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

conservation-lawsmomentum

$$\textbf{My question:}$$

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

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:

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