# What is the "actual" impact velocities of two colliding objects? Answered

Is it relative to the observer, or one of the objects in the collision?  The reason I ask is that  Marilyn Vos Savant answered a question about this in her column and I think she did it an injustice by saying that: an object moving 50 MPH towards another object moving, say 40 MPH (in the exact opposite direction), that the force of the impact will be.....the same as if each had struck a stationary wall at each of those speeds.  I think she is assuming a "perspective" and not taking the relative postion of "observer" or participant into consideration making her answer, inaccurate depending on perspective.  (I've seen figures on the "relative" speed of a baseball leaving a bat)

Any enlightenment on this would be appreciated.

Tags:

## Comments

The forums are retiring in 2021 and are now closed for new topics and comments.

In what the physics textbooks call a perfectly inelastic collision,
http://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision
two masses, m1 and m2, (and initial velocities v1 and v2 respectively) collide and then stick together, and then move together, stuck-together,  with a new shared velocity, v', pronounced  "v-prime".

Assuming the initial speeds of m1 and m2 were v1 and v2 respectively, then conservation of momentum requires

m1*v1 + m2*v2 = (m1+m2)*v'

And as a result:
v' = (m1*v1 + m2*v2)/(m1+m2)

Again v' is the speed at which the two stuck together masses, m1+m2, are moving after this perfectly inelastic collision.

(Sometimes this v' is called the velocity of the center of mass frame, because it is all the momentum in the problem divided by all the mass.)

Now consider the amount by which the velocity of each mass has changed as a result of this collision.  For each, this change in velocity is

For m1:  deltav1 = (v' - v1) = -(v1 - v')
For m2:  deltav2 = (v' - v2) = -(v2 - v')

and I think these are the numbers you want for this game of defining the severity of a collision in terms of a speed.  E.g. saying there is such a thing as  a "10 MPH collision", or a "50 MPH collision", and also for saying that such a v-speed collision is equivalent in severity to a collision with a "brick wall" (for a small mass moving a speed v and a brick wall moving at speed 0).

Note that the "brick wall" case is essentially what you get if you have one of the masses much larger than the other, say m2>>m1. Then v' is approximately v2.  So the two masses smack together, stick, and move together at whatever speed the brick wall was moving at. The small mass m1 sees a velocity change of deltav1 = -(v1-v2), and the big mass, the brick wall does not see its velocity change at all, since deltav2= -(v2-v2) = 0.

Also note that the collision between a bicycle-with-rider and car-with-driver is going to be approximately like bicycle-with-rider hitting a brick wall.  I mean suppose the car-with-driver has m2=1000kg, and moves to the right with v2 = 15m/s,and suppose the  bicycle-with-rider has m1 = 100 kg and moves to the left with v1 = -5 m/s.

Then
v' = (-500 kg*m/s + 15000 kg*m/s)/(1100 kg) = 13.2 m/s

And
deltav1 = - (-5 - 13.2) = 18.2 m/s
deltav2 = - (15 - 13.2) = 1.8 m/s

So in words I could say, to the bicyclist it feels like "a 18.2 m/s collision", and to the driver of the car it feels like "a 1.8 m/s collision".

And that's pretty close to the brick wall approximation, which says it would feel like a 20 m/s collision to the bicyclist, and like nothing to the brick-wall-car.

The numbers are different when m1 and m2 are close to the same size.

For two cars-with-driver with equal mass traveling colliding head on with opposite initial velocities, e.g. m1=m2= 1000kg, and v1 = 20 m/s, v2 = -20 m/s, then obviously it is a made up physics problem, because it has too much symmetry in it.
;-)

But, I do the same math as before.  It turns out v'= 0, and thus the delta v for each car is just

deltav1 = - (20 - 0) = -20 m/s
deltav2 = - (-20 - 0) = 20 m/s

As if each car were colliding with an invisible brick wall sandwiched in between the two cars.

And this pretty much agrees with what MVS wrote in her column, here:
http://www.parade.com/askmarilyn/2012/03/04-sunday-column.html

But I think she could have been more clear.

Although it makes me wonder if the answers I write are clear. Ha!
;-)

Re-figuring v' and deltav1, deltav2, for a car collision with less perfectly symmetric numbers, e.g m1=1000kg, m2=800kg, v1=15m/s, v2=-25m/s, is left as an exercise for the interested reader.

Ok, so maybe I was thinking perceived or "relative motion" and so the peception of impact is different....still the "original" question pitted a bicycle against an automobile, and I can not see a bike, heading into a wall, having the same "results" as one hit by a car going 60 mph.

The force of an object head-on with another equal object at the same speed will be the same as hitting a brick wall to a dead stop. The resultant force is the same, because the result is the same.

