Impulse: Elastic vs Inelastic

[JAH360]

So I'm confused about a question in TBR which goes as follows:

In a head-on collision between two identical cars, which is true when comparing elastic and inelastic collisions? (Assume that the initial momenta and collision times are the same, regardless of the type of collision."

A. An elastic collision is less damaging to the cars than an inelastic collision.
B. An inelastic collision is less damaging to the cars than an elastic collision.
C. Either type of collision is equally damaging.
D. The more damaging collision involves smaller impulses.

TBR says A is the answer, but isn't D also correct? The less damaging collision is the elastic collision, which would have a greater change in momentum for both cars, since they both recoil, causing a greater impulse. It actually says this in the answer explanation. It also says occupants in a vehicle experience smaller impulses in an inelastic collision. So wouldn't the more damaging collision, the inelastic collision, involve smaller impulses? Isn't D also correct? Is this is a mistake on TBR's part, or am I missing something?

This is chapter 4, passage 6, number 39.

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[Thoroughbred_Med]

that's a good question... anyone?

 

[Winterbourne]

I'm not sure of the answer to this question, but this might help…

In the elastic collision, the total KE in conserved and in the inelastic collision total KE is not. This loss in KE is sometimes the energy of deforming the object.


The thing is I am not exactly sure how impulse comes into play:

In a collision, if the impact time is increased, the collision is less damaging. Impulse of a force on something equals the change in its momentum due to the force. So impulse is change the momentum and momentum in the collision is conserved. If you increase the time over which the force acts, the impulse is decreed. So perhaps, in the inelastic collision some of the KE is absorbed in the deforming of the object, which means the impact is short and tis the average force (impulse/time) is large.

Can someone else join in? This is what I know, maybe someone can help connect the dots….

 

[sillyjoe]

"....momenta and collision times are the same, regardless of the type of collision"

The answer to your question is addressed in the question stem. The impulse is the same regardless of the collision. Either way, A is the better answer since kinetic energy is conserved and there are no deformations.

 

[justadream]

@sillyjoe

I thought elastic collisions had greater impulse?

I know this isn't the greatest "source" but:

http://www.vernier.com/innovate/impulse-comparison-for-elastic-and-inelastic-collisions/

 

[sillyjoe]

@sillyjoe

I thought elastic collisions had greater impulse?

I know this isn't the greatest "source" but:

http://www.vernier.com/innovate/impulse-comparison-for-elastic-and-inelastic-collisions/

I don't know the conceptual physics here. I am just referring to the fact that the question STEM specifies that the impulses are the same.

 

[justadream]

@sillyjoe

Which part of the question stem says that?

Part of the answer (taken from the OP) provided by TBR is apparently "The less damaging collision is the elastic collision, which would have a greater change in momentum for both cars, since they both recoil, causing a greater impulse"

 

[sillyjoe]

@sillyjoe

Which part of the question stem says that?

Part of the answer (taken from the OP) provided by TBR is apparently "The less damaging collision is the elastic collision, which would have a greater change in momentum for both cars, since they both recoil, causing a greater impulse"

You're right. I should clarify. They say in the question stem:

(Assume that the initial momenta and collision times are the same, regardless of the type of collision)

If the change in momentum is the same and the collision times are the same, isn't the force the same?

I just figured that an inelastic collision has deformations and damages the car more than an elastic collision where there is 0 deformations the the car. The answer discusses damage to the car.

 

[justadream]

@sillyjoe

I though the force wasn't the same (isn't that why elastic has greater impulse)?
But when I did the question myself, I thought the answer should be elastic because there is greater impulse exchange.

However, TBR's answer also makes sense to me in that damage would be caused by an inelastic collision (some of the energy going into deforming the car).

 

[sillyjoe]

@sillyjoe

I though the force wasn't the same (isn't that why elastic has greater impulse)?
But when I did the question myself, I thought the answer should be elastic because there is greater impulse exchange.

However, TBR's answer also makes sense to me in that damage would be caused by an inelastic collision (some of the energy going into deforming the car).

Honestly, I don't know. I guess it does have a greater impulse. Even so, the damage on the car is less in elastic because it doesn't deform.

