Can I calculate force in the following situation if the following information are given?

Question

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?

Answer 1

If we assume:

• The "collision" takes no time and is perfectly inelastic.
• The first car is friction-less.
• The second car has the parking break engaged.
• The parking break applies constant frictional deceleration once the second car is in motion. (This is a reasonable approximation of how car breaks behave)

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.