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If we have a ball which we throw toward a wall which touches the wall and bounces back then how will you calculate the force applied by the wall on the ball because the the contact time of the ball and the wall is infinitely small so force must be infinitely large?

In a similar case If a ball is dropped from a height of 80cm (10kg ball), What amount of force will the ground apply on the ball(suppose ball comes to halt after touching the ground)?

In both the cases the duration of change of momentum is infinitely small so should the force be infinitely large ??

Note: Pls try and ignore any mistakes in the question because I am new to stack exchange

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    $\begingroup$ Saying the change in momentum occurs over an "infinitely small" time, and thus we get an "infinitely large" force is certainly an excellent approximation. If you want to actually determine a value for the force though you will need to specify a small, yet finite, time. $\endgroup$ – Aaron Stevens Oct 10 '19 at 16:28
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It is not in fact infinitely small. If you look at slow motion video of the collision, you can see that the ball and the wall are in contact for reasonable amount of time. During that time, energy of motion is stored as elastic energy od the ball (some is naturally lost) and then converted back to the kinetic energy of the ball which is now moving in the other direction.

In our idealized model of the collision, force is indeed infinite, but that information is not important if you know that momentum has to be conserved.

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  • $\begingroup$ If in an ideal situation the force is infinite than the energy required also must be infinite ? $\endgroup$ – Yashvik gupta Oct 11 '19 at 4:11
  • $\begingroup$ No, because in ideal situation this force does no work because there is no displacement on which force is acting. In real case $E = W = \int F dr$, and elastic force does work and stores it as the elastic potential energy of the ball. In ideal case $dr$ is zero and you have $\infty \times 0$ situation, but you can be sure that $W = \int F dr$ is finite. $\endgroup$ – Luka Mandić Oct 11 '19 at 7:57
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For something like a tennis ball, one or two assumptions are in order:

1) Assume some amount of "flatness" of the ball when it is most compressed against the wall.

2) Estimate the distance from the center of the ball to the wall at this point. Call this distance "d".

Note that before the collision, the distance from the center of the ball to the surface of the ball is the ball's radius ("r"). Therefore, the center of the ball is one radius away from the wall at the instant when it first touches the wall. One instant later, the center of the ball is a distance "d" from the wall.

Since the velocity of the ball just as it first touches the wall is known ($v_i$), the velocity of the ball at maximum compression against the wall is known ($v_f=0$), and the distance moved during the compression against the wall is known ("r-d"), you can calculate the average acceleration from the formula $v_f^2 = v_i^2 + 2a \Delta x$, where $\Delta x = r - d$. Knowing the average acceleration, you can calculate the force on the ball, which is also the force on the wall due to Newton's 3rd law, from the equation $F=ma$.

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  • $\begingroup$ Thank you for your answer . Although I understood the answer but still had a query in my mind , what if the object is incompressible , like an iron ball ? $\endgroup$ – Yashvik gupta Oct 11 '19 at 4:13
  • $\begingroup$ Even an iron ball is compressible, although to a MUCH less degree. Google Young's Modulus. For something like an iron ball, the amount of compression, and the time for that compression, is MUCH smaller than it is for a tennis ball, so the acceleration and force is much larger. $\endgroup$ – David White Oct 11 '19 at 4:55

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