It's like throwing a rubber ball against the wall (elastic) vs. throwing play-doh against the wall (inelastic)

 

[justadream]

Yeah if someone could chime in on whether

Elastic Collisions (greater impulse) have greater FORCE than do Inelastic Collisions (less impulse)

 

[JAH360]

So hopefully this won't show up on my MCAT, because I'm confused how to grasp impulse conceptually. In an elastic collison, the average force is greater, but the delta t is smaller; with inelastic collisions, the average force is smaller, but the delta t is greater. So perhaps impulse is constant for a collision involving a specific mass and velocity, and thus the average force and change in time are just inversely related? Like, the speed of light in a vacuum is constant, but wavelength and frequency can change as long as their product remains constant. Is this the same idea for impulse? Idk if I'm explaining it well...

 

[sillyjoe]

So hopefully this won't show up on my MCAT, because I'm confused how to grasp impulse conceptually. In an elastic collison, the average force is greater, but the delta t is smaller; with inelastic collisions, the average force is smaller, but the delta t is greater. So perhaps impulse is constant for a collision involving a specific mass and velocity, and thus the average force and change in time are just inversely related? Idk.

Impulse = change in momentum = F*delta(t)

The change in momentum is the same for the collisions I believe. However, the force and time are what is different in each collision as you pointed out.

F = ma = delta(mv)/t

If you multiply the above equation by t you get:

F*t = change in momentum

I am pretty sure the change in momentum for a given collision is the same because momentum is conserved. It is kinetic energy that is either conserved or not conserved and as a result the F*t part of the equation changes with the different collisions.

 

[JAH360]

I don't know the conceptual physics here. I am just referring to the fact that the question STEM specifies that the impulses are the same.

But the STEM merely specifies that the ORIGINAL momenta are the same. Since the delta t for both collisions are the same, the elastic collision, which involves recoil and thus a larger change in momentum, would experience a larger impulse. Right?

 

[sillyjoe]

But the STEM merely specifies that the ORIGINAL momenta are the same. Since the delta t for both collisions are the same, the elastic collision, which involves recoil and thus a larger change in momentum, would experience a larger impulse. Right?

Is the overall change in momentum not the same for both the elastic and inelastic collisions if there initial momenta were the same?

 

[JAH360]

Is the overall change in momentum not the same for both the elastic and inelastic collisions if there initial momenta were the same?

No, because there's a larger change in momentum with recoil, as the delta v for recoil will be larger. If an object collides with a wall with an initial velocity of 10 m/s, and then recoils at -5 m/s, the total magnitude of delta v is 15 m/s (-5m/s - 10m/s). However, if there were no recoil, the total magnitude of delta v would just be 10 m/s. And since the change in momentum is mass x delta v, the change in momentum, and thus the impulse, would be larger for the elastic collision.

Which brings me back to my original question... Why is answer D not also correct?

 

[sillyjoe]

No, because there's a larger change in momentum with recoil, as the delta v for recoil will be larger. If an object collides with a wall with an initial velocity of 10 m/s, and then recoils at -5 m/s, the total magnitude of delta v is 15 m/s (-5m/s - 10m/s). However, if there were no recoil, the total magnitude of delta v would just be 10 m/s. And since the change in momentum is mass x delta v, the change in momentum, and thus the impulse, would be larger for the elastic collision.

Which brings me back to my original question... Why is answer D not also correct?

I thought the change in momentum would be the same since it is conserved and the only difference in the collisions is KE is lost in an inelastic collision.

Guess I was wrong and would like someone to explain as well!

 

[JAH360]

I thought the change in momentum would be the same since it is conserved and the only difference in the collisions is KE is lost in an inelastic collision.

Guess I was wrong and would like someone to explain as well!

Yeah, momentum should be conserved. Blah, I'm confused. No one else seems to be chiming in, so I might need to consult the experts of Yahoo Answers, lol.

 

[sillyjoe]

@gettheleadout would you please be able to lend your expertise to this question/thread?

 

[justadream]

@gettheleadout

I second @sillyjoe !

