9
$\begingroup$

If an object hits a soft surface it will bounce lower compared to the object hitting a hard surface, isn't the impulse in the first case equal to the impulse in the second case, so why does the object bounce lower when it hits a softer surface?

$\endgroup$
  • 4
    $\begingroup$ I do not think you can solve this problem without taking into account the dissipation of energy (and momentum). $\endgroup$ – NDewolf Mar 9 at 16:00
  • $\begingroup$ What do you mean by “soft”? $\endgroup$ – Bob D Mar 9 at 16:02
  • 4
    $\begingroup$ You should be a little more precise about what you mean by "soft". Drop a bowling ball on a foam mattress and it will barely rebound if at all. Drop it on a trampoline, and it will bounce back quite a bit. But, is a trampoline "hard"??? $\endgroup$ – WRSomsky Mar 9 at 16:04
  • $\begingroup$ @WRSomsky i meant a softer surface than the hard surface. $\endgroup$ – Zheer Mar 9 at 16:17
  • 12
    $\begingroup$ If you would change the word "soft" to "inelastic" then your question would practically answer itself. $\endgroup$ – Solomon Slow Mar 9 at 16:26
35
$\begingroup$

You need to be careful as to exactly what you mean by the terms "softer" and "harder". It is rather the degree of inelasticity of the collision that determines how high the object bounces.

For example, you can have a spring with a low spring constant that you might consider "softer" than a spring with a higher spring constant which is "stiffer". If the two springs are ideal (no friction losses) an object bouncing off each of them will reach the same height because the collisions would be considered perfectly elastic. (Assumes the object itself is perfectly rigid).

However, at the macroscopic level all real collisions are inelastic. Kinetic energy will be lost due to friction associated with the inelastic deformation of the colliding objects. Since most softer surfaces will undergo more deformation, it stands to reason that the collision with such surfaces will be more inelastic than harder surfaces as they potentially undergo more deformation.

The degree of inelasticity of a collision is reflected in the coefficient of restitution (COR) of the colliding objects. That is the ratio of the final to initial relative velocity between two two objects after they collide, and is a number less than 1. All other things being equal the COR will likely be lower in a collision with a softer surface depending on how much it deforms and the degree to which the deformation is permanent.

Shouldn't there be an equal force in magnitude and opposite in direction when the object hits the surface in both cases? therefore in both case shouldn't the object bounce back to the same height?

Per Newton's third law the force the surface exerts on the object is equal and opposite to the force the object exerts on the surface. But the magnitude of the equal and opposite forces will not be the same when the object initially contacts the surface as when it comes off the surface after penetrating the surface. That's because energy is lost as heat when the object penetrates the surface. That means there is less kinetic energy of the object when it comes off the surface then when it initially hit the surface. Less kinetic energy means less velocity and a lower rebound height.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Shouldn't there be an equal force in magnitude and opposite in direction when the object hits the surface in both cases? therefore in both case shouldn't the object bounce back to the same height? $\endgroup$ – Zheer Mar 9 at 17:22
  • $\begingroup$ @Zheer I've updated my answer to respond to your follow up question. Hope it helps. $\endgroup$ – Bob D Mar 9 at 18:06
  • $\begingroup$ i know that energy is lost as heat when the object penetrates the surface but since in both cases the ball will have the same impulse, what happens to the force? , Is some of the force really lost or something like that?if the force isn't lost how can the ball bounce lower? $\endgroup$ – Zheer Mar 9 at 18:15
  • 1
    $\begingroup$ @Zheer The action and reaction law tells you, that forces acting on the object and on the surface during the collision are the same size. It does not tell you how long or how big these forces are supposed to be or how big impulse they need to exert. There is no conservation of force, or impulse. $\endgroup$ – Umaxo Mar 9 at 19:15
  • 3
    $\begingroup$ @Zheer Exactly what I have been trying to tell you. The force does work on the wall dissipating the energy behind the force as heat in the wall. When an object bounces off a surface unless the collision is perfectly it loses kinetic energy by vibrating the molecules of the surface material. With less kinetic energy on the rebound, its velocity is less. I'm sorry if you can't get it, but I have no more time to devote to this. $\endgroup$ – Bob D Mar 9 at 21:25
7
$\begingroup$

When an object bounces off of something, the bounce is not instantaneous, and in fact goes through a sequence of changes.

