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My friend is trying to tell me the force of an impact of riding a bicycle into a stationary car at a speed $v$ will be the same force as if the car is traveling at the same speed $v$ and drives into a stationary bicycle? I don’t see how this makes any sense. Can you tell me if this is a true statement? I would imagine you would get much more force if the car drove into the bicycle rather then the bicycle into the car even though they are both going the same speed.

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  • $\begingroup$ The wording suggests a two-body problem, but a real collision with two vehicles also includes the road: Earth, a massive stationary third body. $\endgroup$ – Whit3rd May 11 '18 at 8:50
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Consider the single head-on collision between a car and a bicycle, seen and studied by two observers. The bicycle (with rider) and car remain locked together after the collision.

One observer, (A), is using the frame of reference of the car as it is just before the collision; to him, the car is at rest and the bicycle comes up and smashes into the car. If the car can roll freely, the combined wreckage will roll off slowly.

Another observer, (B), uses the frame of reference of the bicycle as it is just before the collision; to him, the bicycle is standing still, and the car comes up and smashes into the bicycle. The cycle is scooped up and carried off by the car, which slows slightly.

Each observer comes complete with video and sound recorders.

They will certainly disagree as to the initial and final velocities of the various pieces of debris after the collision; that's why different frames of reference are, well, different.

But they certainly will record the same level of damage to the car, bike and, sadly, rider. There's only one collision. The rider can't be slightly bruised as seen by B and mangled as seen by A. The forces must be the same for both observers.

And most importantly, none of this will change if A also observes the fact that the ground is at rest in his frame, or B observes this, or neither one of them observes this. After all, an observer using the ground frame of reference may see a bike racing down the highway and being hit from in front (or even behind) by the car...

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The force of impact is not dependent on the change in speed (i.e., a bike crashing into a car with some initial speed having zero speed after the crash) but rather on the change in momentum. Change in momentum of an object is given by the product of its mass and its change in speed, so because a car has much more mass than a bike, the change in speed will be much less to produce the same change in momentum as a bike at a high speed. Thus, a car traveling at some speed would have a higher impact force than a bike at the same speed.

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  • $\begingroup$ How does this address the original question? $\endgroup$ – DJohnM May 11 '18 at 5:58
  • $\begingroup$ OP is asking about the difference in force of impact between a car and a bike. I try to explain that the force of impact is dependent on the change in momentum and is therefore significantly affected by the mass, contrary to the friend's statement. Edited to be a bit more conclusive. $\endgroup$ – Billy Kalfus May 11 '18 at 6:01
  • $\begingroup$ Are there two collisions: a car-bike collision and a car-car collision? $\endgroup$ – DJohnM May 11 '18 at 6:07
  • $\begingroup$ That is how I interpreted the question. If I misinterpreted it I welcome the OP to correct me. $\endgroup$ – Billy Kalfus May 11 '18 at 6:08
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Your friend is correct. If a bicycle of mass $m$ travelling with speed $v$ runs into a stationary car of mass $M$ then its speed changes to 0 (assuming $m<<M$) and the change in momentum, the impulse, is $mv$. This can also be thought of as the average force times the time of the collision. If the car has speed $v$ and the bicycle is stationary then after the collision the bicycle has speed $v$, momentum $mv$, so the impulse and the average force are the same.

Which is counter-intuitive. As a cyclist I know I'd rather hit a car than be hit by a car, if I had to do one or the other. The difference is that after the first collision I and my mangled bicycle are at rest on the road, whereas after the second I'm traveling at speed, I'm going to fall off into the road and sustain further injures, possibly including going under the wheels of the car.

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  • $\begingroup$ Of course, the situation of the bike hitting the car is not actually equal to the car hitting the bike. In the real world, the bike would stop quickly after hitting the car, whereas the car would tend to keep rolling after hitting the bike. When the car hits the bike, there is a high probability that the bicyclist will get run over by the car. $\endgroup$ – David White Feb 5 at 3:14
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The change in momentum is the impulse $\int_{\rm collision} F\,dt$ and this will be the same for all situations as will the variation with time of the force of interaction between the car and the bicycle.

Let the mass of the car be $M$ and the mass of the bicycle (and rider?) be $m$ and assume that initial speed of the car (Case 1) and the bicycle (Case 2) be $v$ with the other vehicle initially at rest.
Further assume that before the collision the stationary vehicle do not have the brakes on and after the collision the car and the bicycle move off together.

Case 1 - car moving

Using conservation of momentum the final speed of the two vehicles is $\dfrac{Mv}{M+m}$, the change in momentum of the bicycle is $\dfrac{mMv}{M+m}$ and the decrease in kinetic energy of the system (car and bicycle) is $\dfrac12\dfrac{mM}{M+m}v^2$.

Case 2 - bicycle moving

Using conservation of momentum the final speed of the two vehicles is $\dfrac{mv}{M+m}$, the change in momentum of the bicycle is $\dfrac{mMv}{M+m}$ and the decrease in kinetic energy of the system (car and bicycle) is $\dfrac12\dfrac{mM}{M+m}v^2$.

Note that the change in momentum of the bicycle and the loss of kinetic energy is the same in both cases.
What does all this tell us?

Well you could analyse a Case 3 where both the car and the bicycle are moving but their relative speed of approach is $v$ to find that the change of momentum of the bicycle and the kinetic energy loss is the same as before.
A good way of analysing such a situation is to jump into the centre of mass system (car and bicycle)
In the centre of mass frame the velocity of the car will be $\dfrac{mv}{M+m}$ and the velocity of the bicycle will be $\dfrac{Mv}{M+m}$ in the opposite direction to that of the car irrespective of the initial velocities of the car and the bicycle as long as their relative velocity of approach is $v$.

This is a good example of Newton's first law in action.
In case 1 with the car moving you could say that the frame of reference is the bicycle, in case 2 the frame of reference is the car and in case 3 the frame of reference is the Earth or the centre of mass of the system and all of these frames of reference are inertial.


In the real world the collision between a car and a bicycle would have many phases and not just be a two body interaction.

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