Force exerted on a chain of wagons pushed by a car..

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Discussion Overview

The discussion revolves around the forces exerted on a chain of wagons pushed by a car, focusing on the dynamics of collisions, momentum conservation, and the resulting accelerations and forces in a hypothetical scenario involving a car and multiple wagons. Participants explore various aspects of Newtonian mechanics, including acceleration, force calculations, and the implications of collisions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that after a collision, the speed of the car does not decrease, while another argues that momentum is conserved, suggesting the speed of the car will decrease.
  • Calculations for the acceleration of the car and wagons are presented, with one participant stating the new acceleration after a collision is 1.5 m/s² and later 1 m/s² as more wagons are added.
  • There is a discussion about the forces exerted on the wagons, with claims that the force exerted by the car on the first wagon is 1 kN, but another participant challenges this by stating there are additional forces acting on the wagon.
  • A hypothetical scenario involving a 10 kg ball colliding with many smaller balls leads to questions about the forces during the collision and whether any force is exerted on the 10 kg ball.
  • Participants discuss the implications of perfectly elastic collisions and the unrealistic assumptions that lead to infinite forces, emphasizing the need for more information to calculate actual forces.

Areas of Agreement / Disagreement

Participants express differing views on the effects of collisions on speed and force, with no consensus reached on the calculations and implications of forces during collisions. The discussion remains unresolved regarding the exact nature of forces exerted during the hypothetical scenarios presented.

Contextual Notes

Limitations include assumptions about perfectly rigid and incompressible objects, which may not reflect real-world behavior. The discussions also highlight the dependence on definitions of forces and the need for time frames in force calculations.

Karagoz
Newtonpng.png


Imagine a car that weights 1000Kg. Its engine pushes it forward with a force of 3kN. So the car is accelerating at 3m/s^2 (imagine there are no friction).

After 100 seconds (the speed of car at that moment is 300m/s, very realistic), the car hits a wagon that weighs 1000Kg, that is on rest. And imagine both the car and the wagon is so hard that no deformation happens because of the hit.

From my understanding, the speed of the car doesn't decrease after the hit. But the acceleration decreases from 3m/s^2 to 1.5m/s^2. It's because:
The car and the wagon totally weights 2000Kg. The force on the car is still at 3kN.
So then the new acceleration must be: 3000N / 2000Kg = 1.5m/s^2.

Force exerted on the wagon is 1kN:
Since the wagon is accelerating at 1.5m/s^2 and weights 1000Kg, the force exerted on it must be: 1.5*1000 = 1.5kN.

And the force exerted from the wagon to the car must be equal, which is: -1.5kN (negative since the direction is opposite)

The car drives with a wagon on the front side, after some seconds the car and wagon hits another wagon. And from now the car pushes two wagons forward.
The new acceleration is 1m/s^2 (3kN / 3000Kg = 1).

The force exerted on the first wagon from the car is 1kN:
Since the first wagon is accelerating at 1m/s^2 and weighs 1000Kg, the force exerted on the wagon from the car is 1kN.
And the force exerted on the car from the wagon is -1kN.

The same is true for the second wagon too. The force exerted on the second wagon from the first wagon is too 1kN. And the force exerted on the first wagon from the second wagon is
-1kN.

Are the calculations correct?

Why is the force exerted on the car (that makes the car accelerate) is less than the force the car exerts on the first wagon?
When the wagon and the car weighs same and have same velocity and acceleration, shouldn't the force exerted on both be the equal?

And why the force exerted by the car on the first wagon (and the force exerted by the first wagon on the car) decrease when the number of wagons increase?
 

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Karagoz said:
From my understanding, the speed of the car doesn't decrease after the hit.
Momentum is conserved. The speed of the car will decrease after the hit. But that does not affect your acceleration calculations.

[Now the car is pushing one wagon. It hits the second and...]
Karagoz said:
The new acceleration is 1m/s^2 (3kN / 3000Kg = 1).

The force exerted on the first wagon from the car is 1kN:
Yes, the acceleration must be 1 m/s^2. The net force on the first wagon must be 1 kN to fit the calculated acceleration. But that does not mean that the force from the car on the first wagon is 1 kN. There is another force acting on the first wagon. It is the sum of those forces that has to add to 1 kN.
 
jbriggs444 said:
Momentum is conserved. The speed of the car will decrease after the hit. But that does not affect your acceleration calculations.

