Who is up for a challenge?

Chemistry4me · « **on:** 31/05/2009 04:15:59 »

It's a bit quite at the moment so how about having a physics problem of the week? [

]
Here's an easy one for you:
A man spins a 0.5 kg stone on the end of a 2 meter string which has a breaking strain of 25 N. He whirls the stone 2.45 m above the ground. Eventually the string breaks and the stone flies off horizontally. Ignoring ~~gravity~~ [:I] air resistance, how far does the stone land from his feet?

Madidus_Scientia · « **Reply #1 on:** 31/05/2009 07:54:43 »

Ignoring gravity? I guess it would hit a building or a tree or something, don't think it would have enough energy to leave the Earths atmosphere before getting slowed down and stopped by air resistance. So, it wouldn't land at all?

Chemistry4me · « **Reply #2 on:** 31/05/2009 08:03:13 »

OPPPS [:I], I meant air resistance! [

]

syhprum · « **Reply #3 on:** 31/05/2009 08:11:21 »

May we assume the Earth is flat so that the path taken is parabolic not elliptical ?

Chemistry4me · « **Reply #4 on:** 31/05/2009 08:20:11 »

Yes. Parabolic.

syhprum · « **Reply #5 on:** 31/05/2009 09:09:52 »

Ignoring the correction that should be made for the fact that the string is not horizontal when it breaks I calculate that the distance will be 22.66 meters

Chemistry4me · « **Reply #6 on:** 31/05/2009 09:19:19 »

That's not what I got [ $:-\$ ]. Maybe the question will be better illustrated with a diagram.

[diagram=461_0]

The string is parallel to the ground when it breaks.

syhprum · « **Reply #7 on:** 31/05/2009 09:38:01 »

My fault I worked with a one meter string, the way you posed the question was fine its just that I can't read!

PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.

Chemistry4me · « **Reply #8 on:** 31/05/2009 09:43:47 »

Quote from: syhprum on 31/05/2009 09:38:01

PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.

I am not sure what you mean.

syhprum · « **Reply #9 on:** 31/05/2009 10:04:15 »

The string will never be horizontal there will always be a drop from the operators hands to the height of the stone.
This will have a small effect on the breaking strain which I have decided can be ignored but the height of the stone when it is released is highly relevent.
I await some other answers after my wrong result, this is a very commen cause of failure in exams not reading the question

syhprum · « **Reply #10 on:** 31/05/2009 17:29:50 »

Is there no one out there who understands simple physics and can do arithmetic ?.
I will post no further calculations until I see what other solutions are offered.

Chemistry4me · « **Reply #11 on:** 01/06/2009 00:33:25 »

Looks like everyone is on holiday. [

]

erickejah · « **Reply #12 on:** 01/06/2009 04:07:00 »

100m?

Chemistry4me · « **Reply #13 on:** 01/06/2009 04:11:07 »

What? 100m! No. A hammer throw doesn't even go that far. [

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erickejah · « **Reply #14 on:** 01/06/2009 04:14:21 »

ok, yes I am wrong. can u tell me how to calculate the speed of the object just before the string breaks?

Chemistry4me · « **Reply #15 on:** 01/06/2009 04:24:04 »

F_c = mv²
r

erickejah · « **Reply #16 on:** 01/06/2009 04:31:28 »

thanks..

what about 70.7107m

Chemistry4me · « **Reply #17 on:** 01/06/2009 04:34:31 »

Err...that's not what I got. How are you calculating this?

erickejah · « **Reply #18 on:** 01/06/2009 04:40:23 »

i got the final velocity by:

√(F_c*r) = v
m

giving me 10m/s

then i just calculated the time which the object takes to touch the floor by:

Δy=1/2(V_i+V_f)t
Which gave me: .707107s

at the end I just multiply those values... [:-[]

erickejah · « **Reply #19 on:** 01/06/2009 04:42:26 »

by the way, if u Karen see this post. please fix my smiley.

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