First, the object slows down to a stop, and both the object and the surface are deformed. If they are both hard, then the deformation will be very small, but it will still exist.

Second, both the object and the surface reverse their deformations, speeding the object back up in the reverse direction.

The difference that makes a softer surface produce lower bounces is that soft surfaces stop that second stage early, leaving the surface still partially deformed. This reduces the time of contact, and thus the impulse of the bounce. The energy lost to this is converted primarily to heat.

More spring-like surfaces also do it more gradually, leaving some of the un-deformation still in progress when the object loses contact with the surface. This reduces the force during the time of contact, and thus the impulse of the bounce. The energy lost to this is converted to more macro-scale vibration, which may even be readily visible.

Illustrating with an example:

A ball has 1000 kg * m/s momentum when it first touches a soft surface. The ball and surface each exert 100 N force on each other as long as they are in contact.

The first stage, the "impact", lasts 10 seconds. During the impact, the ball slows to a stop and deforms slightly, and the surface deforms a lot. Both the ball and surface cause an impulse of 1000 N * s on each other, during this stage by itself.

The second stage, the "rebound", lasts only 5 seconds. During the rebound, the ball accelerates away and un-deforms, and the surface un-deforms only partially. Both the ball and surface cause an impulse of 500 N * s on each other, during this stage by itself.

The end result is that the ball bounces off with only half the speed it started with. For a hard surface in the same scenario, the rebound would instead last almost as long as the impact, about 10 seconds, the surface would un-deform all the way, and the ball would bounce with almost as much speed as it started with.

P.S. Notes on concepts and units:

Momentum: The product of mass and velocity. It has unit kilograms * meters / second, or kg * m/s. This unit is equivalent to N * s.

Force: How hard you're pushing on something. Its unit is Newtons, or N.

Impulse: The product of force and time, and also the change in momentum. It has unit Newtons * seconds, or N * s. This unit is equivalent to kg * m/s.

Newton's third law applies to force, and indirectly to impulse. It does not apply to momentum.

In typical everyday collisions, gravity applied over some previous time span may be the source of some of the momentum involved, but is inconsequential for the collision itself because it is utterly dwarfed by the forces exerted by the objects on each other. The collision between a ball and a hard floor takes something like a millisecond to finish, but produces an impulse large enough to reverse - not just stop - momentum that took around 500 milliseconds for gravity to build up.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does impulse have to do with the bounce, i mean if i apply 100N in 10seconds to the ball in the first case and 50N in 20 seconds it is still the same total force isn't it? $\endgroup$ – Zheer Mar 10 at 7:59
  • $\begingroup$ @Zheer With a soft surface and a collision where the first stage ("impact") involves 100N for 10 seconds, the second stage ("rebound") might involve 100N for 5 seconds. The object came in with 1000 N * s momentum, lost all of it in the impact, and regained only 500 N * s momentum (in the opposite direction) in the rebound. $\endgroup$ – Douglas Mar 10 at 8:05
  • $\begingroup$ that makes sense but doesn't the surface push back with more force than what the object had, i mean if the ball pushes the surface with 1000N and it takes 1000N for the ball to stop then the surface applies 500N more to push back the ball, doesn't that violate newton's third law since the surface pushed back with 1500N and not 1000N? $\endgroup$ – Zheer Mar 10 at 8:13
  • $\begingroup$ @Zheer Er, no? The ball is pushing back on the surface during the rebound too. Both the ball and surface get a total of 1500 N * s change in momentum from the collision. If the surface didn't apply more total than it takes to stop the ball, then the result would be the ball just stopping with no bounce at all. $\endgroup$ – Douglas Mar 10 at 8:15
  • 1
    $\begingroup$ @Zheer I added an example and some notes, so take another look. $\endgroup$ – Douglas Mar 10 at 9:35
4
$\begingroup$

Let's break down all the forces involved. Note: I've surely missed a bunch, and this got even more complicated than I was expecting when I started writing it.

