A car collides with a bus, who experiences a greater change in momentum?

[Reuben_Leib]

I am sorry I can't seem to get the LaTex to work

$$\textbf{My question:}$$

A car collides with a fast-moving bus, which vehicle experiences the greater change in momentum?

I am seem to get different answers on the internet:
Chegg says both the same,
quora says car,
brainly says car,
some youtube vids say both
Bard says: car has the greater change, but also contradics its self

I just want to confirm for myself: That the change in moment is the same?

$$\textbf{My Proof1}$$

Before collision: let total momentum = $P_{tot}$, and $P_{bus}$ initial momentum of bus, $P_{car}$ initial momentum of car,

$$P_{bus} + P_{car} = P_{tot}$$

Let the change in momentum be $\Delta P_{buss}$ and $\Delta P_{car}$, thus after the collision the: (using conservation of momentum)

$$P_{bus} + \Delta P_{buss} + P_{car} + \Delta P_{car} = P_{tot}$$

Now subtract the first equation from the second we get:

$$\Delta P_{buss} + \Delta P_{car} = 0$$

or

$$\Delta P_{buss} = -\Delta P_{car}$$

$$\textbf{Conclusion}$$

I believe that this proves that the change in momentum is the same in magnitude.

$$\textbf{My Proof2}$$ -by contradiction

suppose that the magnitude $\Delta P_{buss} >\Delta P_{car}$, thus there exists $\alpha > 0$, where $P_{buss} = -(P_{car} + \alpha)$. The negative is needed else momentum not conserved.

but then we get:

$$-\Delta P_{car} - \alpha + \Delta P_{car} = 0$$

but then $\alpha = 0$ which is a contradiction

1. Are my proofs sound?
2. is the change in moment is the same?

Thank you for your answer.

 

[Ibix]

It depends what question you ask. Momentum is conserved, which is what your method 1 uses to show (correctly) that the momentum changes are equal in the collision.

However, on Earth, after a collision both vehicles will come to a stop relative to the surface of the Earth. So apart from the simple "two bodies collide, what happens" calculation you did there's an interaction with a third body, the Earth, and if you factor this in the momentum change of each vehicle is pretty much its initial momentum in the Earth rest frame, which will probably be larger for a fast moving bus than a car moving at unspecified speed.

So your answer is correct for the way you are interpreting the question. But if you assume the vehicles come to a stop due to damage the answer depends on their initial speeds and masses.

 

[PeroK]

A car collides with a fast-moving bus, which vehicle experiences the greater change in momentum?

I am seem to get different answers on the internet:
Chegg says both the same,
quora says car,
brainly says car,
some youtube vids say both
Bard says: car has the greater change, but also contradics its self

First, you can't trust everyone on those sites. So, you need to figure out who is reliable and who is not.

I just want to confirm for myself: That the change in moment is the same?

In any collision between two bodies, Newton's third law applies. A corollary to this is equal change in momentum, as change in momentum is force integrated over time. Technically, therefore, the change in momentum during the collision is equal (in magnitude) and opposite in direction.

However, as pointed out above, if we include the braking force of kinetic friction on the road surface, then both eventually lose all their momentum (relative to the surface of the Earth).

 

[Hornbein]

The momentum is the same but the presumably less massive car experiences greater deceleration, which is what hurts.

 

[Dale]

It is good that you did a lot of research and checked multiple sources. However, never use a large language model AI for learning physics. They are designed to produce well written conversations, not factual conversations. LLM’s will use the right words in a confident and persuasive style, but whether they get the facts right or wrong appears to be completely random.

Momentum is conserved in the collision, so during the collision itself the change in momentum will be equal. Therefore the change in velocity will be greater for the less massive vehicle.

As others have said, after the collision they will both eventually come to rest so the change in momentum depends on whichever had the greatest initial momentum.

 

[kuruman]

It is an accepted fact that when a perfectly inelastic collision takes place between a bus and a car, the car will suffer most of the damage. The erroneous conclusion from this is that more damage implies more force which implies more momentum change over the same time interval. Here is a way to explain the damage imbalance. I will assume that the collision between the car and the bus is perfectly inelastic. I will further stipulate that, after the collision, the two are locked together and move on frictionless ice. This distinguishes the damage done during the collision from the damage done as the vehicles come to rest.

The damage is independent of the inertial frame of reference in which the situation is analyzed. In the CM frame the momenta are equal in magnitude and opposite in direction. Let $~p~$ be the magnitude of the momentum of each vehicle before the collision. This magnitude is reduced during the collision and becomes zero when the collision is complete but the vector sum of the momenta is zero at all times.

Now the kinetic energy of each vehicle (in the CM frame) starts at $~\dfrac{p^2}{2m}~$ and drops to zero when the collision is completed. Since the mass of the bus is 2-3 times the mass of the car, the number of Joules that go into bending metal and breaking plastic is 2-3 times greater in the case of the car than the bus.

Another way to look at the damage imbalance is using momentum conservation. The car and the bus are extended objects but each has its own center of mass. Consider what happens at the onset of the collision when the front of the car just barely touches the front of the bus. The CM of the car and bus will be getting closer to each other while damage is done. This relative motion will cease when the collision is complete. Let O be the final position of the car-bus CM. For every 1 inch that the CM of the bus has moved relative to O since the onset of the collision, the CM of the car will have moved 2-3 inches. More deformation means more damage.

 

[Reuben_Leib]

Thank you for your replies.

 

[jbriggs444]

Since the mass of the bus is 2-3 times the mass of the car, the number of Joules that go into bending metal and breaking plastic is 2-3 times greater in the case of the car than the bus.

This does not follow. At least not without additional (reasonable) background assumptions and careful reasoning.

The number of Joules that go into bending metal and breaking plastic depend on whose metal and plastic is more yielding under a fixed amount of force. If you mount a crush barrier on the front of the car, that is where the Joules will go. If you mount a crush barrier on the front of the bus, that is where the Joules will go instead.

Hence devices like this:

 

[FactChecker]

If you only consider the immediate effect of the impact, you are correct. You will find that a lot of people will intuitively confuse momentum with kinetic energy. That is probably why so many people gave the wrong answer of the car having the greater change.

 

[kuruman]

This does not follow. At least not without additional (reasonable) background assumptions and careful reasoning.

Yes, I did not mention the reasonable assumption that the materials making up each deformable vehicle are comparably yielding. The fact remains, though, that in the CM frame more kinetic energy Joules are available for internal dissipation in the car than in the bus. It follows from the work-energy theorem that the non-conservative work done by the bus on the car is greater than the work done by the car on the bus. Since the forces doing the work are equal in magnitude, the asymmetry of the work done by one vehicle on the other implies an asymmetry in the distance over which each vehicle travels under the action of this force. More distance traveled means more deformation.

 

[Hornbein]

It is worth noting that cars are deliberately designed to crush in order to absorb kinetic energy. This reduces deceleration thus protecting the occupants.

I was once in a small car that was rear ended by a big old school Cadillac. The rigid Caddy was largely undamaged. It's occupant was scarcely affected but this was because the rear of our automobile was crushed, reducing deceleration. This wasn't entirely good. If someone had been sitting in our rear seat they would have been crushed.

Busses don't always win. I saw a video of the interior of a bus that ran into a lighting standard. The passengers were flung forward. A few busses have seat belts but no one but me uses them.