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TrueTears

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Re: TrueTears question thread
« Reply #15 on: January 13, 2009, 07:09:31 pm »
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momentum at the start is equal to the momentum at the end

yeah i know that but u got where is the final speed of the cart and is the final speed of the coal. The momentum is the same, but the speeds could be different, how do u know here the speed of the coal and cart are the same?
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Re: TrueTears question thread
« Reply #16 on: January 13, 2009, 10:12:39 pm »
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also..

2. Two ice skaters, Dean and Melita, are performing an ice dancing routine, in which Dean( with a mass of 70kg) glides smoothly at a velocity of 2 ms^-1 due east towards a stationary Melita(with a mass of 50kg), holds her around the waist and they move off together. During the whole move, no significant friction is applied by the ice. Where is the centre of mass of the system comprising Dean and Melita 3s before impact.
(the answer in the book is, between Dean and Melita, 2.5m from Dean). How do u get that?

3. A car of mass 1500kg travelling due west at a speed of 20 ms^-1 on an icy road collidies with a truck of mass 2000kg travelling in the opposite direction at the same speed, the vehicles lock together after impact.
a) Which vehicle experiences the greatest change in velocity?
b) which vehicle experiences the greatest change in momentum?
c) which vehicle experiences the greatest force?

4. A billard player claims that he can make a red stationary ball move north with a speed of 2.5ms^-1 by striking it with a white ball moving at a speed of 2ms^-1. When challenged he said that according to the Law of Conservation of Momentum the white ball would rebound with a speed of 0.5ms^-1, ie 0.5ms^-1 south. Explain, using calculations, why the player's claim is not correct - even though it is consistent with the Law of Conservation of Momentum.
« Last Edit: January 13, 2009, 11:36:20 pm by TrueTears »
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Re: TrueTears question thread
« Reply #17 on: January 13, 2009, 10:27:07 pm »
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momentum at the start is equal to the momentum at the end

yeah i know that but u got where is the final speed of the cart and is the final speed of the coal. The momentum is the same, but the speeds could be different, how do u know here the speed of the coal and cart are the same?

While the coal is in the cart, it has the same speed as the cart. An instant after coal falls out, there are no forces acting on it apart from gravity, so it must have the same speed as the train.

also..

2. Two ice skaters, Dean and Melita, are performing an ice dancing routine, in which Dean( with a mass of 70kg) glides smoothly at a velocity of 2 ms^-1 due east towards a stationary Melita(with a mass of 50kg), holds her around the waist and they move off together. During the whole move, no significant friction is applied by the ice. Where is the centre of mass of the system comprising Dean and Melita 3s before impact.
(the answer in the book is, between Dean and Melita, 2.5m from Dean). How do u get that?

3. A car of mass 1500kg travelling due west at a speed of 20 ms^-1 on an icy road collidies with a truck of mass 2000kg travelling in the opposite direction at the same speed, the vehicles lock together after impact.
a) Which vehicle experiences the greatest change in velocity?
b) which vehicle experiences the greatest change in momentum?
c) which vehicle experiences the greatest force?

2. The formula for centre of mass is

At 3s before impact, the skaters are apart. At this time, define the position of Dean to be and the position of Melita to be , then

.

Hence, the centre of mass is 2.5m from Dean.

3.
a) Intuitively, the car with less mass should experience a greater change in velocity. You can confirm this by solving for v and noting the differences.
b) The change in momentum can be derived from part c). Since and by Newton's third, the forces felt by each car is the  same, the momentum change will be the same. You can also prove this from manual calculation.
c) By Newton's third law, both vehicles experience the same force.

4. It breaks the law of conservation of energy. You can derive through conservation of kinetic energy that for elastic collisions. It is a tedious derivation, but have a go with:



and



ANYWAY, Back to the problem


BEFORE: be speed of hitting ball, be speed of stationary ball
AFTER: be speed of hitting ball,   be speed of stationary ball

But , so using ,



...[1]

Setting up the standard conservation of momentum equation:



(from the derived expression)







But

So if the masses of the two balls are equal, then the speed of the ball being hit will be equal to the speed of the hitting ball beforehand. The hitting ball loses all its velocity in order to conserve momentum. This is the basis for Newton's pendulum. The reason why, in pool, you sometimes see the hitting ball continue, is because it does not directly hit the stationary ball.

