MHB Non-Impulsive Forces: What is Fair to Assume?

[csmith23]

I was reworking a problem from class today, and made a simple assumption, in regards to the problem, that made this problem much easier.

My question: Is it fair to assume that the velocity of Va = Vb, after impact? The weights are non-impulsive, since the difference in weight is small.

I ask this because my teacher found Vb by using coefficients of restitution. My assumption gave me the same answer as my teacher. there is a picture that goes with the problem, but for my question I believe its irrelevant.

Question: The 0.5 kg cylinder A is released from rest from the position shown and drops distrance h1. It then collides with the 0.4kg block B, the coefficient of restitution is e. Determine the maximum downward displacement h2 of block B. Neglect ALL FRICTION, and ASSUME block B is initially held in place by a hidden mechanism until the collision begins. The two springs of modulus k are initially un-stretched.

There are numbers in this problem but also, its irrelevant to my question.

 

[topsquark]

I was reworking a problem from class today, and made a simple assumption, in regards to the problem, that made this problem much easier.

My question: Is it fair to assume that the velocity of Va = Vb, after impact? The weights are non-impulsive, since the difference in weight is small.

I ask this because my teacher found Vb by using coefficients of restitution. My assumption gave me the same answer as my teacher. there is a picture that goes with the problem, but for my question I believe its irrelevant.

Question: The 0.5 kg cylinder A is released from rest from the position shown and drops distrance h1. It then collides with the 0.4kg block B, the coefficient of restitution is e. Determine the maximum downward displacement h2 of block B. Neglect ALL FRICTION, and ASSUME block B is initially held in place by a hidden mechanism until the collision begins. The two springs of modulus k are initially un-stretched.

There are numbers in this problem but also, its irrelevant to my question.

Generally when two unequal masses collide you can't expect the speeds to be equal. You said that you got the same result as your teacher, but also that the masses of the two objects are almost equal. This is probably why your method works...it's because the two answers are also going to be pretty much the same.

My question is which way does the first object move immediately after impact? With the given information I can't even tell which way object 1 is going to move initially after the collision: up or down? (I haven't tried a "brick wall" frame yet. That would likely give more information here.)

-Dan

 

[csmith23]

Generally when two unequal masses collide you can't expect the speeds to be equal. You said that you got the same result as your teacher, but also that the masses of the two objects are almost equal. This is probably why your method works...it's because the two answers are also going to be pretty much the same.

My question is which way does the first object move immediately after impact? With the given information I can't even tell which way object 1 is going to move initially after the collision: up or down? (I haven't tried a "brick wall" frame yet. That would likely give more information here.)

-Dan

I will post a picture of the problem in a few hours. But to answer your question, Cylinder A is on top of B. Both are intially at rest. Mass A is released and Drops onto B. When the masses make contact, A stops and B has the kinetic energy. Hope that helps

 

[topsquark]

I will post a picture of the problem in a few hours. But to answer your question, Cylinder A is on top of B. Both are intially at rest. Mass A is released and Drops onto B. When the masses make contact, A stops and B has the kinetic energy. Hope that helps

Okay, now I'm not certain what the issue is. Which likely means that I'm missing some detail.

The definition of the COR (which I'll call [math]\epsilon[/math]) is
[math]\epsilon = \frac{v_2 - v_1}{v_{10} - v_{20}}[/math]

Letting [math]v_{20} = v_1 = 0[/math] we get
[math]\epsilon = \frac{v_2}{v_{10}}[/math]

So [math]v_2 = \epsilon~v_{10}[/math]

End of problem? Let me know if I missed something.

-Dan

 

[csmith23]

https://drive.google.com/file/d/0B0sTLxpj9nCcMFZDUVNvN0hGRGs/view?usp=sharing

Hopefully you can access this link.

 

[topsquark]

Okay, then my derivation was correct. Do you have any questions about it?

Do you know how to continue the problem? (Hint: Use the Work-Energy theorem.)

-Dan

 

[csmith23]

I understand the problem, only if my approach was correct. The difference weight off both objects can be ignored when trying to find velocity of B after impact. Is it consider a non-impulsive force? The book we use in class describes an impulsive force in terms of certain forces acting as a very large force of short duration. The opposite of that is considered non impulsive.

m(v1) = m(v2) => v1 = v2

With that being said the weight can be neglected. If the difference in weight was larger then yes the approach of this problem would have to be different.

 

[topsquark]

I understand the problem, only if my approach was correct. The difference weight off both objects can be ignored when trying to find velocity of B after impact. Is it consider a non-impulsive force? The book we use in class describes an impulsive force in terms of certain forces acting as a very large force of short duration. The opposite of that is considered non impulsive.

m(v1) = m(v2) => v1 = v2

With that being said the weight can be neglected. If the difference in weight was larger then yes the approach of this problem would have to be different.

The COR and bodies at rest make it easy to find v_2 from v_10. In this case it makes a very nice little shortcut and we don't need masses for it. But if, as I had originally supposed, v_1 was not equal to 0 then the masses would have to come into play. In this case, thanks to the COR, we didn't have to worry about the masses at all.

And, yes, as with almost any collision problem, we are considering the force to be "non-implusive" so that the collision lasts for only an infinitesimal amount of time.

-Dan