# Work Done/Energy Transferred in One Dimensional Collision

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• MattGeo
In summary, an elastic collision between two objects results in the same displacement being covered by both objects.
MattGeo
I spend a lot of time thinking about collision problems because for me they are both extremely interesting and often very difficult to grasp when one thinks about them beyond the basics we are taught in introductory or even intermediate university courses.

Suppose there is a perfectly elastic collision between 2 objects and kinetic energy of the system is completely conserved. Disregard the specifics of masses and velocities for the time being because all we need to really consider is the fact that for kinetic energy to be conserved in the collision there must be a transfer of energy from one object to the other, so all that is occurring is a redistribution of existing kinetic energy in the system under consideration.

This seems to imply that the work done by one object is equal to the negative work done by the second object. We know that the force generated by the collision is the same on both objects due to Newton's 3rd Law. We also know that the force interaction occurs for a singular duration and so the time of contact is the same for both objects. This assures conservation of momentum.

So, if the force is the same on both objects and the energy lost by one object is the energy gained by another object, does this mean that for perfectly elastic collisions the displacement occurring during contact is always the same for both objects? W = F*d = change in kinetic energy?

This seems somewhat confusing to me because if you have 2 masses where one is very small and the other is very large and you apply the same force to both for the same amount of time, the smaller mass will go through a much larger acceleration, covering a much larger distance in the same amount of time and therefore would have more work imparted to it, causing a greater change in kinetic energy. This is not the case however in an elastic collision.

Assume now that a very massive object is moving at speed toward a stationary object of small mass. They collide perfectly elastically and kinetic energy is conserved. It seems to me that because the moving object already has a starting velocity when it collides with the stationary one that as the stationary object of much lesser mass undergoes a very large acceleration it will end up covering the same displacement that the massive object already moving will cover as it slows down via much slower deceleration. Would this be what is actually happening? At some point they must be moving at equal velocities while still in contact and so the contact interface would be moving as well over some displacement.

I've seen a few authors or online contributors often refer to thinking of collisions in terms of colliding springs or colliding masses which have springs attached. On Physics LibreText, a popular open-source platform, a collision is discussed where 2 unequal masses are attached to a compressed spring and when the spring is allowed to decompress the smaller mass covers a larger distance due to larger acceleration in equal time with equal force. And so it is argued that the "collision" distributes the energy in different ways where the displacements are not the same depending on mass.

Is this a mistake? Or is this just a case that is different from an elastic collision?

If the objects are in contact and exerting forces on each other they must move the same distance. If they moved different distances they would lose contact.

PeroK said:
If the objects are in contact and exerting forces on each other they must move the same distance. If they moved different distances they would lose contact.
So say that two objects collide in an elastic collision. They have different masses and are made of different materials. Upon collision they each deform to some extent and store elastic energy before rebounding. Being that they are different materials with different molecular lattices I imagine it would be like having 2 different springs with 2 different stiffness constants. So they'd compress by different amounts and there'd be different effective displacements internal to each body. Imagine colliding 2 different springs. Different masses and different stiffness constants. How does this square with our saying that the displacements are the same for each?

One more thing. If 2 objects of different composition and therefore different material elasticity collide, then how does the 3rd Law force pair even get set up in the collision? I am trying to think about it as 2 springs crashing together or 2 particles of same charge approaching and then bouncing off but that presupposes that the force experienced is a function of separation/compression of the 2 objects. The more they impress on each other / the closer they get, the larger the force experienced by each. How does it arise electromagnetically (in terms of repulsion at the micropscopic level) that is the SAME force on each, if the 2 objects have different molecular lattices?

Is it the same as 2 horiztonal springs being connected in series having the same force throughout when a force is applied?

MattGeo said:
So say that two objects collide in an elastic collision. They have different masses and are made of different materials. Upon collision they each deform to some extent and store elastic energy before rebounding. Being that they are different materials with different molecular lattices I imagine it would be like having 2 different springs with 2 different stiffness constants. So they'd compress by different amounts and there'd be different effective displacements internal to each body. Imagine colliding 2 different springs. Different masses and different stiffness constants. How does this square with our saying that the displacements are the same for each?
In my first reply, I assumed simply that there was a constant force acting for a time and while the objects moved a common distance while in contact.