« Last Edit: January 13, 2009, 10:55:51 pm by /0 »

TrueTears

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Re: TrueTears question thread
« Reply #18 on: January 14, 2009, 09:43:35 pm »
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thanks ST

« Last Edit: January 15, 2009, 03:06:39 pm by TrueTears »
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Re: TrueTears question thread
« Reply #19 on: January 14, 2009, 10:21:49 pm »
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A road is to be banked so that any vehicle can take the bend at a speed of 30 ms^-1 without having to rely on sideways friction. The radius of the curvature of the road is 12m. At what angle should it be banked?
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Re: TrueTears question thread
« Reply #20 on: January 14, 2009, 10:29:42 pm »
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Find out the centripetal force required which = mv^2/r

Then find out the angle at which when banked where the road will provide a force towards the centre of the circle which will equal this force.
( i think you would have to work it out by doing angle stuffs with the normal reaction force or something)
« Last Edit: January 14, 2009, 10:32:27 pm by Table »

TrueTears

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Re: TrueTears question thread
« Reply #21 on: January 14, 2009, 10:38:45 pm »
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Find out the centripetal force required which = mv^2/r

Then find out the angle at which when banked where the road will provide a force towards the centre of the circle which will equal this force.
( i think you would have to work it out by doing angle stuffs with the normal reaction force or something)
yeah so
where is the frictional force, but it says it doesnt need to rely on it, so

so after subbing in the values.

(round 9.8 to 10)

so



but problem is u cant solve that lol out of range... so it must be wrong equation


so any help?


« Last Edit: January 14, 2009, 10:42:42 pm by TrueTears »
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Re: TrueTears question thread
« Reply #22 on: January 14, 2009, 11:13:02 pm »
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Hmmm. I get the same equation

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Re: TrueTears question thread
« Reply #23 on: January 15, 2009, 12:10:22 am »
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is correct.

Also,


TrueTears

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Re: TrueTears question thread
« Reply #24 on: January 15, 2009, 02:47:02 pm »
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why can't u do ? ie the perpendicular component of the gravity should be equal to the normal force?
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TrueTears

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Re: TrueTears question thread
« Reply #25 on: January 15, 2009, 03:11:22 pm »
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and also

A space craft leaves earth to travel to the moon. How far from the centre of the is the spacecraft when it experiences a net force of zero?

(extra info
Earth mass :
Earth's radius of body:
Earth's radius of orbit around sun:
Moon's mass:
Moon's radius of body:
Moon's radius of orbit around earth: )

Any help would be much appreciated, ive been stuck on these for a while x.x"" lol
« Last Edit: January 16, 2009, 02:55:05 pm by TrueTears »
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Re: TrueTears question thread
« Reply #26 on: January 15, 2009, 09:20:30 pm »
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for number two, a hint would be to equate the equation for the gravitational force acting upon the satellite from the moon to the gravitation force acting upon the satellite from the earth as this would give you the distance where the net force is zero

You would get equations like this
Fnet (from moon) = GM(moon)m(satellite)/r(moon)^2
Fnet (from earth = GM(earth)m(satellite)/r(earth)^2

where r(moon) is the distance from moon centre to the satellite
and r(earth) is the distance from the earth centre to the satellite

hence at the point where net force  on satellite  = 0

m(moon)/r(moon)^2 = m(earth)/r(earth)^2

and since  r(moon) = 3.84 x 10^8 - r(earth)

m(moon)/ (3.84 x 10^8 - r(earth))^2 =  m(earth)/r(earth)^2

sub in relevant numbers to get the answer

p.s. the answer i think i get is 3.46 x 10^8 m. dunno how correct that is
« Last Edit: January 15, 2009, 09:23:36 pm by Table »

TrueTears

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Re: TrueTears question thread
« Reply #27 on: January 15, 2009, 09:33:08 pm »
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yeah i got that answer as well... the book says 3.02 x 10^8 hm...
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Re: TrueTears question thread
« Reply #28 on: January 16, 2009, 02:55:21 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.
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Re: TrueTears question thread
« Reply #29 on: February 24, 2009, 11:22:09 am »
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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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