Solving the Mystery of Two Cart Exploding Force

In summary: V1 = 5*V2V2 is 5 times bigger.In summary, the conversation discusses an experiment involving two carts and how their acceleration is affected by adding mass to one of the carts. The first trial shows that the acceleration of the red cart is decreased by a factor of five when its mass is increased by a factor of five. The question then asks for a prediction of the acceleration of the blue cart in the third trial, which has 1000 g of mass added to the red cart. The correct answer is that the blue cart's acceleration is increased by a factor of five compared to the first trial. The conversation also touches on the concept of Conservation of Momentum and its role in understanding the relationship between mass
  • #1
lolohhi
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Homework Statement


I feel silly posting such a simple problem, but a substantial amount of thinking, Googling, and scouring these forums has not yielded an answer.

An experiment involving the "explosion" of two carts is conducted where an internal plunger in one of the carts is triggered, forcing the two carts apart. Motion sensors on the ends of the track record the acceleration of the carts as they move away from each other.

Three successive trials are conducted. In the first trial, the plunger is triggered in the cart forcing the two 250-g carts away from each other.

This is repeated twice more, first with an additional 500-g mass on one of the carts, then again with an additional 1000 g on one of the carts.

In the first trial (no added mass), the acceleration of the red cart was 6.00 m/s2
.

...

In the third trial (1000 g of mass added to the red cart), the acceleration of the red cart is decreased by a factor of 5 compared to the first trial. What is a reasonable prediction for the acceleration of the blue cart in the third trial?​



Homework Equations


Newton's Second Law: F*_net = ma* (where the asterisk denotes a vector quantity)
Newton's Third Law: F*_(A on B) = F*_(B on A)


The Attempt at a Solution


I reasoned that the acceleration of the blue cart remains constant regardless of the mass of the red cart. Given the statement that the red car's acceleration is reduced by a factor of five when its mass is increased by a factor of five (from 250 g to 1.25 kg), according to (F*_net = ma*), I would expect that F*_net on the red cart remains constant. Since the magnitudes of F*_net_red and F*_net_blue must be equal by Newton's Third Law, I reasoned that the force exerted on the blue cart must be the same regardless of the mass of the red cart, given that the net force exerted on the red cart seems to remain constant regardless of its mass. As the mass of the blue cart remains constant, if the net force exerted on it remains so as well, then its acceleration must remain constant, too. To explain the correct answer in which the blue cart moves with greater acceleration when the mass of the red cart is increased, one must assume that the red cart exerts a greater force on the blue cart when the mass of the red cart is increased, which makes no sense to me whatsoever.

Thinking in less abstract and more intuitive terms, I'm still confounded. On the one hand, increasing the mass of the red cart should not increase the force with which its plunger extends, and thus the blue cart should have the same force exerted on it regardless of the mass of the red cart. On the other hand, if I imagine increasing the mass of the red cart to several thousand times its current value, it seems the reaction force of the blue cart pushing against the red cart would not move the red cart at all; if the red cart didn't move, less motion would be "wasted" moving the red cart (if that makes any sense at all) and the blue cart would move away faster than it did when the red cart had a lesser mass.

The correct answer is that the blue cart's acceleration increased by a factor of five when the mass of the red cart was increased by a factor of five. I still don't understand why the blue cart wouldn't move away at the same acceleration regardless of the mass of the red cart. Any assistance you provide will be much appreciated.​
 
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  • #2


Welcome to PF.

I think you want to think about the Conservation of Momentum.

Before the plunger V = 0 for both.

That means that the momentum of 1 is the opposite of the other after release.

Momentum is M*V so that

M1V1 = M2V2

Masses the same ... Velocities the same.

What happens then when M1 = 5*M2? What must V2 be relative to V1?

5*M2*V1 = M2*V2
 
  • #3




Thank you for sharing your thoughts and reasoning on this problem. It is always important to question and analyze results in scientific experiments, even if they seem simple at first glance. In this case, it seems that there may be some confusion about the forces involved in the experiment.

First, it is important to note that the force exerted by the plunger in the red cart is not the only force acting on the system. In fact, there are multiple forces at play here, including the force of gravity, the force of friction, and the force of the plunger. The acceleration of each cart is determined by the net force acting on it, which takes into account all of these forces.

In the first trial, when there is no added mass, the net force on the red cart is equal to the force of the plunger, which is also the net force on the blue cart due to Newton's Third Law. This results in an acceleration of 6.00 m/s^2 for both carts.

In the second trial, when an additional 500 g mass is added to the red cart, the net force on the red cart is now greater than the force of the plunger. This is because the added mass increases the force of gravity acting on the red cart, which must be overcome by the plunger to produce the same acceleration as in the first trial. As a result, the net force on the blue cart is also greater, resulting in a greater acceleration for the blue cart.

In the third trial, when 1000 g of mass is added to the red cart, the same principle applies. The added mass increases the force of gravity acting on the red cart, requiring a greater force from the plunger to produce the same acceleration as in the first trial. This increased force on the red cart also results in a greater force on the blue cart, and thus a greater acceleration for the blue cart.

In summary, the key concept to understand here is that the net force on each cart is determined by all of the forces acting on it, not just the force of the plunger. As the mass of the red cart increases, the forces of gravity and friction also increase, requiring a greater force from the plunger to maintain the same acceleration. This in turn results in a greater acceleration for the blue cart due to Newton's Third Law.

I hope this explanation helps clarify the results of the experiment. Keep questioning and analyzing results in your scientific pursuits!
 

FAQ: Solving the Mystery of Two Cart Exploding Force

1. What is the main goal of solving the mystery of two cart exploding force?

The main goal is to understand the force that causes two carts to explode when they collide with each other.

2. What are the potential factors that could contribute to the explosion of the two carts?

Some potential factors include the speed and direction of the carts, the materials they are made of, and the angle of impact.

3. How can we conduct experiments to determine the exploding force of two carts?

We can conduct experiments by adjusting the variables mentioned above and recording the results of the collisions. We can also use mathematical equations to calculate the force.

4. What are some real-life applications of understanding the exploding force of two carts?

Understanding this force can help in designing safer transportation systems, improving car safety features, and preventing accidents in industries that use moving carts or vehicles.

5. Are there any ethical considerations in studying this phenomenon?

Yes, there may be ethical considerations in conducting experiments that involve collisions and explosions. Proper safety precautions should be taken, and the well-being of any living beings involved in the experiments should be prioritized.

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