What is the maximum height reached by two masses after an elastic collision?

[Lolagoeslala]

Homework Statement

a 6 kg, originally held at rest in position 1 as shown in the diagram below, slides down a frictionless quadrant of a verticle circle of radius 15 m. It then collides elastically with another 16 kg mass at rest at position 2. They bounce off each other and move on the horizontal frictionless surface and up the frictionless quadrant of a verticle circle on each side. Calculate the verticle height "H" for each ball at which they would momentarily come to rest after elastic collision.

Diagram: http://s1176.beta.photobucket.com/user/LolaGoesLala/media/jkmlkm.jpg.html

The Attempt at a Solution

So i was thinking of starting with the kinetic energy formula btu then i realized i need to include the gravitational potential energy but i need the height for it... so can anyone help me out on starting with this problem?

 

[tiny-tim]

Hi Lolagoeslala!

... so can anyone help me out on starting with this problem?

to start this problem, you need to find the speed of the first mass just before the collision, and then use conservation of energy and conservation of momentum to find the speeds just after the collision

 

[Lolagoeslala]

Hi Lolagoeslala!


to start this problem, you need to find the speed of the first mass just before the collision, and then use conservation of energy and conservation of momentum to find the speeds just after the collision

so to start this problem...
i would use the equation...

1/2m1v1 + mgh = 1/2m1v2
This represents the mass at rest state with the height which equals to the velocity at no height

correct? but what would the height be then?

 

[tiny-tim]

1/2m1v1 + mgh = 1/2m1v2

what is this ??

you can't add a momentum to an energy

 

[Lolagoeslala]

what is this ??

you can't add a momentum to an energy

OOPS sorry I am so sorry...
the equation is all wrong...
its supposed to be like this:

1/2m1v1^2 = mgh + 1/2m1v1`^2

this is for the conservation of energy just before it hits the collision for m1

 

[tiny-tim]

OOPS sorry I am so sorry...
the equation is all wrong...
its supposed to be like this:

1/2m1v1^2 = mgh + 1/2m1v1`^2

this is for the conservation of energy just before it hits the collision for m1

ok, at least that makes sense now!

but v₁' is 0

 

[Lolagoeslala]

ok, at least that makes sense now!

but v₁' is 0

Wait isn;t the v1 zero.. that is the intially and that is the rest position...
so wouldn;t it be

mgh = 1/2m1v1`^2 and we are trying to find the speed with which it hits the second ball?

 

[tiny-tim]

mgh = 1/2m1v1`^2 and we are trying to find the speed with which it hits the second ball?

yes, that's correct …

that gives you the initial speed (just before the collision) of the first ball

(btw, i would have used "u" rather than "v", for reasons which may shortly appear obvious )

now use conservation of energy and conservation of momentum for the elastic collision​

 

[Lolagoeslala]

yes, that's correct …

that gives you the initial speed (just before the collision) of the first ball

(btw, i would have used "u" rather than "v", for reasons which may shortly appear obvious )

now use conservation of energy and conservation of momentum for the elastic collision​

Wait ...

so i would use the equation
m1gh = 1/2m1v1`^2

now what's the height? is that the radius?

 

[tiny-tim]

now what's the height? is that the radius?

yes …

… slides down a frictionless quadrant of a verticle circle of radius 15 m

… it's a quadrant, so it goes from 3 oclock to 6 oclock … that's a height of one radius

 

[Lolagoeslala]

yes …… it's a quadrant, so it goes from 3 oclock to 6 oclock … that's a height of one radius

so...

m1gh = 1/2m1v1`^2
6kgx9.8m/s^2x15m = 1/2(6kg)v1`^2
882 J / 3 kg = v1`^2
√294 J = v1`
17.14 m/s = v1`

Is that correct? :D

 

[tiny-tim]

(try using the X² and X₂ buttons just above the Reply box

yes

(but you could have deleted m₁ from the equation completely, and so not used 6 kg at all )

 

[Lolagoeslala]

(try using the X² and X₂ buttons just above the Reply box

yes

(but you could have deleted m₁ from the equation completely, and so not used 6 kg at all )

Oh ok.. so now would i used the conservation of momentum

so like

m1v1 + m2v2 = m1v1` + m2v2`
m1v1 = m1v1` + m2v2`
but then i have two unknown variables v1` and v2`

 

[tiny-tim]

… but then i have two unknown variables v1` and v2`

yes, so you also need to use conservation of energy (for the collision)

 

[Lolagoeslala]

yes, so you also need to use conservation of energy (for the collision)

umm but wouldn't the energy be lost thenn ...
so they wouldn't be equal to one another would they...
lets see...

