Thread: Elastic and Inelastic Collisions

Hello.

I am a bit confused on a question concerning linear elastic and inelastic collisions. I see that two characteristics of a completely inelastic collision are first that the sum of the initial and final momentums are equal, or:

and second, that after the collision both objects stick together. If we assume that the second object involved in the collision is initially stationary, and we wanted to solve for a final velocity of this second object, we can then write the above formula as:


About elastic collisions we know that kinetic energy before the collision is equal to kinetic energy after the collision. We can write this with the following equation:

Again, assuming the initial velocity of our second object is zero, and solving this for a final velocity of the second object we get:


What I want to figure out is which type of collision will have a higher likelyhood of closing the door. The parameters given are:
= .4 kg
= 30 kg
= 1 m/s
= 0

The solution I come up with is that while the inelastic collision gives a precise value of final velocity, the elastic collision gives a range of values that includes the value for the final inelastic velocity. Using what information I have worked out, I conclude that neither type of collision is more or less likely to close the door since the elastic collision could give both a value of velocity greater and lesser than the velocity given by the inelastic collision.

Would someone be kind enough to go over the work I have shown above and tell me what you think about it? My intuition tells me that if you know that a collision is an elastic or inelastic collision and you knew all the initial variables, you should be able to predict the outcome. This should include the final velocities of both objects.

How are you getting a range of velocities for an elastic collision? I think there'd always be one result.

Also, I think you might want to try this with conservation of momentum instead of conservation of kinetic energy. I could be wrong, but it might make things simpler.

By closing the door, I assume you mean you are throwing something at an open door, either an elastic ball or an inelastic lump of clay, and you want to know which imparts the greater velocity to the door.

In the elastic situation, your formula for v_2f has an unknown v_1f. You can't just plug any arbitrary number in there. To find that you need to apply not just conservation of energy but also conservation of momentum. The velocity of the recoiling rubber ball will be negative. This means the door, being much heavier, will get almost twice the +momentum the ball initially had, in order to compensate for the -momentum of the recoiling ball.

I guess your in University Physics I, It has been a year or so but I think in your equation with the radical yielding either a positive or negative answer will just tell you the vector direction that m1 will travel. You can have a negative velocity but not a negative speed so if you just want to know which V2 will close the door the fastest it seems like it would be the elastic collision because when you compute its value neglecting the fact that the numerator under the radical will be a negative number and produce unreal roots, It has the larger magnitude and therefore the larger velocity.

Isn't the result that the elastic equation should yield a larger magnitude? Inelastic equation's are non-conservative so it seems that their final velocity should be less. Because the inelastic impact absorbs some of the energy.

Elastic equation's can be simplified to u1 + v1 = u2 + v2 or at least that was what my professor told me.

Anyways just using tuition I think that the Elastic equation should produce a greater V2 velocity.

In inelastic equations the right hand portion of the equations only have 1 velocity but a greater mass. this add mass increases the total momentum but decreases the final velocity.

If you have a stationary object and it is hit by some other object of any mass. The resulting Vfinal should be less than object 2's initial velocity.

I don't know if I have talked in circles or not and I didn't actually compute any values I just used some logic that I hope isn't faulty. I hope that some how helped and wasn't a waste to read.

Originally Posted by GenerationE

Isn't the result that the elastic equation should yield a larger magnitude? Inelastic equation's are non-conservative so it seems that their final velocity should be less. Because the inelastic impact absorbs some of the energy.

That would give you the right answer but for the wrong reason. Energy is conserved in the elastic collision, but most of that energy is carried away with the rubber ball in the other direction. Demen is only interested in the momentum/energy of the door.

Demen has come up with 1 equations with 2 unknowns. The solution is to write another equation - conservation of momentum.

As I see it the inelastic impact should impart the most velocity to the door. The ball recoils because some of the energy of the impact was stored in the elastic compounds of the ball, which was released as it went the other way. So the closer the return velocity of the ball is to the initial velocity, the less the door would move, with the door not moving and the lost energy being converted to heat in the ball and a very small amount in the door at maximum recoil speed. Some energy would go into deforming the clay, but most would be transferred to the door IMO.

This site has an applet.
http://www.walter-fendt.de/ph14e/collision.htm
Plugging in the values m1= 0.4, m2=30, and v1=1, it turns out that the final velocities of m2 are .143 inelastic and .286 elastic which is double as I predicted. For purposes of closing the door, though, you have to add the momentum of both masses in the inelastic case. This makes it a bit closer with momentum of .200 inelastic and .286 elastic.

Originally Posted by Demen Tolden

Hello.