Stage 1: Ball is in your hand before you drop it.

  • Force 1: Gravity on the ball (Opposite force is the ball pulling on the earth, but this is inconsequential to this problem)
  • Force 2: Depending on how you're holding the ball, either normal force or gravity. The sum of both which will equal the force of gravity on the ball (if they didn't the ball would move)

Stage 2: Ball is let go, and begins falling.

  • Force 1: Gravity on the ball
  • Force 2: Air resistance on the ball
  • Force 3: Force applied by the ball on the surrounding air to move it out of the way so that it can take that new place. Force 3 == Force 2
  • Net force down as acceleration is observed

Stage 3: ball begins contact with ground.

  • Force 1: Gravity on ball.
  • Force 2: Normal force applied by ground to ball.
  • Force 3: Force applied by ball to ground. Force 3 == Force 2
  • Force 4: Anti-compression force. Couldn't think of a better name here, but as you compress a ball, it has a tendency to return to its ball shape. This force gets higher and higher as you compress the ball. This happens with the ground as well, but at a different "rate". I'll get into what I mean by rate in a bit
  • Force 5: Going to group a lot of things here, but the ball is pushing against the ground, which is rippling vibrations through the ground and air.
  • Net force: Well, very very very small amount of net force down at part of the ball that has impacted the ground. The top of the ball has a much higher net force down at this point.

Stage 4: Compression Equilibrium. This is the point at which the ball stops accelerating downward. This is a very complicated system to pin down, but this is the point where the anti-compression force is equal to the normal force applied to the ground. I.E the moment before the ball starts bouncing back

  • Force 1: Gravity
  • Force 2: Anti-compression force. The ball is now returning itself to a ball. As a consequence of this the ball will apply a force to the ground.
  • Force 3: The ground also pushes back against this anti-compression force to apply an upward force to the ball
  • Force 4: Air resistance above the ball fighting against the anti-compression.
  • Force 5: The ground may or may not be bouncing back at this point. It depends on the hardness/softness/elasticity of the ground relative to the ball. This will greatly change how much of the normal force is applied to the ball at this step. This is what I was referring to by a different rate. There's no guarantee that the bounce back from the ground and the anti-compression of the ball will be in sync.
  • Force 6: More vibrations through the ground
  • Net force: Up (or at least up the moment after equilibrium. There is a moment of time where the net force on the ball is actually 0!)

Stage 6: Ball leaving the ground

  • Force 1: Gravity
  • Force 2: Air resistance, the ball is no longer impacted by the ground anymore now

As you can see from these forces, a lot of force has gone to places that isn't propelling the ball back to its original place.

But regardless of that, look how complicated this analysis is. There's too many little interactions between forces to make good sense of what's happening. This is precisely why we use energy to reason about these things. Because we have a conservation of energy law that greatly simplifies all calculation, and we have a solid understanding of kinetic energy, and heat. Our understanding of elastic collisions is basically just experimentation, then building theories around the system as a whole, rather than trying to derive it from more primitive/fundamental physics.

The most important thing here is that there is no such equivalent with forces. The closest thing to a conservation of force is that forces must have an equal an opposite force at all times. But do not misunderstand that, that tells you nothing in this example, as you can see from all the forces listed, they all have opposite forces, but the net effect is still lost energy.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, but what does softness of material mean? Is it the opposite of hardness? $\endgroup$ – Zheer Mar 10 at 21:00
  • 1
    $\begingroup$ That's a good question, and it's fairly loosely defined in this post. But what matters in the example is how much the material deforms, and how quickly it snaps back to original shape $\endgroup$ – Cruncher Mar 10 at 23:26
2
$\begingroup$

