Exploring Momentum: Two Skaters in Motion

[Help_Me_Please]

I'm a little bad with putting physics into words. The math doesn' phase me .. but these conceptual questions are a different story.

First one.

A pendulum swings back and forth. Does the tension force in the string do work on the pendulum bob?Does the force of gravity do work on the bob?Explain your answers.

I know that the force of gravity does work on the bob but I don't know how to explain it, and I have no idea about the tension force.

Second one.

A karate expert can break a stack of boards in half by hitting the boards with the side of one bare hand. Will the boards break more easily if they are struck swiftly or slowly.

The answer is swiftly. Now .. what I have is that if he strikes swiftly he has more velocity and therefore more kinetic energy. Is that an ok explanation?

Last one. I know I know ... I'm dumb xD.

Two skaters initially at rest push against each other so that they move in opposite directions. what is the total momentum of the two skaters when they begin moving?

Now I'm taking a guess here, but I have that the total momentum would be zero because they are moving in opposite directions and the momentum on one side cancels out the momentum on the other? Don't the masses of the skaters have something to do with it though? Obviously not becuase it doesn't say that they are equal in mass or what either masses are.. but why doesn't that have anything to do with it?

 

[quasar987]

I'm a little bad with putting physics into words. The math doesn' phase me .. but these conceptual questions are a different story.

First one.

A pendulum swings back and forth. Does the tension force in the string do work on the pendulum bob?Does the force of gravity do work on the bob?Explain your answers.

I know that the force of gravity does work on the bob but I don't know how to explain it, and I have no idea about the tension force.

The explanation is different for both.

For the tension: Since the bob is moving on a circle, you can see that, wherever the bob is, a tiny displacement of the bob is approximately tangent to the circle, and thus perpendicular to the tension. So the angle between the vector displacement and the tension is 90°. And since cos90° = 0 we have dW = (Tension)*0*(tiny displacement) = 0, where dW is the tiny amount of work done by the tension during the tiny displacement we talked about. Now the total work is just the sum of these tiny pieces of work: W=0+0+0+...+0=0. (Note that the rigourous solution to this problem requires the calculus-based definition of work, which I assumed you didn't know because at least when I was in grade K-12, they didn't taught it to us.)

For gravity: The question is a bit unclear. It is unclear wheter they ask for the work done by gravity in general or just during one cycle (i.e. one "back and forth" of the pendulum). In either case, I assume you are familiar with the relation $W_{gravity} = -\Delta U_{gravitational}=-mg\Delta y$. Where $\Delta y$ represents the vertical displacement. So the answer depends on the starting height and ending height of the bob. However, it it is interesting to notice that since the bob gets back to its starting point after a complete cycle, we have $\Delta y=0$ and it follows that for this particular displacement, W = 0.

 

[dextercioby]

I'm a little bad with putting physics into words. The math doesn' phase me .. but these conceptual questions are a different story.

First one.

A pendulum swings back and forth. Does the tension force in the string do work on the pendulum bob?Does the force of gravity do work on the bob?Explain your answers.

I know that the force of gravity does work on the bob but I don't know how to explain it, and I have no idea about the tension force.

Yes.Only gravity does work on the bob,because the tension force would have to compress the wire,which i think it doesn't.Gravity does work,because the system converts potential energy into kinetic one.Potential energy is minus the wotk done by gravity.

Second one.

A karate expert can break a stack of boards in half by hitting the boards with the side of one bare hand. Will the boards break more easily if they are struck swiftly or slowly.

The answer is swiftly. Now .. what I have is that if he strikes swiftly he has more velocity and therefore more kinetic energy. Is that an ok explanation?

The breaking of the boards is determined by the quantity of momentum and the force of the impact.He would have to hit the top board with a great speed and at the same time the duration of impact be very small.It's momentum transfer that's important,because it's linked with the force via the second law if dynamics of Newton.

Last one. I know I know ... I'm dumb xD.

Two skaters initially at rest push against each other so that they move in opposite directions. what is the total momentum of the two skaters when they begin moving?

Now I'm taking a guess here, but I have that the total momentum would be zero because they are moving in opposite directions and the momentum on one side cancels out the momentum on the other? Don't the masses of the skaters have something to do with it though? Obviously not becuase it doesn't say that they are equal in mass or what either masses are.. but why doesn't that have anything to do with it?

Apply the law of momentum conservation.You'll find your answer immediately.

Daniel.

 

[Help_Me_Please]

Thanks SO much! I think I actually understood that. =)

 

[quasar987]

Apply the law of momentum conservation.You'll find your answer immediately.

Yes, indeed. First step is to label your two skaters as a "system". It is probably written somewhere in your textbook that no force exerted upon the particles of a system by the particles of the system can alter the total momentum of the system*. There you go.


* Again, this is only fully explained by the calculus-based version of mechanics but mainly, it is because of Newtons third law: whenever a particle exerts a force on another, the "another" exerts an equal and opposite force of the first, such that the total force on the system will always be 0, and so will the change of momentum, according to the second law.