1/2m1v1`^2 = 1/2m1v1`^2 + 1/2m2v1`^2

i still have two missing variables..

 

[tiny-tim]

umm but wouldn't the energy be lost thenn ...
so they wouldn't be equal to one another would they...

but the question says it's elastic (ie, no energy is lost) …

… It then collides elastically with another 16 kg mass at rest at position 2 …

lets see...

1/2m1v1`^2 = 1/2m1v1`^2 + 1/2m2v1`^2

i still have two missing variables..

yes, you now have two equations and two unknowns …

so solve it​

 

[Lolagoeslala]

but the question says it's elastic (ie, no energy is lost) …



yes, you now have two equations and two unknowns …

so solve it​

v1`=(m1-m2/m1+m2)v1
V1`=(6 kg-16kg/22kg)17.14 m/s [E]
v1`= 7.79 m/s [W]

v2` = (2m1/m1+m2)v1
v2`= (12kg/22kg)17.14m/s[E]
v2`=9.34m/s[E]

 

[Lolagoeslala]

but the question says it's elastic (ie, no energy is lost) …



yes, you now have two equations and two unknowns …

so solve it​

YOU THERE?

 

[tiny-tim]

(just got up :zzz:)

v1`=(m1-m2/m1+m2)v1
…
v2` = (2m1/m1+m2)v1

yes, those equations are correct …

but how did you get them? ​

 

[Lolagoeslala]

(just got up :zzz:)


yes, those equations are correct …

but how did you get them? ​

school lesson we had to derive them so i had them.. now what should i do ?? :O?

 

[Lolagoeslala]

(just got up :zzz:)yes, those equations are correct …

but how did you get them? ​

ARE YOU SURE IM ON THE RIGHT TRACK?!? i asked this tutor about it, he was like no you don't need to find the 17.14 m/s, all you need is the energy .. and work with that. I am so CONFUSED

 

[haruspex]

i asked this tutor about it, he was like no you don't need to find the 17.14 m/s, all you need is the energy .. and work with that.

Your tutor is wrong. Without using conservation of momentum during the collision process you have no way of knowing how the kinetic energy is apportioned thereafter.
(Note that momentum is not conserved while the balls are traveling around the quadrants.)

 

[tiny-tim]

Hi Lolagoeslala!

(just got up :zzz:)

school lesson we had to derive them so i had them..

You won't have them in the exam.

It would be good practice to solve the equations directly.

(and yes, you do need to use energy conservation during the collision)

Now use energy conservation for the arcs.

 

[Lolagoeslala]

Hi Lolagoeslala!

(just got up :zzz:)


You won't have them in the exam.

It would be good practice to solve the equations directly.

(and yes, you do need to use energy conservation during the collision)

Now use energy conservation for the arcs.

WHATS THAT?
sorry for the caps..
whats the energy conservation for the arcs
is there some new equation for that?

 

[tiny-tim]

whats the energy conservation for the arcs
is there some new equation for that?

no, it's just the usual mgh stuff!

(i'm saying "energy conservation for the arcs" to distinguish it from "energy conservation for the collision" )

 

[haruspex]

(and yes, you do need to use energy conservation during the collision)

The confusion came from the tutor's statement that energy conservation was all that was needed, implying there was no need to use conservation of momentum during the collision.

 

[Lolagoeslala]

no, it's just the usual mgh stuff!

(i'm saying "energy conservation for the arcs" to distinguish it from "energy conservation for the collision" )

Alright so i did this calculation.