An inelastic collision is an example of friction: One of the objects, in your example the surface, will plastically deform while its innards are rubbing on each other, producing heat. If there is no elasticity at all, say with a heap of soft sand, no energy other than heat is stored in the ground during the impact. If the ball hitting it is sufficiently hard, e.g. made of steel, no significant amount of energy is stored in it on impact, either. Because friction only exerts a force on the moving objects as long as they move, the steel ball will simply form a dent in the sand (or lump of clay, or whatever) until its momentum and energy is used up resp. transferred, and then sit there, with no repellent force acting upon it. That's why it does not jump back up. Depending on the mass of the second object some or most of the original kinetic energy has been converted into heat, while the overall momentum of the ball-surface system is of course unchanged: The collision consisted of only inner forces which do not change the momentum of the overall system.

Note that when the collision partner is the Earth, like when we bounce a Super Ball off the tarmac or throw it into a heap of sand, we must be careful with momentum considerations: Because Earth's mass is so huge. It is not obvious that momentum is conserved during the fall or during the bounce because the ball accelerates and even completely inverts direction seemingly violating the law that momentum be conserved.

But it is important to understand that the attraction between Super Ball and Earth, as well as all forces during a collision, are entirely mutual. The Earth starts moving towards the ball when the ball falls; both experience exactly the same change of momentum in exactly the opposite directions (which must be so because the attraction is an inner force and cannot change the overall momentum...). It's just that the Earth weighs 6 million billion billion kilograms as opposed the the Super Ball's 0.02, so that when the ball moves with a speed of 5 m/s towards the Earth before impact, the Earth moves with only 1/60 million trillion trillionth m/s towards the ball which is only noticeable if you look really close. Joke aside, this effect is measurable or even pronounced with moons orbiting planets, or planets orbiting the sun: Being equally attracted by the orbiting bodies, the central body "wobbles" sometimes so much that it's entirely outside the system's center of gravity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In an inelastic collision the shape of the clay changes right? If i bend a spoon enough it will break then it loses its elasticity,what about the clay if i form a dent why it still has some elasticity? $\endgroup$ – Zheer Mar 10 at 18:57
  • 1
    $\begingroup$ @Zheer Reality is messy: Any deformation, even of very inelastic substances like sand or clay, begins with an ever so slight elastic deformation until some threshold is exceeded, and then continues into the plastic realm while some energy is still stored in a (minuscule) elastic deformation. At the same time of course the steel ball will also undergo an ever-so-slight elastic deformation. This energy will be recovered later, but it is small. Clay or play-do adhesion also masks it, which is why they are good to demonstrate inelastic collisions. $\endgroup$ – Peter - Reinstate Monica Mar 11 at 8:46
1
$\begingroup$

The collision with the surface lasts much longer than it would if the surface was hard. The force applied on the object by the surface is given by the formula: $\sum F=\frac{\Delta p}{\Delta t}$ where $\Delta p$ is the change of momentum of the object ($p_{final}-p_{initial}$ where $p_{initial}$ is $mu$ and $p_{final}$ is $0$). From this formula it is obvious that when $\Delta t$ is getting smaller the total force is increasing.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does delta(t) Really mean, i mean is it the time required for the force to be applied to the object or how much force is applied each second to the object? $\endgroup$ – Zheer Mar 9 at 15:55
  • $\begingroup$ Shouldn't the total force have the same value in both cases, but in one case the force is applied in a short time and in the other case the same force is applied in a long period of time, therefore they should bounce back at the same height, am i right? $\endgroup$ – Zheer Mar 9 at 15:58
  • 1
    $\begingroup$ What does the rate of change in momentum have to do with the reason the ball bounces lower? Momentum is always conserved but kinetic energy is not for an inelastic collision. $\endgroup$ – Bob D Mar 9 at 16:01
  • 1
    $\begingroup$ @user254610 Drop the object on two ideal springs, one with a lower spring constant than the other, so one is "harder" than the other. You will get different forces but the object will bounce off at the same speed and reach the same height. So I don't think your answer addresses the OP question, but that's just my opinion. $\endgroup$ – Bob D Mar 9 at 16:44
  • 1
    $\begingroup$ @Zheer "the same force is applied in a long period of time." I think you are thinking of the concept of "impulse." And there is no guarantee that two collisions will result in the same impulse. The impulse leading up to the object reaching a standstill must be the same (since we're decelerating the same body by the same amount), but what happens after that point is a matter of how the energy got stored in the materials. $\endgroup$ – Cort Ammon Mar 9 at 18:27
1
$\begingroup$

