A moving car collides with a stationary car, after which they move at the same velocity. After going some distance they stop. The following information is given:

  1. mass of the 2 cars
  2. the velocity of the 1st car before collision
  3. the distance the 2 cars covered together after the collision

Will it be possible to calculate the force the stationary car exerted on the moving one?

My friend used the equation $$v^2=u^2+2as$$ and the 3rd information stated above to calculate acceleration caused by the force exerted by the stationary car.
But, had there not been the friction of the road, wouldn't the 2 cars be moving forever after the collision?

Is it my friend or I am incorrect? If I am right, would I be able to calculate it if the time the two vehicles were in collision were given?

  • $\begingroup$ In your friend's equation, what is $u$? $\endgroup$ May 28, 2019 at 16:39
  • $\begingroup$ whats the exact problem? i wanna have a go $\endgroup$
    – Hisham
    May 28, 2019 at 16:51
  • $\begingroup$ You need to use conservation of momentum because this is an inelastic collision. However, you don't know how long the collision took, so you don't have enough information to calculate the force of the collision. $\endgroup$ May 28, 2019 at 16:59

1 Answer 1


If we assume:

then the question becomes

"What's the breaking force exherted by the second car as they're coming to a stop?"

We can find the velocity at the beginning of the breaking period using conservation of momentum: $$v_{0+} = v_{a}\frac{m_{a}}{m_{a}+m_{b}}$$

Let's call the breaking force $f$, the total mass $m$ and the (negative) acceleration is $a$ as usual.
We know that the velocity during the whole breaking period is $$v_{t} = v_{0+} + ta = v_{0+} - t \frac{f}{m}$$ Turn that around to get time as a function of velocity: $$t = \frac{m(v_{0+} - v_t)}{f}$$ Now we can say how long the breaking takes ($t_{end}$), by setting $v_t = 0$ $$t_{end} = \frac{m\,v_{0+}}{f}$$ Which is hardly surprising, and might not seem useful because it's an expression in terms of $f$, which is exactly the value we ultimately want to know.

We know the position $x_{end}$ at $t_{end}$, and we can express the position as a function of time: $$x_{t} = -\frac{f}{2m}t^2 + v_{0+}t$$ $$\therefore\quad x_{end} = -\frac{f}{2m}\left(\frac{m\,v_{0+}}{f}\right)^2 + v_{0+}\frac{m\,v_{0+}}{f} = \frac{3m\,v_{0+}^2}{2f} $$ $$\therefore\quad f = \frac{3}{2} \frac{m\,v_{0+}^2}{x_{end}} $$

Double-check my work of course.

  • $\begingroup$ If the collision takes no time the force exerted would be infinite. $\endgroup$
    – nasu
    May 28, 2019 at 20:09
  • $\begingroup$ Shape of matter,do u mean the cars stopped because of the 2nd car's exerting force on the 1st one? If that is so then I'm wrong about the fact that since both cars would be moving at the same speed after colliding, one of them won't be able to exert force on the other to cause that to accelerate $\endgroup$
    – user459284
    May 28, 2019 at 22:03
  • $\begingroup$ Shape of matter,do u mean that the 1st car stopped because of the 2nd car's exerting force on the 1st one even after the 2 cars have started moving at the same velocity? $\endgroup$
    – user459284
    May 28, 2019 at 22:19
  • $\begingroup$ @nasu; There are 2 things that could be meant by "collision": There's the transition period, during which car #2 is Accelerating. And there's the breaking period during which the two cars are DEcelerating together. With the information given, there's no way to calculate the forces during that first period (and no reason to suppose the force would be constant). Above I've assumed (among other things) that the first period is of negligible duration (as described here), and solved for the force in the second period. $\endgroup$ May 28, 2019 at 23:26
  • $\begingroup$ @user459284, Two objects can certainly exert force on each other while moving at the same velocity. To take the trivial case: If I push as hard as I can against a brick wall, the wall and I will both have velocity 0, but we'll be exerting huge forces on each other. $\endgroup$ May 28, 2019 at 23:28

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