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I am reading an elementary mechanics book and I have the following question now.

When a ball hits the wall, the higher the speed of the ball is, the stronger the ball exerts a force on the wall.

Why?

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    $\begingroup$ Think of the fact that: $F\approx m\dfrac{\Delta v}{\Delta t}$ $\endgroup$ – Syrocco Dec 7 '19 at 13:26
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    $\begingroup$ Because force is what slows down the ball from its high speed to zero. Force is what causes the deceleration, and a higher speed requires larger deceleration at impact, thus larger force. $\endgroup$ – Steeven Dec 7 '19 at 13:51
  • $\begingroup$ @Syrocco Thank you very much. $\endgroup$ – tchappy ha Dec 7 '19 at 13:57
  • $\begingroup$ @Steeven Thank you very much. $\endgroup$ – tchappy ha Dec 7 '19 at 13:57
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One way of seeing it is by using the initial state and the final state as important markers of what went on.

I'm going to assume the wall is a way to talk about an object that remains fixed in position (perhaps because it has a huge mass compared to the objects you are throwing at it, for example). Force is the rate of change in the quantity of motion over time. Between the time the ball is coming towards the wall and the time the ball is going away from it after the collision something in the quantity of motion has change. Is not the intensity of the motion that has changed (only) but the actual direction of that quantity of motion.

If there is a change in motion $\Delta \vec{p}$ over a timespan $\Delta t$ then there is a force $\vec{F} = \Delta \vec{p}/\Delta t$. This is Newton's second Law. If you throw a ball with a higher speed it means that $\Delta \vec{p}$ is larger since the wall has to decelerate the ball across many more intermediate velocities during the collision (to prevent the ball crossing the wall's position or to prevent the wall from moving out of position). Thus with a larger $\Delta \vec{p}$ you have a larger $\vec{F}$.


Mathematical approach

Your question can be mathematically sumarized as

$F_{b,w} \propto v_i$

Where $F_{b,w}$ is the force exerted on the wall by the ball and $v_i$ is the initial speed of the ball, in a time $t_i$ before it hits the wall.

We are going to proof that the statement $F_{b,w} \propto v_i$ is indeed correct in this situation.

  1. $p_i \propto v_i$ is true, because $p_i = mv_i$ is the momentum of the ball at time $t_i$ and because $m$ is independent of $v_i$ (both the ball's mass and the initial velocity can be selected independently of one another) and because we are assuming the mass of the ball remains constant.

  2. $\Delta p \propto p_i$ is also true. The change in momentum $\Delta p = p_f-p_i$ is examined between a time $t_i$ before the ball touches the wall and a time $t_f$ after the ball detaches from the wall. Our constraint is the fact that the ball can't traverse the wall, which means that $p_f \leq 0$ (if we use the convention that the positive direction of the axis is towards the wall). For example, $p_f = 0$ would be the situation where the wall stops entirely by hitting the wall and $p_f = - p_i$ would be another situation where the ball perfectly bounces off the wall. We don't care if the collision is elastic or inelastic, the fact is that for the ball not entering the region behind the wall we must state that $p_f \leq 0$. Therefore $\Delta p = p_f-p_i \leq -p_i \propto p_i$.

  3. $F_{w,b} \propto \Delta p$ is also true. Newton's II Law of motion states that $F_{w,b} = dp/dt$ is the force excerted by the wall on the ball for any instant $t$. Because before and after hitting the wall the ball was in constant rectilinear motion (because there where no forces applied on it until it touched the wall and because of Newton's I Law of motion), we can be confident that $F_{w,b} = \Delta p/\Delta t$. If we fix $t_i$ and $t_f$ then $\Delta t = t_f-t_i = constant$, which proofs that indeed $F_{w,b} \propto \Delta p$.

As long as these 3 statements hold true the certainty of $F_{w,b} \propto v_i$ is established. Because of Newton's III Law, which states that $F_{w,b} = -F_{b,w}$, we finally conclude that $F_{b,w} \propto v_i$.

These are our assumptions to make this proof work:

  • Newton's laws of motion hold true.
  • The ball can't be located behind the wall.
  • The ball doesn't loose nor gain any mass in the process.
  • There are no other forces applied on the ball than the one excerted by the wall.

I think these assumptions are implicit in the question and thus the answer is complete.

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  • $\begingroup$ but, you are assuming $\Delta p$ grows faster than $\Delta t$ $\endgroup$ – Wolphram jonny Dec 8 '19 at 1:45
  • $\begingroup$ @Wolphramjonny No, $\Delta t$ doesn't have to grow when $\Delta \vec{p}$ grows. They are totally independent. You can fix a $\Delta t$ by examining the situation on a time $t_1$ before the ball hits the wall and a time $t_2$ after the ball hits the wall. You don't really care about where $t_1$ and $t_2$ are located as long as they are before and after the collision respectively. Once you have fixed your points of examination, the force will be proportional to the change in momentum, and that change in momentum will have to be proportional to the initial momentum, thus the initial velocity. $\endgroup$ – Swike Dec 8 '19 at 11:49
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    $\begingroup$ .True, thanks!. $\endgroup$ – Wolphram jonny Dec 8 '19 at 15:29
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$$\pmb {\underline {\text {Short Intuitive Answer}}}$$ The answer that this is due to $F=\frac {\Delta p}{\Delta t}$ is true but the force you get is an average one. But if you consider the maximum force that would be imparted on the ball then also the dependence on $v_0$ is true. Here it goes:

This is related to the elastic property of the material. Since the more speed an object has the more kinetic energy it has.

Ball hitting a wall

Higher kinetic energy it has it requires more compression to store it in the form of elastic potential energy of compression. Now higher compression causes higher strain($\Delta l/l$) and hence higher stress($F/A$)(under the elastic limit of the material. This means thathigher velocity means higher force.


$$\pmb {\underline {\text {Long Computational Answer}}}$$

Since $$Y= \frac {\sigma}{\epsilon}$$ , $$\frac {E_{absobed}}{V}=\frac {1}{2}\sigma \epsilon$$

Toughness of a material

and $$E_{Kinetic}= \frac {1}{2}mv_0^2$$ Now since $E_{absobed}=E_{Kinetic}$ $$\Rightarrow \frac {1}{2}mv_0^2= V \frac {\sigma^2}{2Y}$$ Now since $\sigma = F_{max}/A$

$$\Rightarrow F_{max} =v_0A \sqrt {\frac {mY}{V}}$$


$$\pmb {\underline {\text {Suggested Readings}}}$$

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The Force on the wall is given by change in Momentum of the ball, Thus Force = Change in momentum F = mv Thus greater the speed, large the force will be.

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    $\begingroup$ Your statement $F=\Delta p$ is wrong. Instead it is $F=\Delta p/\Delta t$ as explained in Swikes' answer. $\endgroup$ – Thomas Fritsch Dec 7 '19 at 16:11

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