Is this correct- momentum question

[klm]

A 20 g ball of clay traveling east at 2.0m/s collides with a 30 g ball of clay traveling 30 degree south of west at 1.0m/s.
What are the speed of the resulting 50 g blob of clay?

so this is what i i did:

mivi=mfvf
vx= (.02)(2) / (.05) = .8
vy= (.03)(-1)/ .05 = -.6

is that much correct so far? and then i can use vy= v sin theta to find v?

 

[mjsd]

hard to understand what you have done with your notations.
but this is clearly a conservation of momentum question. so remember this is a 2D problem, so both total momentun in x and y direction must be the same before and after.

 

[klm]

sorry for the messy calculations!

i am basically finding the Vx = ( mass of ball 1)( velocity of ball 1) / ( total mass )
so Vx = ( 0.02 kg) ( 2 m/s ) / ( 0.05 kg ) = 0.8

and Vy = (.03 kg) ( -1 m/s) / (.05 kg) = -.6
(is the bold part in this equation correct? )
but i am a little confused if i can use this equation to find V now, Vy= v sin(theta)

 

[mjsd]

As far as i understand it. you have two momentum vectors to begin with and you need to find the resultant vector. so due to momentum conservation, you get
\vec p_1 + \vec p_2 = \vec p_f
where \vec p_i = m_i v_i
the speed that you are after is really |\vec v_f|

 

[klm]

so i don't need the components then? or did i find the components of Vf..?

 

[mjsd]

so i don't need the components then? or did i find the components of Vf..?

do you know how to handle vectors? do u know how to add them?

first resolve everything into component forms and get two equations: one for each x and y

 

[klm]

okay for the first ball: Vx= 2 and Vy= 0
second ball: Vx= -1cos(30)
Vy= -1sin(30)

would the be correct?

 

[HallsofIvy]

Let me see if I can give more detail about what you have done.
You have one ball of 20 g (0.02 kg) moving east at 2 m/s. It's scalar momentum is 0.02(2)= 0.04 kgm/s so the vector momentum is 0.04\vec{i}. I can see no reason to divide by total mass.

You have one ball of 30 g (0.03 kg) moving south at 1 m/s. Its scalar momentum is 0.03(1)= 0.03 kgm/s so the vector momentum is -0.03\vec{j}.

The total momentum vector before collision is 0.04\vec{i}- 0.03\vec{j}.

The combined balls must have the same momentum vector, so letting \vec{v} be its velocity vector, (0.02+ 0.03)\vec{v}= 0.04\vec{i}- 0.03\vec{j}.

NOW you can get the velocity vector by dividing that equation by the total mass 0.05 kg: \vec{v}= 0.80\vec{i}- 0.06\vec{j}

The speed is the length of that vector.

 

[klm]

okay i think i understand what you did...but i thought for some reason you should divide by the total mass, but i think we somehow got to same place either way. but one question, should that 0.06 be 0.6 ( just want to make sure, not trying to be picky :)! so the speed would just be .8^2 + .6^2 and then the square root of that number.

 

[klm]

because i seem to be getting 1 and that is incorrect...

 

[saket]

okay for the first ball: Vx= 2 and Vy= 0
second ball: Vx= -1cos(30)
Vy= -1sin(30)

would the be correct?

That looks good to me! (Assuming +x to be along east direction and +y along north.)
Go ahead...

 

[klm]

ok so for ball 2 Vx= -.866 and Vy= .5 and then the magnitude of that is .999
and ball 1 the magnitude is just 2

 

[klm]

so if i have to do m1v1+ m2v2 :
(.02)(2) + (.03)(1)

 

[saket]

no... u have to conserve linear momentum in each direction seperately! (remember, momentum is a vector quantity.. so treat it as a vector... in 1-D, since it can be only positive or negative direction, you used to get result even considering as scalar -- but with apprpriate positive and negative signs.)

 

[saket]

ok so for ball 2 Vx= -.866 and Vy= .5 and then the magnitude of that is .999
and ball 1 the magnitude is just 2

Of course it would be 2 for 1st ball, and 1 for 2nd ball. From this only you got these components!
Maybe it's a typing mistake.. but v_y = -.5. Check it.

 

[klm]

oh so is it -1
so .04i - .03j

 

[klm]

haha yeah sorry. it is -.5, i forgot the neg sign sorry!

 

[saket]

Conservation of linear momentum
In x-direction: m₁*v_1,x + m₂*v_2,x = (m₁ + m₂)*v_f,x
In y-direction: m₁*v_1,y + m₂*v_2,y = (m₁ + m₂)*v_f,y

 

[klm]

okay sorry this might be a dumb question, but why do we need to break up the conservation into components?

 

[klm]

okay so i got .2804 for vfx and -.3 for vfy and then i just do square them both and take the square root and get .411 which should be the speed correct?

 

[saket]

okay sorry this might be a dumb question, but why do we need to break up the conservation into components?

we are not breaking conservation.. we are breaking momentum -- a vector quantity .. in two components.. and then applying conservation on each of these components individually, because we know that.. at 90 degrees, things do not interfere. (Angle between x and y-axis 90 degree) .. now if you will ask why it is so.. very simple, a vector quantity can affect another, only if it has a component in the latter direction.. and Cos 90 = 0!

 

[saket]

okay so i got .2804 for vfx and -.3 for vfy and then i just do square them both and take the square root and get .411 which should be the speed correct?

seems good