Especially with regard to whether the following is true:

"Elastic Collisions (greater impulse) have greater FORCE than do Inelastic Collisions (less impulse)"

 

[gettheleadout]

To respond to the OP, yes, @JAH360, Answer D is also correct. The question appears odd because it specifies that collision times remain the same between the elastic vs. inelastic collision scenarios, but regardless, if the two objects collide and recoil elastically the impulse is greater than if they stuck together (inelastically).

To @sillyjoe, conservation of momentum only ensures that the total momentum of the system is constant. For two equal mass objects approaching each other in completely opposite directions at the same speed, the initial momentum of the system is zero. If the objects collide elastically and recoil, the final net momentum of the system is still zero. If the objects collide and stop, they end at rest and the net momentum of the system at the end is more obviously zero (system is at rest). The impulses experienced by the objects are different between the two cases, but conservation of momentum makes no statement about those impulses directly (although indirectly, CoM means that for a system with net zero initial momentum, the impulses experienced by all components in a collision must sum to zero).

To @justadream, we've established, as you said, that elastic collisions involve greater impulses than inelastic collisions. Your question concerns the force involved in these collisions. In the question stem posted by the OP, collision times are specified as invariable between the elastic vs. inelastic collisions, but in reality deformation takes time. Drawing out the time over which a collision occurs actually implies deformation on its own, and we can connect this increased collision time and the inelasticity of the collision to reduced force in two ways. One is mathematical, where we can look at two identical, symmetrically moving objects colliding perfectly inelastically and deforming. We see deformation, longer collision time vs. elastic, and lower impulse. Since, assuming force is constant throughout a collision, impulse is J = F*t, and we know that deformation increases t but decreases J, we must have F decreasing substantially. The second method is intuitive; this is how seatbelts work.

Now, to answer your question directly, about the statement, "Elastic Collisions (greater impulse) have greater FORCE than do Inelastic Collisions (less impulse)," I want to be careful with terminology here. I don't mean to nitpick, but with force in particular extra care should be taken in describing what's happening. Force itself is not a property of a collision as a whole nor of an object in motion or during collision. For objects of constant mass (i.e. the colliding objects don't shatter), and in this case, force describes a change in kinetic energy experienced by an object as a result of contact with another object (F = ma, m is constant so F is proportional to acceleration, which is change in velocity, and F describes this change in velocity relative to the object's mass; this relative relationship is most easily understood in terms of kinetic energy). For these symmetrical collisions we're discussing, not only are the forces experienced by the objects equal in magnitude (Newton's Third Law), but the objects' changes in velocity are equal as well (because their masses are equal). Since we've established that for calculations of impulse in these collisions, J = F*t and relative to an elastic collision, F decreases substantially for an inelastic collision, we can make a definitive statement that, "the force experienced by an object in an elastic collision is greater than that experienced in the same, inelastic collision."

 

[JAH360]

To respond to the OP, yes, @JAH360, Answer D is also correct. The question appears odd because it specifies that collision times remain the same between the elastic vs. inelastic collision scenarios, but regardless, if the two objects collide and recoil elastically the impulse is greater than if they stuck together (inelastically).

To @sillyjoe, conservation of momentum only ensures that the total momentum of the system is constant. For two equal mass objects approaching each other in completely opposite directions at the same speed, the initial momentum of the system is zero. If the objects collide elastically and recoil, the final net momentum of the system is still zero. If the objects collide and stop, they end at rest and the net momentum of the system at the end is more obviously zero (system is at rest). The impulses experienced by the objects are different between the two cases, but conservation of momentum makes no statement about those impulses directly (although indirectly, CoM means that for a system with net zero initial momentum, the impulses experienced by all components in a collision must sum to zero).

To @justadream, we've established, as you said, that elastic collisions involve greater impulses than inelastic collisions. Your question concerns the force involved in these collisions. In the question stem posted by the OP, collision times are specified as invariable between the elastic vs. inelastic collisions, but in reality deformation takes time. Drawing out the time over which a collision occurs actually implies deformation on its own, and we can connect this increased collision time and the inelasticity of the collision to reduced force in two ways. One is mathematical, where we can look at two identical, symmetrically moving objects colliding perfectly inelastically and deforming. We see deformation, longer collision time vs. elastic, and lower impulse. Since, assuming force is constant throughout a collision, impulse is J = F*t, and we know that deformation increases t but decreases J, we must have F decreasing substantially. The second method is intuitive; this is how seatbelts work.