Find the total energy for m1 at height 15 m
ET1 = mgh
= (6kg)(9.8m/s)(15m)
= 882 J

Then i found the velocity of m1 as it will touch the ball
mgh = 1/2mv1^2
882 J = 1/2(6kg)v1^2
294 J = v1^2
17.146 m/s = v1

Then i used this equation to find the v1` final
v1` = (m1 - m2/m1+m2)v1
v1` = (-10/22) (17.146 m/s)
v1` = 7.79 m/s

then i used this equation to find the v2` final
v1+v1` =v2`
17.146 m/s

+ 7.79 m/s

= v2`
9.356 m/s = v2`
Then i used this equation to find the height reached by v1` and v2`

mgh = 1/2mv1`^2
mgh = 1/2mv2`^2

mgh = 1/2mv1`^2
(6kg)(9.8m/s^2)h = 1/2(6kg)(7.79m/s)^2
h = 3.096 m

mgh = 1/2mv2`^2
(16kg)(9.8m/s^2)h = 1/2(16kg)(9.356m/s)^2
h = 4.46 m

BUT why does the second ball increase in the height? ​

 

[Lolagoeslala]

The confusion came from the tutor's statement that energy conservation was all that was needed, implying there was no need to use conservation of momentum during the collision.

Alright so i did this calculation.

Find the total energy for m1 at height 15 m
ET1 = mgh
= (6kg)(9.8m/s)(15m)
= 882 J

Then i found the velocity of m1 as it will touch the ball
mgh = 1/2mv1^2
882 J = 1/2(6kg)v1^2
294 J = v1^2
17.146 m/s = v1

Then i used this equation to find the v1` final
v1` = (m1 - m2/m1+m2)v1
v1` = (-10/22) (17.146 m/s)
v1` = 7.79 m/s

then i used this equation to find the v2` final
v1+v1` =v2`
17.146 m/s

+ 7.79 m/s

= v2`
9.356 m/s = v2`
Then i used this equation to find the height reached by v1` and v2`

mgh = 1/2mv1`^2
mgh = 1/2mv2`^2

mgh = 1/2mv1`^2
(6kg)(9.8m/s^2)h = 1/2(6kg)(7.79m/s)^2
h = 3.096 m

mgh = 1/2mv2`^2
(16kg)(9.8m/s^2)h = 1/2(16kg)(9.356m/s)^2
h = 4.46 m

BUT why does the second ball increase in the height? ​

 

[haruspex]

then i used this equation to find the v2` final
v1+v1` =v2`

That's conservation of momentum? I thought the masses were different.

BUT why does the second ball increase in the height?

You mean, why does it go higher than the first ball? I think that will come right if you put in the two different masses.

 

[Lolagoeslala]

That's conservation of momentum? I thought the masses were different.

You mean, why does it go higher than the first ball? I think that will come right if you put in the two different masses.

Hi .. nope that's not the conervation of mometum..
that is a equation for an elastic collision...

 

[haruspex]

Ah, ok.
I calculate the ratio of the heights, h₂/h₁ = (2m₁/(m₁-m₂))². That agrees with your final answer. It follows that for m₁ to climb higher than m₂ it needs to have at most 1/3 of the mass of m₂.
Consider the case where the masses are the same. m₁ would stop dead and m₂ would climb to height h.

 

[Lolagoeslala]

Ah, ok.
I calculate the ratio of the heights, h₂/h₁ = (2m₁/(m₁-m₂))². That agrees with your final answer. It follows that for m₁ to climb higher than m₂ it needs to have at most 1/3 of the mass of m₂.
Consider the case where the masses are the same. m₁ would stop dead and m₂ would climb to height h.

but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?

 

[haruspex]

but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?

Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.

 

[Lolagoeslala]

Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.

Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?

 

[haruspex]

Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?

The scenario is not symmetric. As I suggested, consider two equal masses. The first (initiating) mass will stop dead and transfer all its KE to the other. Now make the first mass a bit smaller. It will bounce back some, but still most of the energy will be transferred. It won't achieve its 'fair' share of energy (i.e. in proportion to mass) until it's down to one third the mass of the other. And to get equal energy it would have to be smaller still.