How can momentum but not energy be conserved in a collision, if objects keep moving in the same direction?

Question

For example, there are 2 objects with the same mass, moving in the same direction, one with speed 5 m/s and the other 1 m/s. They collide and keep moving in the same direction, now one with speed 4 m/s and the other 2 m/s. Momentum is conserved because

$$5\cdot mass + 1\cdot mass = 4\cdot mass + 2\cdot mass$$

but kinetic energy is not, because

$$\frac12 \cdot mass \cdot 5^2 + \frac12 \cdot mass \cdot 1^2 \neq \frac12 \cdot mass \cdot 4^2 + \frac12 \cdot mass \cdot 2^2$$

How, in this scenario, can momentum be conserved but not the kinetic energy? I understand the case when 2 objects move towards each other, collide, and then keep moving in the opposite directions - because momenta are vectors and energy is a scalar, so the values of momenta change, but the vector sum stays constant. However, in the aforementioned scenario, because objects keep moving in the same direction, the vector sum of momenta equals to the sum of their values.

If both momentum and kinetic energy depend only on mass and velocity, how can the value of one change, but not the other? If the kinetic energy was lost, shouldn't velocity decrease, and because momentum depends on velocity, shouldn't it change as well?

Answer 1

The thing to remember is that there is essentially only one kind of linear momentum, but many forms of energy. When you say energy is not conserved, you are evidently only considering translational kinetic energy.

When two objects collide they cause each other to vibrate. Some of the vibration causes pressure waves in air which is the sound of the collision and that is acoustic energy leaving the scene. The vibration eventually damps out increasing the internal energy of the objects. Energy is conserved, but some of it is converted from kinetic energy to these other forms.

The direction of motion is irrelevant to conservation laws. Energy is a scalar and knows nothing of direction anyway.

Answer 2

Kinetic Energy and momentum are always conserved in elastic collisions. As a result, it is relatively easy (and pedagogically instructive) to measure the kinetic energy and momentum before and after the collision in an elastic collision to compute the post-collision velocities of two masses. In this special/idealized case, it is possible to solve for the velocity and direction of both masses after collision by using the simple $p=mv$ and $KE= \frac {1}{2} mv^2$ conservation equations.

But, in an inelastic collision, the number of masses to measure and compute in the before and after-collision state makes the numerical/formulaic computation of the post-collision direction and velocity of the masses impractical. Still, the conservation of total energy and momentum is hidden behind the complexity of the motion and numerous entities involved after the collision.

Since this problem stipulates that the collision is inelastic, we shall attempt to illuminate how both momentum and total energy are conserved amidst the confusion of kinetic energy conversion to numerous forms.

Obviously, in an inelastic collision, some of the kinetic energy of the masses will be lost to various other forms of energy. The total system energy will always be conserved, it is just complex to track the details of every molecules conversion of kinetic energy into another form of energy after the collision.

Listed explicitly, the kinetic energy of the pre-collision masses will be converted into a variety of post-collision energy-species, such as: 1) the residual kinetic energy of the masses, 2) the kinetic energy of the thermal energy of molecular motion, 3) the acoustic energy of molecular air vibration, 4) the photonic energy of IR photons released from the higher than ambient temperature masses, and 5) the energy held in activated atomic orbitals, etc.

But, the fact that kinetic energy converts into many kinds of energy whose total energy is conserved, does not intuitively resolve the apparent discrepancy between kinetic energy and momentum. In other words, what is the nature of momentum that it is apparently conserved in an inelastic collision, while kinetic energy is not?

Regarding momentum, if we calculate the vector sum of the momentum of the two masses pre- and post-inelastic collision, the sum of the momentum vectors will be equal before and after the collision.

To intuitively understand the distinction between momentum and kinetic energy that produces this apparent difference in conservation, we note that the collision of every molecule involves an equal and opposite action and reaction. This principle of interaction is at the heart of the conservation of momentum.

Energy is a different aspect of the properties of various types/configurations of space (e.g., mass is a type of energy, as is the motion of a mass, as is the electromagnetic packet of a photon, as is the attraction of charged bodies, etc.). Energy reflects the magnitude of an (unknown, not fully understood, but real) property of space. Energy is conserved because that property cannot be created or destroyed, only converted.

Example: When two equal masses collide and rebound, in the center of mass frame, their momentum are opposite and equal, and the net momentum of the system is zero, both before and after the collision, whether elastic or inelastic. The total energy associated with the two masses before the collision is equal to the total energy after the collision.

Momentum is conserved on the molecular level in every interaction/collision between molecules. Thus, even though an inelastic collision is complex, the momentum changes of all the constituent molecular collisions all cancel each other out. The result is a net zero change in momentum as measured before and after the macroscopic collision.

The total energy is conserved because the energy associated with the moving bodies is neither created nor destroyed in this, or any interaction. Thus, the magnitude of the total energy is conserved.

Answer 3

You already have your counterexample. Another simple one is two things which collide and stick together. Like if they are identical masses, and we choose the reference frame where they have equal and opposite momentums, the net momentum is zero before and after, but the net kinetic energy is non-zero before and zero after.

See, you are entirely right, for one particle. With one particle, there is only one unknown velocity, so if you know its mass and you know its momentum, then you know its velocity, and then you know its energy. But then when you have two particles, there are two unknown velocities. The conservation of momentum only tells you that a certain mass-weighted sum of the two velocities must remain constant. But then you have one more degree of freedom, essentially the difference between the velocities, that conservation of momentum does not determine.

So if two equal masses collide, and I choose the frame of reference where their momentums are opposite, conservation of momentum just says that the sum of the velocities must always be zero, $v_1+v_2=0$ so therefore $v_1=-v_2$. But it says nothing about the difference in velocities $v_1 - v_2$.

The key is, I am not saying that each individual particle keeps the same momentum. If I were saying that, I could never explain how a car accelerates, because then it doesn't keep the same momentum. So that's not a valid restatement of Newton's third law. But conservation of momentum is Newton's third law, and it has to do with how particles interact, not individual particles all alone: I push you, you push me, the transfer of momentum is equal and opposite.

Now if I add on conservation of energy, I find that this gives me another equation which also constrains $v_1$ and $v_2$. So then I can solve for both. Or if I make an assumption that kinetic energy goes to zero afterwards, like I did with the particles which stick together, then I can solve for both. But notice that if I had a collision among three particles, there might be one other velocity in the picture, in which case, I still don't have enough information with momentum and energy, and I might need to include something like angular momentum.

Answer 4

If you do the problem in the center of mass, then the momentum of the system will be zero, but the kinetic energy is not. It will depend on the reduced mass, $$\mu = \frac{m_1m_2}{m_1+m_2},$$ and the relative velocity of one particle to the other: $$K_{cm}=\frac{1}{2}\mu\left((\vec{v}_2-\vec{v}_1)\cdot (\vec{v}_2-\vec{v}_1)\right).$$

The particles are moving opposite each other in the center of mass. When they collide, they both reverse directions (or both stop and stick together). If the relative velocity magnitude is smaller after the collision, the kinetic energy has decreased, probably due to deformation, heating, etc. If the relative velocity magnitude is the same, but they have collided, then they have exchanged velocities (relative to the CM frame).

Mathematically, KE and momentum are different quantities because KE depends on the square of the speed and is a scaler, whereas momentum depends linearly (in Newtonian physics) on the velocity, a vector.

Conceptually, the two quantities arise from different perspectives. KE arises from a force-differential distance relationship combines the vector behaviors to get a scaler, and momentum arises from a force-time relationship which preserves the vector behavior.