I am a bit confused on a question concerning linear elastic and inelastic collisions. I see that two characteristics of a completely inelastic collision are first that the sum of the initial and final momentums are equal, or:

and second, that after the collision both objects stick together. If we assume that the second object involved in the collision is initially stationary, and we wanted to solve for a final velocity of this second object, we can then write the above formula as:


About elastic collisions we know that kinetic energy before the collision is equal to kinetic energy after the collision. We can write this with the following equation:

Again, assuming the initial velocity of our second object is zero, and solving this for a final velocity of the second object we get:


What I want to figure out is which type of collision will have a higher likelyhood of closing the door. The parameters given are:
= .4 kg
= 30 kg
= 1 m/s
= 0

The solution I come up with is that while the inelastic collision gives a precise value of final velocity, the elastic collision gives a range of values that includes the value for the final inelastic velocity. Using what information I have worked out, I conclude that neither type of collision is more or less likely to close the door since the elastic collision could give both a value of velocity greater and lesser than the velocity given by the inelastic collision.

Would someone be kind enough to go over the work I have shown above and tell me what you think about it? My intuition tells me that if you know that a collision is an elastic or inelastic collision and you knew all the initial variables, you should be able to predict the outcome. This should include the final velocities of both objects.

There is no such thing as an inelastic collision. To create tension something has to give. The hardest substance on earth, will create a tension. Or destruction.

Sincerely,


William McCormick

Originally Posted by Harold14370

By closing the door, I assume you mean you are throwing something at an open door, either an elastic ball or an inelastic lump of clay, and you want to know which imparts the greater velocity to the door.

In the elastic situation, your formula for v_2f has an unknown v_1f. You can't just plug any arbitrary number in there. To find that you need to apply not just conservation of energy but also conservation of momentum. The velocity of the recoiling rubber ball will be negative. This means the door, being much heavier, will get almost twice the +momentum the ball initially had, in order to compensate for the -momentum of the recoiling ball.

Harold the clay will only stick because it creates a vacuum against the door. If you wrap the clay with aluminum foil or wax paper, it will in fact not stick to the door.

As the clay reaches the point it will not compress against the door, it will push the door away, and fall. Because clay does have some elasticity. All things do. Even if it is the tension in a drop of water.

If the door is hard enough, and free standing, it will fly off, if you hit it with enough velocity. The clay object wrapped in foil, will impact the door, smash the clay, and drop. Throwing the door a good distance. Or with more velocity the clay will vaporize. And dent a metal door, and throw it.

I do not like the terminology of elastic and inelastic. It hides what is taking place.

According to these terms the door is elastic, and the clay is inelastic. So the collision is what? Only the velocity will tell.

Two hammers smacking together to me seem like an inelastic collision. Even though they bounce off one another, rather violently. If you look at the compression of each hammer, it is very small.

Or two billiard balls hitting one another. Again the compression is small. But the bounce is big. Elastic really is hard to swallow. High tension objects or surfaces seem more real to me.

A rubber ball, compressing does seem like an elastic collision. But I think the clay has an elastic collision, because of the time it takes to compress the clay. This can create massive energy. Killer force to a human head. The human head will bounce off of it.

If you smash a forty five round against a steel plate, from a nickle auto forty five, the round vaporizes. Dents the 3/16" plate, but does not scratch the paint. And moves the plate well.

I do not see this terminology as anything but confusing.





Sincerely,


William McCormick

Originally Posted by Harold14370

This site has an applet.
http://www.walter-fendt.de/ph14e/collision.htm
Plugging in the values m1= 0.4, m2=30, and v1=1, it turns out that the final velocities of m2 are .143 inelastic and .286 elastic which is double as I predicted. For purposes of closing the door, though, you have to add the momentum of both masses in the inelastic case. This makes it a bit closer with momentum of .200 inelastic and .286 elastic.

There is something wrong with the applet. After you input m2 it automatically reverts to 1kg for some reason. An applet I found HERE gives values for m2 of 0.026 elastic and 0.013 inelastic. I made the mistake of equating velocity with energy, instead of thinking in terms of conservation of momentum. I feel like an idiot, but now I know better! Good one Harold.

In case Demen hasn't solved this one yet, the solution to the simultaneous equations for head-on elastic collision is given here
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html

The result is





For Demen's example, the results are
v_2f (inelastic)=.01316
v_2f (elastic)=.02632
momentum (inelastic)=0.4
momentum (elastic)=0.789
KE of door plus projectile (inelastic)=.00263
KE of door (elastic)=.0104

Thanks Harold. With the help of you guys and others, I figured it out two days ago and had to cram it into a paper in a way that wasn't very satisfactory to me. Anyway, my major mistake was thinking that while conservation of kinetic energy was something unique to perfectly elastic collisions, conservation of momentum was likewise unique to inelastic collisions.

(conservation of momentum)
(conservation of kinetic energy)

Dividing the two formulas I posted above (which are basically the formulas you posted) seems to be the key to understanding elastic collisions with a second stationary object.