Newton's Third Law and Car Crashes

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SUMMARY

The discussion centers on the application of Newton's Third Law in car crash scenarios, specifically comparing a head-on collision between two cars of equal mass traveling at 50 km/h to a single car traveling at 100 km/h crashing into a wall. It concludes that the forces experienced during the collision are less severe in the case of two colliding cars, as each car exerts an opposing force on the other, resulting in a lower net force compared to the wall's reaction force on the faster car. The fundamental equation F=ma illustrates that greater velocity results in a higher force when brought to a stop over the same time period.

PREREQUISITES
  • Understanding of Newton's Third Law
  • Familiarity with the equation F=ma (Force equals mass times acceleration)
  • Basic knowledge of collision physics
  • Concept of impulse in physics
NEXT STEPS
  • Study the implications of Newton's Laws in real-world crash scenarios
  • Research the concept of impulse and its relation to collisions
  • Explore the effects of varying mass in collision dynamics
  • Learn about energy transfer during car crashes and its safety implications
USEFUL FOR

Physics students, automotive safety engineers, and anyone interested in understanding the dynamics of car collisions and the application of Newton's Laws in real-world situations.

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Homework Statement


In terms of Newton's Third Law, why is it better for 2 cars of the same mass, both going 50km/h to crash into each other than for 1 car going 100km/h to crash into a wall?

Homework Equations


Newton's Third Law
F=ma

The Attempt at a Solution


I have no idea. I'm really confused
 
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what do you mean in better
 
If there are two cars, each one will be subjected to the other's opposing force. If there is one car, and you assume the wall is strong enough to fully stop the car, it will be subjected to the larger resisting force of the wall.

Let's say that ##F_1## is generated by car 1 and ##F_2## is generated by car 2, then ##F_1= -F_2## for a perfect head-on crash ending in full stop.
Let's say that ##F_3## is generated by the fast (100km/hr) car 3 and ##F_4## is generated by the wall, then ##F_3= -F_4## for a car-on-wall crash ending in full stop.
 
giokrutoi said:
what do you mean in better

Less damaging
 
RUber said:
If there are two cars, each one will be subjected to the other's opposing force. If there is one car, and you assume the wall is strong enough to fully stop the car, it will be subjected to the larger resisting force of the wall.

Let's say that ##F_1## is generated by car 1 and ##F_2## is generated by car 2, then ##F_1= -F_2## for a perfect head-on crash ending in full stop.
Let's say that ##F_3## is generated by the fast (100km/hr) car 3 and ##F_4## is generated by the wall, then ##F_3= -F_4## for a car-on-wall crash ending in full stop.
Of the 2 cars, each exerts a lesser force since their collision velocity is less. Whereas if the car crashed into the wall, the wall would exert a stronger reaction force on the car because of the increased velocity at the point of impact? Is this true?
 
That's about right. I would assume that in all cases, it takes the same time for the cars to come to a stop. Thus, acceleration would be proportional to velocity.
More velocity brought down to zero velocity over the same time results in more force.
There are finer points when you deal with impulses and collisions, but as it refers to F=ma, this is the most fundamental explanation.
More velocity, same mass, same time => more force.
 
RUber said:
That's about right. I would assume that in all cases, it takes the same time for the cars to come to a stop. Thus, acceleration would be proportional to velocity.
More velocity brought down to zero velocity over the same time results in more force.
There are finer points when you deal with impulses and collisions, but as it refers to F=ma, this is the most fundamental explanation.
More velocity, same mass, same time => more force.

What if the masses were different for the 2 cars? I'm not sure how mass affects this situation
 
Then you would not expect both cars to come to a full stop.
 

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