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

In summary, the scenario discussed involves a car with a weight of 1000Kg and an engine that exerts a force of 3kN, accelerating the car at 3m/s^2 with no friction. After 100 seconds, the car hits a wagon with the same weight and no deformation occurs. The acceleration decreases to 1.5m/s^2 due to the combined weight of the car and wagon being 2000Kg. The force exerted on the wagon is 1kN and the force exerted on the car by the wagon is -1kN. When the car hits another wagon, the acceleration decreases to 1m/s^2 and the force exerted on the first wagon
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
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  • #6
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.
 
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  • #7
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.
 

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

1. How does the force exerted on a chain of wagons pushed by a car affect the speed of the wagons?

The force exerted on the chain of wagons by the car determines the acceleration of the wagons, which in turn affects their speed. The greater the force, the faster the wagons will accelerate and reach a higher speed.

2. Does the length of the chain of wagons affect the amount of force needed to move them?

Yes, the longer the chain of wagons, the more force will be required to push them. This is because the longer the chain, the more mass it contains, and more force is needed to overcome the inertia of the entire system.

3. How does the friction between the wheels of the car and the ground affect the force exerted on the chain of wagons?

The friction between the car's wheels and the ground is what allows the car to exert a force on the chain of wagons. Without friction, the car's wheels would spin without moving the wagons forward.

4. Is the force exerted on the chain of wagons by the car constant or does it change over time?

The force exerted on the chain of wagons by the car is not constant and can change over time. It depends on various factors such as the weight of the wagons, the terrain, and the speed of the car.

5. Can the force exerted on the chain of wagons by the car be increased by adding more wagons to the chain?

Yes, adding more wagons to the chain will increase the force required to move them. This is because more mass is added to the system, and more force is needed to overcome the inertia of the entire chain.

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