Now, to answer your question directly, about the statement, "Elastic Collisions (greater impulse) have greater FORCE than do Inelastic Collisions (less impulse)," I want to be careful with terminology here. I don't mean to nitpick, but with force in particular extra care should be taken in describing what's happening. Force itself is not a property of a collision as a whole nor of an object in motion or during collision. For objects of constant mass (i.e. the colliding objects don't shatter), and in this case, force describes a change in kinetic energy experienced by an object as a result of contact with another object (F = ma, m is constant so F is proportional to acceleration, which is change in velocity, and F describes this change in velocity relative to the object's mass; this relative relationship is most easily understood in terms of kinetic energy). For these symmetrical collisions we're discussing, not only are the forces experienced by the objects equal in magnitude (Newton's Third Law), but the objects' changes in velocity are equal as well (because their masses are equal). Since we've established that for calculations of impulse in these collisions, J = F*t and relative to an elastic collision, F decreases substantially for an inelastic collision, we can make a definitive statement that, "the force experienced by an object in an elastic collision is greater than that experienced in the same, inelastic collision."

Thanks for the clarification! Very helpful!

 

[sillyjoe]

To respond to the OP, yes, @JAH360, Answer D is also correct. The question appears odd because it specifies that collision times remain the same between the elastic vs. inelastic collision scenarios, but regardless, if the two objects collide and recoil elastically the impulse is greater than if they stuck together (inelastically).

To @sillyjoe, conservation of momentum only ensures that the total momentum of the system is constant. For two equal mass objects approaching each other in completely opposite directions at the same speed, the initial momentum of the system is zero. If the objects collide elastically and recoil, the final net momentum of the system is still zero. If the objects collide and stop, they end at rest and the net momentum of the system at the end is more obviously zero (system is at rest). The impulses experienced by the objects are different between the two cases, but conservation of momentum makes no statement about those impulses directly (although indirectly, CoM means that for a system with net zero initial momentum, the impulses experienced by all components in a collision must sum to zero).

To @justadream, we've established, as you said, that elastic collisions involve greater impulses than inelastic collisions. Your question concerns the force involved in these collisions. In the question stem posted by the OP, collision times are specified as invariable between the elastic vs. inelastic collisions, but in reality deformation takes time. Drawing out the time over which a collision occurs actually implies deformation on its own, and we can connect this increased collision time and the inelasticity of the collision to reduced force in two ways. One is mathematical, where we can look at two identical, symmetrically moving objects colliding perfectly inelastically and deforming. We see deformation, longer collision time vs. elastic, and lower impulse. Since, assuming force is constant throughout a collision, impulse is J = F*t, and we know that deformation increases t but decreases J, we must have F decreasing substantially. The second method is intuitive; this is how seatbelts work.

Now, to answer your question directly, about the statement, "Elastic Collisions (greater impulse) have greater FORCE than do Inelastic Collisions (less impulse)," I want to be careful with terminology here. I don't mean to nitpick, but with force in particular extra care should be taken in describing what's happening. Force itself is not a property of a collision as a whole nor of an object in motion or during collision. For objects of constant mass (i.e. the colliding objects don't shatter), and in this case, force describes a change in kinetic energy experienced by an object as a result of contact with another object (F = ma, m is constant so F is proportional to acceleration, which is change in velocity, and F describes this change in velocity relative to the object's mass; this relative relationship is most easily understood in terms of kinetic energy). For these symmetrical collisions we're discussing, not only are the forces experienced by the objects equal in magnitude (Newton's Third Law), but the objects' changes in velocity are equal as well (because their masses are equal). Since we've established that for calculations of impulse in these collisions, J = F*t and relative to an elastic collision, F decreases substantially for an inelastic collision, we can make a definitive statement that, "the force experienced by an object in an elastic collision is greater than that experienced in the same, inelastic collision."

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