Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum  3Likes
Mar 23rd 2018, 04:02 AM

#1  Junior Member
Join Date: Mar 2018
Posts: 2
 Energy in a crash
Hello! I've been intrigued by the following question for a while now and I was hoping you could help me out
Two cars are driving 36 km/h and are coming from opposite directions.
(like this: car 10m/s > < car 10 m/s)
They each weigh 1000 kg. Then they crash. What is the relative speed and energy of this crash?
My thoughts:
I think most people will say the relative speed of the crash is 20 m/s, and therefore the energy in the crash is according to E=0.5*m*v^2,
= 0,5*1000*20^2 = 200000
But if you calculate the energy that each car individually has, it's only E = 0,5*1000*10^2, = 50000. So with two cars the total energy can only be 100000?

 
Mar 23rd 2018, 04:28 AM

#2  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 424

Yes, you're correct. It makes sense to consider the individual contributions of kinetic energy made by both cars using an external observer as the main reference frame.
Note that there's nothing wrong with calculating the kinetic energy using the relative speed and using one of the cars as a reference frame, but then all of the resulting dynamics (e.g. debris, deformation, etc.) would have to be calculated in the same moving reference frame using that energy budget, which might look a bit odd since the collision is likely going to decelerate both cars rapidly to near zero velocity after collision. For a headon collision it's probably better to use the ground as the frame of reference.
Last edited by benit13; Mar 23rd 2018 at 04:59 AM.

 
Mar 23rd 2018, 04:42 AM

#3  Senior Member
Join Date: Apr 2017
Posts: 498

They both travel at 10m/sec , if everything comes to rest your calculation is correct and the total energy expended is 100,000J
But you have raised an interesting point ... if your frame of reference is located on one of the cars , this car can be seen as stationary , the other car impacts at 20m/sec ...
So why isn't the energy released 200,000?? ...
Because your frame of reference is not allowed to change velocity , so the cars will crash and come to rest , but not rest with reference to your frame of reference , from that perspective both crashed stationary cars will be moving 10m/sec...
From this frame of reference the two cars still have kinetic energy but negative 100,000
So from the moving frame of reference energy is +200,000 and  100,000 which makes 100,000 the same as from a stationary perspective..
Last edited by oz93666; Mar 23rd 2018 at 04:49 AM.

 
Mar 23rd 2018, 05:05 AM

#4  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 424

Originally Posted by oz93666 Because your frame of reference is not allowed to change velocity , so the cars will crash and come to rest , but not rest with reference to your frame of reference , from that perspective both crashed stationary cars will be moving 10m/sec...
From this frame of reference the two cars still have kinetic energy but negative 100,000 
You're quite right Oz, but I think you mean + 100,000 J, not 100,000 J.
Before: 200,000 J
After: 100,000 J
so 100,000 J has been exchanged between the two cars in the moving reference frame, just like the case for the stationary reference frame.

 
Mar 23rd 2018, 05:22 AM

#5  Senior Member
Join Date: Apr 2017
Posts: 498

Originally Posted by benit13 You're quite right Oz, but I think you mean + 100,000 J, not 100,000 J.
Before: 200,000 J
After: 100,000 J
so 100,000 J has been exchanged between the two cars in the moving reference frame, just like the case for the stationary reference frame. 
Well , I'm not sure ...I was making it up as I went along ....
The energy in one car moving at 20 is 200,000J
It has to be 100,000J I agree rather a strange idea negative kinetic energy , but it does make the numbers work out ....
I think I stand by my original post .

 
Mar 23rd 2018, 05:43 AM

#6  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 424

Originally Posted by oz93666 Well , I'm not sure ...I was making it up as I went along ....
The energy in one car moving at 20 is 200,000J
It has to be 100,000J I agree rather a strange idea negative kinetic energy , but it does make the numbers work out ....
I think I stand by my original post . 
Negative energy is impossible. If the cars have 100,000 J after the crash, the total difference is 300,000 Joules. Furthermore, the equation for kinetic energy is
$\displaystyle KE = \frac{1}{2} m v^2$
So the only way to get negative kinetic energy is to have negative mass, which is not the case for cars.

 
Mar 23rd 2018, 06:42 AM

#7  Senior Member
Join Date: Apr 2017
Posts: 498

Originally Posted by benit13 Negative energy is impossible. If the cars have 100,000 J after the crash, the total difference is 300,000 Joules. Furthermore, the equation for kinetic energy is
$\displaystyle KE = \frac{1}{2} m v^2$
So the only way to get negative kinetic energy is to have negative mass, which is not the case for cars. 
I think if you read my post carefully it's OK ....
If the frame of reference is on one car we could be forgiven for thinking the energy at impact is 200,000 , but it can't be , and this is explained because the frame of reference is moving which gives an apparent KE of the crashed stationary cars of 100,000 in the other direction ...we have to subtract it from the 200,000
Hopefully other members will give their thoughts.
Let's look at things in another way .... our frame of reference is on one car , the other impacts at 20m/sec , but they will not come to rest in our frame of reference the whole crashed mass is travelling at 10m/sec
energy in colliding car at 20m/sec = 200,000
energy in moving mass of crashed two cars = 100,000
these two have to be subtracted to get the correct figure of 100,000
It's a mathematical convenience and is valid to have one figure negative .
Last edited by oz93666; Mar 23rd 2018 at 06:55 AM.

 
Mar 23rd 2018, 07:15 AM

#8  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 424

Your post was perfect until this point:
It's a mathematical convenience and is valid to have one figure negative .

It's neither mathematically convenient nor valid to have a negative kinetic energy, it's incorrect. If you were talking about potential energy (e.g. electron energy levels in atoms) then you'd have a point, but not here.
Sorry to be pedantic about this, but I want to make sure the OP understands that you can't just go making energy values negative (at least without consequences), especially in macroscopic motion problems. They can lead to errors. For example, a confused student might try to find the velocity of an 1000 kg object with kinetic energy equal to 100,000 J...
Last edited by benit13; Mar 23rd 2018 at 07:22 AM.

 
Mar 23rd 2018, 07:49 AM

#9  Senior Member
Join Date: Apr 2017
Posts: 498

Originally Posted by benit13 It's neither mathematically convenient nor valid to have a negative kinetic energy, it's incorrect. 
Let's use an analogy ..... with potential energy ....
I define my table as the reference point ...
Anything on the table has zero PE . If I lift an object from the table to the ceiling it has positive PE ...
If I put something on the floor it has negative potential energy within the frame of reference I defined ...
Of course there is no such thing in reality , but within the system I have created it is valid to say the object on the floor has negative energy ....
Interesting thread .... time for sleep where I am (thailand ) ....will sleep on it all and check in tomorrow ....

 
Mar 23rd 2018, 08:02 AM

#10  Senior Member
Join Date: Oct 2017 Location: Glasgow
Posts: 424

Negative potential energy, yes, but not negative kinetic energy.

  Thread Tools   Display Modes  Linear Mode  