Simple collision between two moving bodies.

  • Thread starter Thread starter whammer
  • Start date Start date
  • Tags Tags
    bodies Collision
AI Thread Summary
In a collision between a truck and a car moving at the same velocity, each vehicle exerts a force on the other, consistent with Newton's third law, which states that forces are equal in magnitude and opposite in direction. The discussion highlights the misconception of treating the forces acting on each vehicle as separate when they are actually interdependent. The forces exerted by the truck on the car and vice versa are not independent; they are part of a single interaction. The analogy of colliding with a wall is questioned, as it introduces confusion regarding the nature of forces in motion versus stationary objects. Understanding these dynamics clarifies the mechanics of collisions in one dimension.
whammer
Messages
1
Reaction score
0
Truck and car collide in one dimension. Assume same velocity. I see this as having 4 forces, two on each body.

Each vehicle exerts its own force on the other. The truck exerts a force on car and vice versa. In addition, the truck's exertion of force on the car is reacted to via Newton's third law. and the car's exertion of force on the truck is reacted to via Newton's third law.

Taking the car individually, it has the force from the truck acting on it. Plus it has the reaction force from having acted upon the truck.

I suspect that my logic is flawed but I cannot see why. Should I treat each body's forces as though it slammed into a wall? You slam into the wall, exerting a force on it, whereupon the wall returns the favour. It just seems different to me. The vehicle has velocity whereas a wall does not.

Confused
 
Physics news on Phys.org
Hi whammer, welcome to PF.

When two bodies A and B interact, moving or not moving, A exerts a force FAB to B, and B exerts the force FBA to A. The forces are of equal magnitude and opposite direction. So there is only one force acting on a body. The reaction force a body exerts acts on the other body, not on itself.

ehild
whammer said:
Truck and car collide in one dimension. Assume same velocity. I see this as having 4 forces, two on each body.

Each vehicle exerts its own force on the other. The truck exerts a force on car and vice versa. In addition, the truck's exertion of force on the car is reacted to via Newton's third law. and the car's exertion of force on the truck is reacted to via Newton's third law.

Taking the car individually, it has the force from the truck acting on it. Plus it has the reaction force from having acted upon the truck.

I suspect that my logic is flawed but I cannot see why. Should I treat each body's forces as though it slammed into a wall? You slam into the wall, exerting a force on it, whereupon the wall returns the favour. It just seems different to me. The vehicle has velocity whereas a wall does not.

Confused
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Car Crash Analysis: Analyzing Friction & Speed
Replies
9
Views
4K
Why doesn't Newtons third law also apply to the frame?
Replies
2
Views
2K
Truck carrying a box, friction problem
Replies
10
Views
5K
Solving a Head-on Collision: Qs 1-4
Replies
5
Views
3K
Help understanding this collision question
Replies
3
Views
2K
Problem on Mechanics (Newton's law of motion)
Replies
4
Views
2K
Momentum: why don't the two carts become one?
Replies
32
Views
1K
How Does Gear Size Affect Torque in a Truck Transmission?
Replies
1
Views
1K
B Confused about force body diagram for 2 body collision
Replies
6
Views
2K
Force on the car in a collision
Replies
38
Views
3K

Hot Threads

Recent Insights

Back
Top