# Physics teacher needs help with lab on Newton's laws

## Main Question or Discussion Point

Hello! I am a high school teacher and I am doing a lab on Newton's laws. I need help interpreting part of the lab because the results did not come out to what the laws would have predicted.

The lab consisted of setting up two carts, one with a spring which can be compressed and one without. The first part of the experiment says the students should depress the spring and put the carts together. When released, they both go opposite ways with the same acceleration. So far so good, Newton's 3rd law tells us there will be an action-reaction pair and the 2nd law tells us that a=F/m.

Next, a 1 kg mass is added to the cart with the spring. Again the spring is compressed, they are put together, and the spring is released. Uh-oh. Now here is the problem. We know that again, there is an action and reaction pair, so the force will be the same on both carts. Since the mass of the second cart is greater, its acceleration will be smaller. But what about the cart with no weight. Correct me if I am wrong, but the acceleration here is again equal to F/m where F = -kx (Hooke's law). The force has not changed, the mass has not changed, so the acceleration should be the same as it was in the first trial. Right? All of my students reported a greater acceleration in the second trial for the cart with no added mass than in the first trial. Also, as the mass on the other cart increased to 2kg, the acceleration on the cart with no mass added increased again. Am I missing something, or are there real world factors that would make account for these observations? I considered conservation of momentum, but this does not change anything as the increase in mass of the first cart should be offset exactly by the decrease in its velocity

Thank you so much if you took the time to read this and reply.

Peter

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nrqed
Homework Helper
Gold Member
Hello! I am a high school teacher and I am doing a lab on Newton's laws. I need help interpreting part of the lab because the results did not come out to what the laws would have predicted.

The lab consisted of setting up two carts, one with a spring which can be compressed and one without. The first part of the experiment says the students should depress the spring and put the carts together. When released, they both go opposite ways with the same acceleration. So far so good, Newton's 3rd law tells us there will be an action-reaction pair and the 2nd law tells us that a=F/m.

Next, a 1 kg mass is added to the cart with the spring. Again the spring is compressed, they are put together, and the spring is released. Uh-oh. Now here is the problem. We know that again, there is an action and reaction pair, so the force will be the same on both carts. Since the mass of the second cart is greater, its acceleration will be smaller. But what about the cart with no weight. Correct me if I am wrong, but the acceleration here is again equal to F/m where F = -kx (Hooke's law). The force has not changed, the mass has not changed, so the acceleration should be the same as it was in the first trial. Right? All of my students reported a greater acceleration in the second trial for the cart with no added mass than in the first trial. Also, as the mass on the other cart increased to 2kg, the acceleration on the cart with no mass added increased again. Am I missing something, or are there real world factors that would make account for these observations? I considered conservation of momentum, but this does not change anything as the increase in mass of the first cart should be offset exactly by the decrease in its velocity

Thank you so much if you took the time to read this and reply.

Peter

Just a thought. The force does not act at a single time. It keeps acting as long as the two objects are in contact. Do you have any way to estimate the amount of time the lower mass cart remains in contact with the spring?

So my intuition tells me that when you increase the mass of cart A, because it accelerates less the other cart remains on contact with the spring for a longer period of time and that ends up creating a larger force on it.
This makes sense if you think of the limit as the mass of cart A goes to infinity. Then basically it's as if cart A is a wall. Try it: connect cart B to the spring connected to a wall and measure its acceleration. It should be even larger than when it was connected to cart A with 2 kg on it.

Thanks!! That makes sense. I appreciate the help.

Peter

Like nrged said, the force acts for longer than an instant. As a physics teacher, you should be familiar with the idea the the impulse of a system is equal to the momentum change.
Try applying the equation: $F{\Delta}t={\Delta}mv$. Let us know how that works.

Yes, that works too. I am of course familiar with the impulse-momentum equlivalency, just an oversight on my part to take it in to consideration.

nrqed
Homework Helper
Gold Member
Thanks!! That makes sense. I appreciate the help.

Peter
You are very welcome.
Glad I could help a fellow teacher

LURCH