The difference is, there is 2x as much momentum, and 2x as much force, but 2x as much 'stuff' to absorb that force in the headon so people think it has more force. It just has more energy.

"equal object at the same speed'

no. it won't be.

sum of the forces of 2mph and -6mph are -4mph

if they were the same mass, it would split evenly, both would travel -2mph after the crunch

A car weighs approximately 20-50x what a bicyclist does...so their mass/momentum would undoubtedly steamroll (pardon the pun) through the 'comparatively not there' bicycle.
...or if you prefer
the sum of the momentums
150lbsx2mph vs 4000lbs x -6mph
300 forward mph-pounds, versus 24000 negative mph-pounds

the resultant vector would be -23700 mph-pounds of the combined 2...

and as I said, the resultant direction/speed will be near -5.9mph, and mass of 4150lbs
meaning the bicyclist accelerated from 2mph to -5.9mph, an acceleration of almost 8mph in a very short time...while the car went from 6 to 5.9, and acceleration of 1/80th what the cyclist went through.

If the cyclist just hit a wall and stopped dead, (again, pun not intended), they would go from 2 to 0.

You keep replying that you cannot find the original question.  I thought this was it:
http://www.parade.com/askmarilyn/2012/03/04-sunday-column.html

i.e.

Ned Radich, Fresno, Calif., writes:

Marilyn: When you wrote about why bicyclists should ride with ­traffic (and not against it), you neglected to mention the main reason: decreased ­impact speed in an accident. For example, if you’re riding a bike at 15 mph and you’re struck head-on by a car traveling at 35 mph, you’re exposed to a 50 mph (35 plus 15) impact. But if you’re struck from ­behind, you get a 20 mph (35 minus 15) impact.

Marilyn responds: ...

Just wanted to make sure everybody is on the same page here.

BTW, you noticed that Ned's math is almost right, or rather the Δv for the bicyclist  in Ned's statement  is approximately that predicted for the small mass in a  perfectly inelastic collision between two masses, with one much, much, larger than the other... or um, well I don't know if you noticed that or not.  But I think it's all there in the math in my previous answer.

For the problem above, the impact speed relative to one of the objects would be the sum of the absolute value of each speed.

So for the example posted, the impact speed would be

Abs(50) + Abs(-40) =
50 + 40 = 90mph.

If the objects collide not in a straight line, you would have to extract the composite vectors - that is, the value of speed in X and Y in case of a 2D situation.

For dummies:

Marilyn Vos Savant is saying that the force from a collision exerted on a person resulting from walking (5mph) towards a speeding train (250mph) would be equivalent for the same person to walk (5mph) into a wall.

Walking towards a speeding train is not the same as walking into a wall.

Could you link to or quote the column in question? The impact speed is the sum of the two, unless we're playing with relativity or something that weirds it up like that. Knowing the original statement and wording could bring out some nuance that makes her statement correct.

It doesn't matter what the objects are, the laws of physics still apply. Yes a car crash is much different than hitting a bike but the actual question is about the misconception that the impact speed of any two objects is the sum of the speeds of the objects and thus the impact is the same as a single impact at that sum velocity. You have to take into account mass and the conservation momentum.

maybe an experiment will help clear this up. http://www.dpccars.com/car-videos-10/05-08-10page-MythBusters-Crash-Force.htm

If you don't believe the maths check out Mythbusters 2010 Season (Season 8 I believe) , episode 7 ( ep 143).

Part of the confusion is that people don't know/remember that force = mass x acceleration, not velocity. Also the stationary wall example only holds true when the objects have the same mass and speeds (opposite velocities) but the principle still holds.

Marilyn is exactly right.
Think of it this way; two identical cars weighing 2000 pounds each are travelling at exactly 25 miles per hour and collide dead center, head on. The force each car receives from the impact is the same.
Each car imparts a force in an amount equivalent to its weight (mass) times its speed. If one car travelling at 25 mph hit a concrete wall it would generate 50,000 pounds of force (F=MV).
The two cars in the above example together generate 100,000 lbs of force in a head on collision. Since each car is identical and the speed is the same, each car would impart 50,000 lbs. of force to the other car, exactly the same as colliding with the stationary wall at 25 mph.
In order for both of them to receive an amount of force equivalent to travelling 50 mph there needs to be an additional 100,000 lbs of force. Where would that additional force come from? It doesn't and therefore, the head on collision is the same as each car hitting a brick wall at 25 mph.
In the example of the train and the bicycle colorex posed, the impact force of the train on the bicyclist would be massive, the bicyclist's impact force on the train is almost negligilble.