If you model the collision as two springs, then by Newton's third law each spring exerts and equal and opposite force on the other. If the spring constants differ, then one spring will deform more than the other. As the forces are equal, the exchange of momentum is equal and opposite. Where's the problem?

MattGeo said:
One more thing. If 2 objects of different composition and therefore different material elasticity collide, then how does the 3rd Law force pair even get set up in the collision? I am trying to think about it as 2 springs crashing together or 2 particles of same charge approaching and then bouncing off but that presupposes that the force experienced is a function of separation/compression of the 2 objects. The more they impress on each other / the closer they get, the larger the force experienced by each. How does it arise electromagnetically (in terms of repulsion at the micropscopic level) that is the SAME force on each, if the 2 objects have different molecular lattices?
Newton's third law is built into Coulomb's law for the force between two charged particles, ##q_1, q_2##. The force has the same magnitude for each particle, in opposite directions:$$F = \frac{q_1q_2}{4\pi \epsilon_0r^2}$$

MattGeo
PeroK said:
In my first reply, I assumed simply that there was a constant force acting for a time and while the objects moved a common distance while in contact.

If you model the collision as two springs, then by Newton's third law each spring exerts and equal and opposite force on the other. If the spring constants differ, then one spring will deform more than the other. As the forces are equal, the exchange of momentum is equal and opposite. Where's the problem?Newton's third law is built into Coulomb's law for the force between two charged particles, ##q_1, q_2##. The force has the same magnitude for each particle, in opposite directions:$$F = \frac{q_1q_2}{4\pi \epsilon_0r^2}$$
I wanted to bring up the deformation and elasticity because isn't it necessary anyway for the interaction? We treat objects as rigid but if they actually were fully rigid wouldn't that prevent the Coulombic interaction as they encroach on each other? There's a nonzero interaction in the EM field at any distance I suppose but to actually exchange or store energy you have to affect the potential energy of inter atomic distances to experience the Coulombic repulsion.

To simplify let's just say each object is an identical spring of identical length. They collide. Would this be modeled like 2 springs in series which now have half the spring constant of 1 of the originals because the length is doubled?

MattGeo said:
To simplify let's just say each object is an identical spring of identical length. They collide. Would this be modeled like 2 springs in series which now have half the spring constant of 1 of the originals because the length is doubled?
Two springs is two springs. It's not one spring with double the spring constant.

Sorry, I misread that. Yes, like half the spring constant. Except, the springs will separate if not fastened together.

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## 1. What is work done in a one-dimensional collision?

Work done in a one-dimensional collision is the energy transferred from one object to another as a result of their interaction. It can also be described as the force applied over a distance, resulting in a change in kinetic energy.

## 2. How is work done calculated in a one-dimensional collision?

The work done in a one-dimensional collision can be calculated using the formula W = Fd, where W is work done, F is the force applied, and d is the distance over which the force is applied. This formula assumes that the force and displacement are in the same direction.

## 3. What factors affect the amount of work done in a one-dimensional collision?

The amount of work done in a one-dimensional collision is affected by the mass and velocity of the objects involved, as well as the angle and magnitude of the force applied. The type of collision (elastic or inelastic) also plays a role in determining the work done.

## 4. How does work done relate to energy transferred in a one-dimensional collision?

Work done and energy transferred are closely related in a one-dimensional collision. The work done is equal to the change in kinetic energy of the objects, which is the amount of energy transferred between them. This relationship is described by the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

## 5. Can work be negative in a one-dimensional collision?

Yes, work can be negative in a one-dimensional collision if the force and displacement are in opposite directions. This means that the energy is transferred from the object to its surroundings, resulting in a decrease in its kinetic energy. Negative work can also occur in an elastic collision, where the objects bounce off each other and the force is in the opposite direction of the displacement.

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