Why do you assume that the impulse is the same in both cases? Impulse is equal to the change of momentum, so clearly it is not the same. There is an equal and opposite force in both cases, but these forces are not the same in one case as in the other (indeed the meaning of "soft" is a smaller equal and opposite force during collision as compared to "hard"). Also the time during which the force acts is different, so there really is no reason to think the impulse would be the same.

You really can only think of the difference in terms of the work done by the equal and opposite forces leading to energy being transformed into heat.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't understand, in both cases doesn't the object have the same mass and velocity? $\endgroup$ – Zheer Mar 9 at 22:19
  • $\begingroup$ You have to work from Newton's Laws. Force is proportional to acceleration, not to velocity. $\endgroup$ – Charles Francis Mar 9 at 22:35
  • $\begingroup$ i was talking about the change in momentum and impulse part,"Why do you assume that the impulse is the same in both cases" . $\endgroup$ – Zheer Mar 9 at 22:55
  • $\begingroup$ Yes, and I was telling you that you have to work from Newton's laws. Impulse is the integral of force with respect to time. You are starting from an assumption which is clearly false. The initial velocity may be the same, but the change in velocity/momentum is not. $\endgroup$ – Charles Francis Mar 10 at 7:34
1
$\begingroup$

No, the final speed is lower with a softer surface, so the impulse is lower.

If you throw a ball at a hard surface, the ball will not stop instantly. It will compress as it slows down. The energy of that compression will then be released, pushing the ball back in the direction opposite of the original motion. In the theoretical case of perfect elasticity, all of the energy will be retained by the ball, and the final speed will be the same as the original.

When a ball hits a soft surface, some of the energy is absorbed by the surface, rather than the ball, deforming. If the surface is elastic, then that energy will be restored to the ball. But if the surface is inelastic, that energy will be dissipated into heat. The ball will leave with less energy, and thus less speed.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What i don't get is that there should be an equal total normal force on the ball when the ball hits the surface, i mean if in one case the ball hits the surface with (1000N) andbounces higher but in the other case the ball bounces lower this violates newton's third law since there isn't an equal normal force on the ball, right? $\endgroup$ – Zheer Mar 10 at 7:51
  • $\begingroup$ he explains perfectly why this isnt the case @Zheer - even though you try to explain away energy - the process involves energy - and in case of an soft surface a substantial part of the energy of the ball is transferred over into the surface deformation - which happens to a much less degree with a hard surface - this surface deformation energy simply is missing when the ball accelerates back after surface contact - and for Newton .. you seem to forget the "force" that was needed to shove the surface particles out of the way in the soft surface ... $\endgroup$ – eagle275 Mar 10 at 10:34
  • $\begingroup$ @eagle275 what i don't get is that when you compress a spring, the spring will push pack with the same force you used to get it there for example if i use 100N to move it over 1meter it will push back with the same force until its in equilibrium, however why don't surfaces push back with the same force like springs?assuming for the purpose of this question that If throw a ball that pushes the surface with 1000N when the ball rebounds it rebounds with only 500N, why doesn't the surface push back the ball with 1000N just like springs? $\endgroup$ – Zheer Mar 10 at 11:00
  • $\begingroup$ that's because the spring is elastic - your soft ground is not completely elastic - but has a plastic "component" - deforming this plastic component (make the dent in the surface) will remove energy from your system .. for your spring .. simply provide pliers that clamp together the first 5 "rings" of your spring AFTER you compressed the spring .. then see how far the spring expands after releasing the spring but leave the pliers there ... $\endgroup$ – eagle275 Mar 10 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.