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Author Topic: TrueTears question thread  (Read 68017 times)  Share 

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TrueTears

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Re: TrueTears question thread
« Reply #30 on: February 24, 2009, 11:28:49 am »
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and also...
A 100g glider on a linear air track is moving at 0.5ms^{-1} towards a stationary 200g glider. Magnets mounted on the glides prevent them from actually colliding, but they undergo an elastic interaction. Determine the velocity of each of the gliders after this interaction.
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Re: TrueTears question thread
« Reply #31 on: February 24, 2009, 04:38:05 pm »
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Elastic means energy is conserved: and momentum is conserved:

TrueTears

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Re: TrueTears question thread
« Reply #32 on: February 24, 2009, 04:52:34 pm »
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Elastic means energy is conserved: and momentum is conserved:

Elastic means kinetic energy is conserved.
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TrueTears

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Re: TrueTears question thread
« Reply #33 on: February 24, 2009, 04:53:11 pm »
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any ideas on the 2 q's? XD
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kamil9876

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Re: TrueTears question thread
« Reply #34 on: February 25, 2009, 03:39:53 pm »
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and also...
A 100g glider on a linear air track is moving at 0.5ms^{-1} towards a stationary 200g glider. Magnets mounted on the glides prevent them from actually colliding, but they undergo an elastic interaction. Determine the velocity of each of the gliders after this interaction.

conservation of momentum:

0.1*0.5=0.1v+0.2v'  (v' is the velocity of the 0.2kg glider, whereas v is that of the 0.1kg one)
    0.05=0.1v+0.2v'
         5=10v+20v'
         1=2v+4v'

conservation of total kinetic energy:

0.5*0.1(0.5^2)=0.5*0.1*(v^2)+0.5*0.2*(v'^2)
              0.25=(v^2)+2(v'^2)
  sub in the equation derived from the momentum thingy:

0.25=(v^2)+2((1-2v)/4)^2

Some quadratic will come out from this. (Note v and v' can be positive or negative because of direction). I think this is the right idea so far.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

kamil9876

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Re: TrueTears question thread
« Reply #35 on: February 25, 2009, 04:17:56 pm »
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In an elastic collision between 2 objects of mass and show that the speeds of the approach is equal to the speed of the seperation . The symbols represent speeds not velocities.

I did this question twice and I got the result u2+u1=v1+v2. On the third attempt I got the result that the question gives since I realised that this question is worded very shittly(adverb of the adjective "shit"?). Anyways, what's happening is that initially both objects are travelling in the same direction, but once they collide, they move apart. I deduced this from the equations that were given for speed of approach and seperation. At first, I assumed that they move towards eachother and then move apart, but in that case the two equations given for speed of speration and approach would be different, and that's the shitty wording i was refering to earlier.

Take the initial direction to be positive. Hence the conservation of momentum implies:



The negative sign comes from the assumption that object 2 is the one that rebounds back while object 1 continues going in the positive direction. This is because the expression  u2-u1 implies that u2>u1 since this is a speed(always positive) Hence object 2 must be comming from behind, catching up to object 1 in order to make the collision. Because object 2 is behind, it is the one that rebounds back.

Kinetic Energy:





momentum equation implies that:





By doing a similair thing to the kinetic energy equation, u can get the result:



By equating the two expressions for and factorising using difference of squares, then cancelling out factors etc. you will arrive at the result :)
« Last Edit: February 25, 2009, 04:43:13 pm by kamil9876 »
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TrueTears

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Re: TrueTears question thread
« Reply #36 on: February 26, 2009, 10:57:00 am »
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thanks kamil !!
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TrueTears

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Re: TrueTears question thread
« Reply #37 on: February 26, 2009, 09:57:27 pm »
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and...
1. A block of wood of mass 1.5kg is suspended from a fixed point by two light strings so that it is able to swing through the arc of a circle. A bullet of mass 25g, travelling horizontally at 300ms^{-1} embeds in the block. The block swings and rises through a vertical height of h before coming to rest.

a) Find h
b) calculate the percentage of energy lost during the collision

thanks
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TrueTears

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Re: TrueTears question thread
« Reply #38 on: February 26, 2009, 09:59:21 pm »
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2. A 9.6 cm spring is hung vertically and a 50g mass carrier is hung from it. This extends its length to 10.2 cm. When an additional 50 g mass is added to the mass carrier, the new length of the spring is 10.8cm.

Calculate the spring constant for this spring.
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TrueTears

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Re: TrueTears question thread
« Reply #39 on: February 26, 2009, 10:03:18 pm »
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3. A 1.5 tonne pile-driver falls from a distance of 5.0m onto a pile. Each blow drives the pile 10cm into the earth. If 15% of the energy of the pile-driver is lost during the collision with the pile, calculate the resistance the earth provides against the movement of the pile.

4. Predict the effect on the gravitational force between 2 objects of
a) Halving the distance between them
b) Doubling both masses
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Re: TrueTears question thread
« Reply #40 on: February 27, 2009, 12:28:07 am »
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1. Not entirely sure... mainly because of part b. there's only enough info to work out part a with conservation of energy, which will make the answer to b an obvious 0%, which sounds ridiculous...

2. you can work this out from the first piece of info alone:







3. (again, not entirely sure, but here's a crack at it, hopefully it's the right answer)

Ug = 1500 * 5 * g = 75000 J

Elost = 0.15 * 75000 = 11250 J = W = F * d

F = 11250 / 0.1 = 112,500 N

(I think...)


4.


halving R:

Doubling both m:
« Last Edit: February 27, 2009, 12:40:49 am by Mao »
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TrueTears

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Re: TrueTears question thread
« Reply #41 on: February 28, 2009, 01:07:48 pm »
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thanks mao
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TrueTears

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Re: TrueTears question thread
« Reply #42 on: February 28, 2009, 01:11:16 pm »
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Also this question:

A 50kg passenger in a car is held firmly by her seatbelt during a collision. The seatbelt, in stopping her, puts an average force of 12000N on her for s. Calculate the impulse of the force if the collision time was 0.01s.

So... Isn't

I = = 120Ns.

But my book says its = 12Ns, Impulse doesn't change? But the time changed... So shouldn't the impulse also change?
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TrueTears

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Re: TrueTears question thread
« Reply #43 on: February 28, 2009, 02:10:37 pm »
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Calculate how far an astronaut would need to be away above the Earth in order for his weight to be 0.01 his weight on the Earth's surface.
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Re: TrueTears question thread
« Reply #44 on: February 28, 2009, 02:14:21 pm »
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last one: Three moons around planet X have masses M, 4M and 9M
If the distances of these moons from the planet centre are R, 4R and 9R respectively, calculate the ratio of their orbital speeds.
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