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# Momentum

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1) A 1.0x10^3 kg plane is trying to make a forced landing on the deck of a 2.0x10^3 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck; this braking force is constant and is equal to one-quarter of the plane's weight. What must the minimum length of the barge be for the plane to stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 5.0x10^1 m/s toward the front of the barge?

2) A 0.25kg tennis ball is placed right on top of a 1kg volleyball and dropped. Both balls hit the ground at a speed of 3 m/s simultaneously. Find the (upward) velocity of the tennis ball right after it bounces up from the volleyball. Assume elastic collisions. (HINT: the tennis ball will move faster than 3m/s)

I don't have any idea on how to solve this problems. Can someone give me some hints? I would really appreaciate!:cup:

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I thought I answered this one. Oh well, I'll go again, but a little shorter.

Q1)

This is conservation of energy, pure and simple. KE = 1/2 mV^2. Work done = F d = 1/4 mgd. equate the two, and find d.

Or it could be done by using the equations of motion, s = u + at or something. F = ma, F = 1/4 mg. So we have a. We also have u (initial speed = V), v = 0, a, so we can find d.

2) CoEnergy when volley ball hits ground, so it moves 3ms^-1 upwards after collision with ground. Now, tennis ball is moving 3 ms^-1 downwards, volleyball is moving 3ms^-1 upwards. Use CoE and CoM equations to find speed of tennis ball (and volley ball).

Hope helps.

If you want to be advanced

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oh, and in Q1) for some reason, the mass drops out in the answer. Wierd, questions don't normally give superflous information, but the mass of the plane was irrelevant.

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nononono,

for number 1:

since the barge will move relative to the "clam" sea, it is not necessary to "eliminate" all the kinetic energy of the plane.

actually,three steps to find the answer:

1. use conservation of momentum, calculate the final velocity of the whole system (plane + barge, the velocity should be the same since the plane is stationary relative to the barge)

2. from this calculated velocity, calculate the change of kinetic energy, use the work formula, W=fd1=change in kinetic energy. from that, calculate d1.

3 but remember, the barge was moving when the plane was slowing down! calculate how much the barge moved, and substract it from d1 (use similar equations in 2, find kinetic energy change for the barge, and find d2)

well, kinda lengthy... there might be a better way, this is just a quick reply

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To question one, I've got another idea, although I can't figure out why it's a barge and all that...

The braking force is friction, so you should avoid using the concept of energy. (That's because friction is dissipative. I'ts energy goes to heat!)

Now a constant force is on the plane.So there'll be a uniform acceleration.

Now simple!

Just find out how far the plane will move, until it's velocity becomes zero.

Use the equation:

[math]v^2 = u^2 + 2aS[/math]

where v= final velocity, = zero

u = initial velocity, = [math]50 {ms^{-1}}[/math]

a = what you find from mass of plane and uniform force.

The second is kinda tough. I'll see if I can figure it out...

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the tricky part about number one is that the Barge also moves (it is on water!)..... just keep that in mind....

the second one simply requires conservation of momentum and energy.

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This is conservation of energy, pure and simple. KE = 1/2 mV^2. Work done = F d = 1/4 mgd. equate the two, and find d.

- me

nononono,

for number 1:

since the barge will move relative to the "clam" sea, it is not necessary to "eliminate" all the kinetic energy of the plane. - Tim_Lou

Bollocks to you for being right at my expense.

However painful for me to admit this, Tim_Lou is right. I thought it looked wierd that the mass of the plane was irrelevant. Harder question than I thought. Besides, it was really late, and I was tired, and my calculator broke and lots of other really good excuses that I can't quite think of right now.

Ronthpon, I'm afraid you went for the same approach as me. What you are saying is the same as my second method of answering the same question and have therefore made the same mistake I did. The barge will be moved by the plane as Tim_Lou said and the final velocity of the plane will not be 0 but the same as the final velocity of the barge. It's a good point though, why a barge? Cargo ships arn't normally used for landing planes. Perhaps an aircraft carrier would have been better.

Neverthess I think we should just concede defeat and let Tim_Lou have his day.

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Defeat?! What defeat? I never fought!

In any case it was a matter of understanding the question for me, I guess...

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Defeat?! What defeat? I never fought!

- ronthepon

If I'm going down, I'm taking you down with me :eek_big:

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nononono,

for number 1:

since the barge will move relative to the "clam" sea, it is not necessary to "eliminate" all the kinetic energy of the plane.

actually,three steps to find the answer:

1. use conservation of momentum, calculate the final velocity of the whole system (plane + barge, the velocity should be the same since the plane is stationary relative to the barge)

2. from this calculated velocity, calculate the change of kinetic energy, use the work formula, W=fd1=change in kinetic energy. from that, calculate d1.

3 but remember, the barge was moving when the plane was slowing down! calculate how much the barge moved, and substract it from d1 (use similar equations in 2, find kinetic energy change for the barge, and find d2)

well, kinda lengthy... there might be a better way, this is just a quick reply

Hi,

What does d1 represent?

And for your step 3, "calculate how much the barge moved", how can I calculate it? I am not too sure...

Will the barge be moving forward or backward relative to the water?

For the plane's initial velocity of 50m/s, I think it's relative to the BARGE. Which frame of reference should I consider when using the law of conservation of momentum? (is it the barge?)

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d1 is the distance moved for the plane relative to the water.

d2 is the distance moved for the barge relative to the water.

(the barge will move forward, since it is assumed that it will eventually have the same velocity as the plane, which is in the forward direction)

d2 can be calculated by:

use the final velocity of the system you calculated from 1, find the change in kinetic energy of the barge. [math]\Delta KE = 1/2 m*v^2[/math]

since [math]\Delta KE=f*d_2[/math] (force is the friction), [math]d_2[/math] can be found.

relative to the barge, the plane moved a distance of [math]d_1-d_2[/math] which is the answer.

even if the barge is initially moving relative to the water, we can still pick a reference such that the barge is stationary initially and all these equations will still apply

(if we neglect the dragging of the water...)

or you can just make everything relative to the water and everything will still work out similarly (the kinetic energy change will be final - inital).

however, if you pick the barge as a reference... things can be very complicated, since the barge is accelerated. You can still get the same answer though.

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I can see your confusion Kingwinner. If found the energy arguments hard to follow to.

We know the velocity of plane + barge. Therefore we know the kinetic energy of the plane and barge.

We know the velocity of the plane before the landing. Therefore subtract the energy of the plane and barge after landing from the energy of the plane before landing.

This represents the energy lost by the breaking system.

We also know the force of that breaking system. Therefore, Work = Force x distance (d) so d = W/F.

The question is, what d is this? Is this the distance that the plane has moved over the water or the distance the plane has moved over the barge? These distances will be different since the barge is moving. The answer depends on how the breaking system is defined. I reckon, the breaking force is constant relative to the barge. Therefore d should represent the distance travelled over the barge. If the breaking force is instead relative to the sea, then you would need to subtract from the answer to d the distance the barge has moved.

However, I think the barge will be travelling so slowly that it doesn't make much difference which d is being used.

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Having said that, my answer I think is a rephrasing rather than a different method of the other answers. So take your pick which one is easier to follow. Don't say we don't spoil you here.

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