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

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$\textbf{My question:}$

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

But when I googled I got conflicting answers from "expert answers": Chegg says both the same, quora says car, brainly says car, some youtube vids say both the same.

I tried to prove that it they both have the same change in momentum. I came up with these 2 "proofs" which "prove" my idea, but I would like a second opinion as the "different" answers are casting some doubt.

$\textbf{My Proof1}$

Before collision: let total momentum = $P_{tot}$, and $V_i$ initial velocity of buss, $U_i$ initial velocity of car,

$M_{bus}V_i + M_{car}U_i = P_{tot}$

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

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

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.

$\textbf{My Proof2}$ 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

Answer 1

Forget velocity, it is unnecessary. Call the initial bus (car) momentum $\vec p$ ($\vec k$).

The bus applies a force to the car $\vec F_{c, b}(t)$. From that, the final momentum of the bus is:

$$ \vec p' = \vec p + \int_{t=0}^{\tau} \vec F_{c, b}dt = \vec p + \Delta \vec p $$

or:

$$ \Delta \vec p = \int_{t=0}^{\tau} \vec F_{c, b}dt $$

The car applies a force to the bus $\vec F_{b, c}(t)$, so $$ \vec k' = \vec k + \int_{t=0}^{\tau} \vec F_{c, b}dt = \vec k + \Delta \vec k $$

and

$$ \Delta \vec k = \int_{t=0}^{\tau} \vec F_{c, b}dt $$

Now we can invoke Newton's third (?) law of equal and opposite force, meaning:

$$ \vec F_{c, b}(t) = - \vec F_{b, c}(t) $$

with which you can derive:

$$\Delta \vec k = -\Delta \vec p $$