: you obtain that the momenta should be summed by going to reference frame of one of the cars. Incidentally, if you consider the answer correct, remember to mark it accordingly. :) $\endgroup$
    – stafusa
    Sep 13, 2017 at 15:07

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