Is this correct- momentum question

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  • #1
klm
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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?
 
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  • #2
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.
 
  • #3
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)
 
  • #4
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
[tex]\vec p_1 + \vec p_2 = \vec p_f[/tex]
where [tex]\vec p_i = m_i v_i[/tex]
the speed that you are after is really [tex]|\vec v_f|[/tex]
 
  • #5
so i don't need the components then? or did i find the components of Vf..?
 
  • #6
klm said:
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
 
  • #7
okay for the first ball: Vx= 2 and Vy= 0
second ball: Vx= -1cos(30)
Vy= -1sin(30)

would the be correct?
 
  • #8
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 [itex]0.04\vec{i}[/itex]. 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 [itex]-0.03\vec{j}[/itex].

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

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

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

The speed is the length of that vector.
 
  • #9
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.
 
  • #10
because i seem to be getting 1 and that is incorrect...
 
  • #11
klm said:
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...
 
  • #12
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
 
  • #13
so if i have to do m1v1+ m2v2 :
(.02)(2) + (.03)(1)
 
  • #14
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.)
 
  • #15
klm said:
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 vy = -.5. Check it.
 
  • #16
oh so is it -1
so .04i - .03j
 
  • #17
haha yeah sorry. it is -.5, i forgot the neg sign sorry!
 
  • #18
Conservation of linear momentum
In x-direction: m1*v1,x + m2*v2,x = (m1 + m2)*vf,x
In y-direction: m1*v1,y + m2*v2,y = (m1 + m2)*vf,y
 
  • #19
okay sorry this might be a dumb question, but why do we need to break up the conservation into components?
 
  • #20
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?
 
  • #21
klm said:
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!
 
  • #22
klm said:
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
 

1. What is momentum?

Momentum is a physical quantity that measures the motion of an object. It is the product of an object's mass and its velocity.

2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass (m) by its velocity (v). The formula for momentum is p = mv.

3. What are the units of momentum?

The units of momentum depend on the units used for mass and velocity. In the SI system, momentum is measured in kilogram meters per second (kg⋅m/s).

4. Does momentum have direction?

Yes, momentum has direction. It is a vector quantity, meaning it has both magnitude and direction. The direction of an object's momentum is the same as its velocity.

5. How is momentum conserved in a system?

In a closed system, the total momentum remains constant. This means that the momentum before an event (such as a collision) is equal to the momentum after the event. This is known as the law of conservation of momentum.

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