When a ball of mass $m$ collides elastically having velocity $v$ with a wall, then it retraces itself with the same velocity. Impulse on the ball due to wall is $2mv$ and since there is no external force the same impulse must be exerted on the wall due to ball. But the impulse on the wall is 0 as it has 0 momentum both before and after the collision. So, where is the momentum going


5 Answers 5


It's not going anywhere. The wall must be exerting some force somewhere else (for instance, on the ground), which in turn introduces an external force (like friction) that keeps the wall in place when the ball hits it (relative to the Earth, we consider the case where the Earth is moving due to the force below below).

However, oftentimes we approximate physical systems. Consider the case when we assume to have an elastic collision between a ball of mass $m$ and a much larger mass $M \ \gg \ m$. We have:

$$mv_i \ = \ mv_f \ + \ Mv_{f2}$$

Which implies that:

$$v_{f2} \ = \ \frac{m(v_i \ - \ v_f)}{M}$$

We also have:

$$mv_i^2 \ = \ mv_f^2 \ + \ Mv_{f2}^2$$

We then have:

$$m(v_i^2 \ - \ v_f^2) \ = \ \frac{m^2(v_i \ - \ v_f)^2}{M} \ \Rightarrow \ \frac{v_i \ + \ v_f}{v_i \ - \ v_f} \ = \ \frac{m}{M}$$

If we take the limit:

$$\lim_{M \ \rightarrow \infty} \ \frac{m}{M} \ = \ 0$$

Which implies that as the mass gets larger, $v_i \ + \ v_f \ = \ 0$, meaning that $v_i \ = \ -v_f$, and the ball rebounds with the same velocity that it collided with.

Thus we can simply assume that the wall (and I guess the Earth, since the wall is attached to the Earth!) does not move (velocity is incredibly small) and that the momentum is conserved by simply changing the direction of the ball.

  • $\begingroup$ I really hope this derivation is correct, please let me know if I messed up! $\endgroup$ Sep 29, 2019 at 3:48

The wall has a large mass, and is attached to an Earth with a Very Large mass. That means that it can absorb momentum without any noticeable velocity.

In numbers, the mass of the Earth is about $6 \times 10^{24}$ kg. A velocity of 1 micron per year is about $0.3 \times 10^{-13}$ m/s; even that tiny speed can absorb a momentum of $10^{11}$.


So, where is the momentum going

A force is a transfer of momentum. So if you want to know where the momentum is going, just trace the forces.

The ball exerts a force in the direction of $v$ on the wall, so the ball’s momentum goes into the wall. The base of the wall exerts a force in the direction of $v$ on the earth, so the wall’s momentum goes into the earth.

The earth is so massive that although it gains momentum, the change in its velocity is too small to detect.


Elaborate in detail the conservation of momentum of an elastic ball striking a wall embedded in the earth.

The system includes both the ball and wall-earth. Choose the wall-earth as the reference frame, and assume the ball is moving right, toward the wall-earth, so the velocity will have a positive sign (by convention).

The sum of the momentums of all the elements in the system, in relationship to the reference frame, is constant/unchanged before and after the impact.

The ball, with mass $m$ and velocity $v$ has a momentum of $+mv$ before the impact, and a momentum of $-mv$ after impact. Thus, because the system had a total momentum of $+mv$ before the impact, the wall-earth must have a momentum of $2mv$ after impact.

  • Note: The above explanation, and following equation, are an approximation to reality. Still, it illustrates the principle of momentum conservation (before we complicate the problem and solve it more accurately by considering energy conservation also).

$$p = mv = -mv + 2mv$$

While the wall-earth is massive, it is not infinitely massive. Thus, when the ball hits the wall-earth and rebounds, the earth and wall both move with a finite, albeit extremely small, velocity.

