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Ok, so here are the rules:

-Keep the questions very basic (PHYS 101 & 102); this is, afterall, a computer forum.
-Wait for confirmation that your answer is correct before posting your question.
-If 24 hours passes and still no confirmation, post a new question to keep the game going.


** Please follow the rules! **

I'll start:

Question:

A car (mass = 1100kg) is traveling at 32 m/s when it collides head-on with a SUV (mass = 2500kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the SUV traveling?

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Thanks, great game idea -Josh
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Last Post by Mushy-pea
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    The answer, of course, is 14.08 m/s in the opposite direction of the first car. Read More

  • 1

    um.. It's really not that hard. Involves some knowledge of angular momentum.. and calculus Read More

  • 1

    [QUOTE=vishesh;343442]Not studied Rotational Mchanics yet, but have a bit of knowledge. Is the torque equal 72Nm and angular momemtum equals 72t Js.[/QUOTE] wow, good job. They are vectors, though.. the torque is 72Nm in the k direction, and the angular momentum is 72t Js in the k direction. nice, your … Read More

  • 1

    The conservation of energy/mass is the exact same law!! All the other laws are similar in that they involve that some quantity cannot be created or destroyed.. Read More

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    I got 5,484.375 J and 1 310.79708 calories ? Read More

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That depends: did the passengers come to a halt too, or did one or more of them fly out the windshield?

Does that even matter?

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The answer, of course, is 14.08 m/s in the opposite direction of the first car.

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your answer is correct. -Duki
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yay! My question is:

The position vector of a particle of mass 3kg is given by r(vector)= 4i+3t^2j, where r is in meters and t is in seconds. Determine the angular momentum and torque acting on the particle about the origin.

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Does that even matter?

If you were worried about details, it would matter. The passengers were part of the energy system, and stopping them would require energy as well. If passengers from one vehicle were to continue flying through the windshield, it would change the amount of energy involved in the collision, and would change the result a bit.

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Yes, but the change you are talking about is irrelevant. It would not change the SUV's velocity by much.

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Hm... I got 21.22m/s... then again, I haven't done physics in a long time and may have mis-remembered something along the way... (that was for the first one btw)

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Infarction: we know the momenta of the cars sum to zero because the combined mass has no velocity. So 32 * 1100 = v * 2500. Then 35200 / 2500 = v, i.e. 352 / 25, which is 14 + 2/25 i.e. 14.08.

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Hm... I got 21.22m/s... then again, I haven't done physics in a long time and may have mis-remembered something along the way... (that was for the first one btw)

You use conservation of energy.. the fact that the cars will stick after the collision. Therefore you know that the collision is perfectly inelastic, so the maximum kinetic energy was lost. So, m1*V1 + m2*V2 = (m1+m2)*V

You solve the equation 1100kg * 32 m/s + 2500kg * X = [(1100+2500)*0]kg m/s

therefore x=-14.08 m/s.. which means the SUV had a velocity of 14.08 m/s in the opposite direction of car 1.

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Your explanation does not use the principle of conservation of energy.

Um.. actually it does. I was referring to the law of conservation of energy.. Kinetic energy is not conserved in inelastic collisions, therefore you know that the maximum kinetic energy is lost. However, The law of conservation of energy is true for the system. The energy is transferred from kinetic energy to different kinds of energy (sound, heat, etc.)

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But isn't kinetic energy 1/2 mv^2?

Maybe this is why I did so bad in physics years ago... :sad:

This problem does not require you to find the kinetic energy of anything. You are simply looking for the velocity. Since it is an inelastic collision, all the kinetic energy is lost. So if you found the initial kinetic energy and the final kinetic energy (1/2mv^2), you would find the total change in kinetic energy to be (-initial kinetic energy) b/c the final kinetic energy is 0...

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Isn't that right? It better be.. I'm taking the Physics 2 "Mechanics w/ Calculus" AP test next month..