So this means:
1000 Kg * 100m/s = 2000Kg * x
x = 50m/s.
The new speed after the hitting monent will be 50m/s right? (and it'll incrrase since it accelerates).

jbriggs444 said:
Yes, the acceleration must be 1 m/s^2. The net force on the first wagon must be 1 kN to fit the calculated acceleration. But that does not mean that the force from the car on the first wagon is 1 kN. There is another force acting on the first wagon. It is the sum of those forces that has to add to 1 kN.

So this means:
Force on first wagon from the car:
1m/s2 * 2000Kg = 2kN.

Force on first wagon from the second wagon is:
1m/s2*1000Kg = -1kN.

Sum of the forces on the first wagon is
2kN - 1kN = 1kN.

And sum of the forces on the car is:
3kN - 2kN = 1kN.What about this hypothetical scenario:

A ball made if steel weighing 10 Kg is shot with a force of
1kN out to space where there's no friction or any other force that affect the ball.

So the ball has constant velocity of 100m/s.

Then ball hits another 1 million small balls made of steel at rest on space that weighs each 1g. Sum weigh of them is 1000kg.

The new speed is:
(10kg*100m/s)/1000kg = 0.11m/s.

So the first ball will still move (keep the momentum) but just at slower and slower speed, and it means the ball will never stop even if it hits 1018 of other 1g balls?What's the force exerted on 10kg ball?

Since it slows down from 100m/s to 0.11 m/s, the force exerted on the 10kg ball is:
-99.89m/s * 10kg = -998.9 N.

Or no force is exerted on the ball when hitting?
 
Karagoz said:
So this means:
1000 Kg * 100m/s = 2000Kg * x
x = 50m/s.
The new speed after the hitting monent will be 50m/s right? (and it'll incrrase since it accelerates).
Exactly right.
Karagoz said:
So this means:
Force on first wagon from the car:
1m/s2 * 2000Kg = 2kN.

Force on first wagon from the second wagon is:
1m/s2*1000Kg = -1kN.

Sum of the forces on the first wagon is
2kN - 1kN = 1kN.

And sum of the forces on the car is:
3kN - 2kN = 1kN.
Right you are.
Karagoz said:
The new speed is:
(10kg*100m/s)/1000kg = 0.11m/s.
Arithmetic error?
Karagoz said:
Since it slows down from 100m/s to 0.11 m/s, the force exerted on the 10kg ball is:
-99.89m/s * 10kg = -998.9 N.
Huh? Force is a rate of change of momentum per unit time. You never mentioned a time frame, so you cannot compute a force.
 
The new speed should be this:
(10kg*100m/s) / 1010kg = 0.99 m/s.

jbriggs444 said:
Huh? Force is a rate of change of momentum per unit time. You never mentioned a time frame, so you cannot compute a force.

So, since momentum is the same before and after the 10kg ball hits 106 x 1g balls, it mean no force is exerted on the 10kg ball when it hits one million 1g balls?
 
Last edited by a moderator:
Karagoz said:
So, since momentum is the same before and after the 10kg ball hits 106 x 1g balls, it mean no force is exerted on the 10kg ball whe it hits one million 1g balls?
No, there's definitely a force between the balls. However, we don't have enough information to calculate what it is. If the colliding objects are perfectly rigid and incompressible and the collision is perfectly elastic the calculation will come out to an infinite force acting for zero time - and we know that's not what's really happening, it's an artifact of the unrealistic assumptions that went into the calculation. With any rea objects made of real materials that are not perfectly tried and incompressible we'll calculate a very high force acting for a very short time, but we need these details to do the calculation.
 
Last edited:
Karagoz said:
So, since momentum is the same before and after the 10kg ball hits 106 x 1g balls, it mean no force is exerted on the 10kg ball when it hits one million 1g balls?
There is no external force on the set of one million and one balls. The total momentum of the combined system remains unchanged.

There is a [series of] [impulsive] force(s) on the 10 kg ball. Its momentum changes as a result.
 

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