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So...i suck at physics. Actually, i just sometimes don't go to class because i can't stand the way my teacher teaches. (he doesn't give notes, he just writes what he says on the board and we somehow have to make sense of it. Oh, and he doesn't follow the text book so you can't even teach urself)

 

but yea, i got this take-home test, cuz technically we're not allowed to have a test the week before exams. So he sent it home cuz...don't know. but there's this question i need help with:

 

A comet of mass 6000.kg travelling at 30,000.m/s splits into three pieces that move apart with velocities that make angles of 10degrees above, 20degrees above, and an unknown angle below, all relative to the original trajectory. If the pieces have masses of 1000.kg, 2000.kg, and 3000.kg respectively and the first two pieces are observed to have velocities of 10,000.m/s and 2,000.m/s respectively, find the velocity of the third piece.

 

Needs to be answered using energy and momentum. and there's this random table

P x( ) y( )

- - -

- - -

- - -

Pt - -

- - -

- - -

- - -

Pt' - -

*( - ) represent spaces

 

 

i know its last minute but any help..will help.

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^ lets say the comet originally was travelling along x-axis .

It breaks upto 3 pieces, A(m1) , B(m2) , C(m3).

A's new direction makes an angle of 10 deg above x axis.... etc.

C's new direction makes an angle of theta below x axis. Say its new speed is V3.

Lets call the original mass m0, and original speed V0. (you plug in the numbers later)

 

 

There was no external force in the comet system, so no change in momentum.

So You need to write conservation of momentum in x and y directions.

 

x-direction

original momentum = total momentum of the system after explosion

M V0 = m1 V1 cos(10) + m2 V2 cos(20) + m3 V3 cos(theta) --------- eqn 1

 

 

y direction

original momentum = total momentum of the system after explosion

0 = m1 V1 sin(10) + m2 V2 sin(20) + m3 V3 sin(theta) --------- eqn 2

 

 

you have two equations and two unknowns ; V3 and theta. Solve the equations and you get the speed and direction.

 

Forget the useless table. its not important. just a method for solving equations.

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^ lets say the comet originally was travelling along x-axis .

It breaks upto 3 pieces, A(m1) , B(m2) , C(m3).

A's new direction makes an angle of 10 deg above x axis.... etc.

C's new direction makes an angle of theta below x axis. Say its new speed is V3.

Lets call the original mass m0, and original speed V0. (you plug in the numbers later)

 

 

There was no external force in the comet system, so no change in momentum.

So You need to write conservation of momentum in x and y directions.

 

x-direction

original momentum = total momentum of the system after explosion

M V0 = m1 V1 cos(10) + m2 V2 cos(20) + m3 V3 cos(theta) --------- eqn 1

 

 

y direction

original momentum = total momentum of the system after explosion

0 = m1 V1 sin(10) + m2 V2 sin(20) + m3 V3 sin(theta) --------- eqn 2

 

 

you have two equations and two unknowns ; V3 and theta. Solve the equations and you get the speed and direction.

 

Forget the useless table. its not important. just a method for solving equations.

 

 

THank you!

 

but i have to do the table, thats what i hate about my teacher. He organizes the tests into different boxes, and labels them so you answer it the way he wants it done. its annoying really. but yeah.

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I have a GCSE Physics Exam in 5 days. Does anyone know what is likely to come up?

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Fenstermacher hits a golf shot that travels at 114.8km/hr. The shot strikes a tree and bounces back to the tee at 25% of its original speed. The ball required 3.82 seconds for the round trip. How far away is the tree?

 

The answer is 24.4m. Can someone (ehem, Haku? :ph34r) explain how that's the answer.

 

114.8km/hr is ~31.88m/s. 7.97m/s is 25% of that.

 

I think the avg speed would be 19.875m/s. And the total distance the ball travelled is 75.9m. I just don't know how to figure out how far the tree was!

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Fenstermacher hits a golf shot that travels at 114.8km/hr. The shot strikes a tree and bounces back to the tee at 25% of its original speed. The ball required 3.82 seconds for the round trip. How far away is the tree?

 

The answer is 24.4m. Can someone (ehem, Haku? :ph34r) explain how that's the answer.

 

114.8km/hr is ~31.88m/s. 7.97m/s is 25% of that.

 

I think the avg speed would be 19.875m/s. And the total distance the ball travelled is 75.9m. I just don't know how to figure out how far the tree was!

This is a very poorly constructed question. Don't worry if you can't solve it. There is nothing of use here. It won't be in the exam.

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Fenstermacher hits a golf shot that travels at 114.8km/hr. The shot strikes a tree and bounces back to the tee at 25% of its original speed. The ball required 3.82 seconds for the round trip. How far away is the tree?

 

The answer is 24.4m. Can someone (ehem, Haku? :ph34r) explain how that's the answer.

 

114.8km/hr is ~31.88m/s. 7.97m/s is 25% of that.

 

I think the avg speed would be 19.875m/s. And the total distance the ball travelled is 75.9m. I just don't know how to figure out how far the tree was!

 

the distance should be half that, so 24.4m is wrong.

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the distance should be half that, so 24.4m is wrong.

It should be half of the total distance? Yeah that makes sense. I probably calculated the total distance wrong, could you double check that?

 

But yeah, if the shot travels from A -> B, then B -> A, the distance between A and B is half the total distance the shot travelled. Sweet!

 

I must have made a mistake calculating the total distance the ♥♥♥♥ travelled then.

 

To calculate total distance I thought you find the average speed, and multiply that with time.

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You can't just add the 2 speeds and divide by two to get average speed! that only works when an object undergoes constant acceleration. Also this is not a home work thread. Go else where. Then maybe I'll help.

 

ps> all attempts above are wrong.

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You can't just add the 2 speeds and divide by two to get average speed! that only works when an object undergoes constant acceleration. Also this is not a home work thread. Go else where. Then maybe I'll help.ps> all attempts above are wrong.

I'd like to smack Fenstermacher for sucking at golf and hitting that tree.

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Total distance the what?! traveled then? lol

 

Boy, I am so glad I never took physics.

Physics is cool, I respect it. It is just IMO the toughest to master.

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You can't just add the 2 speeds and divide by two to get average speed! that only works when an object undergoes constant acceleration. Also this is not a home work thread. Go else where. Then maybe I'll help.

 

ps> all attempts above are wrong.

if acceleration wasnt constant, it would be impossible to answer. so you must assume that it is. (for a homework sheet i mean)

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It should be half of the total distance? Yeah that makes sense. I probably calculated the total distance wrong, could you double check that?

 

But yeah, if the shot travels from A -> B, then B -> A, the distance between A and B is half the total distance the shot travelled. Sweet!

 

I must have made a mistake calculating the total distance the ♥♥♥♥ travelled then.

 

To calculate total distance I thought you find the average speed, and multiply that with time.

is there a diagram or something you aren't showing, because what you did sounds right to me.

 

perhaps the solution is wrong or you missed part of the question.

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