It's wrong actually. The correct answer is that you can't tell based on the data. It doesn't state where the combined masses come to a halt for example, nor the ground friction component acting on them. As such there's just one vector out of several you need to compute the entire energy distribution of the system before and after the collision (and that's when seeing it as a 2 dimensional solid body problem, it could be a 3 dimensional problem with other factors coming into play like the bodies dividing up their masses on collision causing multiple bodies to leave the scene afterwards, dispersing the energy).

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It's wrong actually. The correct answer is that you can't tell based on the data. It doesn't state where the combined masses come to a halt for example, nor the ground friction component acting on them. As such there's just one vector out of several you need to compute the entire energy distribution of the system before and after the collision (and that's when seeing it as a 2 dimensional solid body problem, it could be a 3 dimensional problem with other factors coming into play like the bodies dividing up their masses on collision causing multiple bodies to leave the scene afterwards, dispersing the energy).

I think the objects are supposed to be assumed as point particles, rather than 3-dimensional objects. Also, other velocity variables (such as air resistance and static friction) should not be considered.

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It's wrong actually. The correct answer is that you can't tell based on the data. It doesn't state where the combined masses come to a halt for example, nor the ground friction component acting on them. As such there's just one vector out of several you need to compute the entire energy distribution of the system before and after the collision (and that's when seeing it as a 2 dimensional solid body problem, it could be a 3 dimensional problem with other factors coming into play like the bodies dividing up their masses on collision causing multiple bodies to leave the scene afterwards, dispersing the energy).

um yea.. based on the information given, the answer I gave was correct. Unless he wanted friction/air resistance and such considered in the problem he would have said so. He also mentions that this thread is not meant for really difficult problems.. So, perhaps you shouldn't be such a :twisted:

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Um.. actually it does. I was referring to the law of conservation of energy.. Kinetic energy is not conserved in inelastic collisions, therefore you know that the maximum kinetic energy is lost. However, The law of conservation of energy is true for the system. The energy is transferred from kinetic energy to different kinds of energy (sound, heat, etc.)

No, you do not know the maximum kinetic energy was lost from the fact that kinetic energy is not conserved in inelastic collisions; you know that the maximum possible amount of kinetic energy was lost because the problem statement tells you that the vehicles end up at rest.

You did not use the principle of conservation of energy anywhere in your solution to the problem; you used the law that momentum is conserved.

Votes + Comments
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can i make a chemistry question?

What mass of calcium carbonate (CaCO3) reacts completely with 25 cm3 of 2 mol/dm3 hydrocholric acid (HCL)?

CaC03 (s) + 2HCL (aq) -> CaCl2 (aq) + CO2 (g) + H2O (l)

For those of you who dont do chemistry heres a clue:

Its a 1:2 Ratio and n (moles) = m (mass) x Mr (Relative molecular mass)

This is simple really (i did it in school when i was 16 - all you need is a calculator and a periodic table)

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No, you do not know the maximum kinetic energy was lost from the fact that kinetic energy is not conserved in inelastic collisions; you know that the maximum possible amount of kinetic energy was lost because the problem statement tells you that the vehicles end up at rest.

That's the EXACT same thing. All perfectly inelastic collisions involve the objects coming to rest, and therefore, losing the maximum kinetic energy.

You did not use the principle of conservation of energy anywhere in your solution to the problem; you used the law that momentum is conserved.

All conservation laws are basically the same. The law of conservation of energy/mass is the exact same b/c energy and mass are related in the equation E=MC^2... The law of conservation of momentum involves torque, angular momentum,.. and basically the other laws..

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yay can i make a chemistry question?

Actually, no one has answered my question yet:


"The position vector of a particle of mass 3kg is given by r(vector)= 4i+3t^2j, where r is in meters and t is in seconds. Determine the angular momentum and torque acting on the particle about the origin."

But, It might be better to make a new thread titled "Chemistry Game" anyways...

This topic has been dead for over six months. Start a new discussion instead.
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