Because the earth is so large, and its velocity is so small, the amount of energy lost to the earth-wall is very small. Therefore, in most practical/real-world situations, we assume the kinetic energy of the earth-wall is zero, and that the velocity of the ball after impact is $-v_{ball.initial}$. But in fact, the magnitude of the ball's final velocity, $v_{ball.final}$, is incrementally smaller than $v_{ball.initial}$, its incoming velocity.

We can compute the velocity of the earth-wall and ball after impact by using the equations of both energy and momentum conservation. Momentum quantifies the vector aspect of a moving mass and Kinetic Energy quantifies the scalar aspect of moving mass. Considering both the energy and the momentum aspect of a system gives an additional unique/non-redundant piece of information about the system. Therefore, having two equations, allows us to solve for two unknowns.

1) Using the principle of energy conservation, the kinetic energy of the ball before impact is equal to the sum of the kinetic energies of the ball and the wall-earth after impact. Note that energy is a scalar, so we add the absolute value of the energies to compute the total $KE$ after impact.

$$KE = \frac {1}{2}m_{ball}v^2_{ball.initial}= \frac {1}{2}m_{ball}v^2_{ball.final} + \frac {1}{2}m_{wall-earth}v^2_{wall-earth.final}$$

2) Using the principle of momentum conservation, we construct an equation, a relationship, to quantify the fact that the total momentum is the same before and after the impact. Note that momentum is a vector, so the sign of the velocity changes when its velocity reverses, and this must be included in the equation to represent this vital piece of relationship-information.

$$m_{ball}v_{ball.initial} = m_{ball}v_{ball.final} + m_{wall-earth}v_{wall-earth.final}$$

Summary: Every physical interaction obeys the conservation of momentum and energy. And, every physical interaction involves/includes an aspect of the transformation of both momentum and energy in the process of a collision of some type.

Note: When the mass of one of the two interacting bodies is $10$ times larger, its velocity after impact will be on the order of $100$ times smaller (because of the $v^2$ relationship to kinetic energy). Thus, when interacting with planet-sized objects, we typically simplify our computation by ignoring the energy and momentum contribution of the planet. Still, to establish a firm conceptual understanding of the physics behind all interactions, we must at least implicitly acknowledge the miniscule correction to the velocity of the smaller body required by the conservation of kinetic energy and momentum.

Of course, problems/physical-assemblies may be made arbitrarily complex. Energy may be subdivided into many compartments/types, such as: kinetic, potential, thermal, and work. Each of these complexifications of the system adds another degree of freedom. To allow quantification of the final state after an interaction, an additional equation must be constructed for each additional degree of freedom. This allows us to specify the relationship between the before-after/initial-final states and compute an expected quantitative result.


Great answers from the other posters- the 'lost' momentum is indeed conserved by a tiny recoil of the wall and the Earth.

You might also be interested in the following observations.

Loosely speaking, when objects interact momentum and energy is transferred between them in such a way that their combined momentum and their combined energy remain the same.

In a simple case of two bodies of identical mass colliding, the effect of the transfer of momentum on their speeds is the same- one body will slow down and the other will speed up by the same amount. (Incidentally, the question of which body gains momentum and which loses it depends on the frame of reference in which the collision is modelled.)

If the two bodies have differing masses, the effect of the collision on their speeds will be inversely proportional to their masses- the speed of the lighter body will change more than the speed of the heavier one. In an extreme case, such as a ball bouncing on the ground, the speed of the Earth's recoil is negligible.

Kinetic energy is also transferred in collisions. Since the change in KE is proportional to a change in velocity squared, you will see that in the case of a ball bouncing on the Earth effectively no KE can be transferred from the ball to the Earth, as the square of the Earth's recoil velocity is vanishingly small.

From those ideas you should be able to see that in a head-on collision between two bodies the transfer of momentum is minimised when they have the same mass, but the transfer of KE is maximised.

The overlooked recoil of the Earth figures in all everyday phenomena involving moving masses. When you start to walk down the street you gain momentum and the Earth in recoiling from the thrust of your step loses the same amount.


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