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Find the momentum using i +j

  • Thread starter Sneakatone
  • Start date
  • #1
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it ask for the format in the picture provided. I dont know what is i or j. The equation for momentum is P=mv .

a)Im guessing initial momentum is 26*0.43=11.18.
unless its 26m/s at max height then initial momentum would be zero.
 

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  • #2
461
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The question asks you to decompose the momentum vector in its horizontal i and vertical j directions.The notation (i ,j ,k ) represents the units vectors in a along the x ,y z direction.
 
  • #3
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so 11.18 is the hypotenuse making a 90,27, 63 triangle. how would I find the sides?
 
  • #4
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so 11.18 is the hypotenuse making a 90,27, 63 triangle. how would I find the sides?
Trigonometry
 
  • #5
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I ended up with x=5.08, y=9.96 ,
would that be the values for initial or maximum momentum?
 
  • #6
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These are the components of the initial momentum.
Can you work out the ball's momentum at its maximum height?
 
  • #7
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wouldnt momentum at max height be zero because velocity would also be zero?
 
  • #8
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Well, the y-component of the velocity will be zero.
What about the x-component?
 
  • #9
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I used the equation t=(vf-vi)/g
(0-26)/9.82=2.6s
5.08/2.6=1.9m/s * 0.43kg=0.84 kg* m/s
 
  • #10
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What are the forces in the x-direction?
What does this tell you about how the x-component of the velocity changes?
 
  • #11
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Are there no forces that act on it?
 
  • #12
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After it leaves the foot, the ball only has the gravitational force on it acting vertically down.
There are clearly no horizontal forces on it, so the horizontal component of the velocity will be constant.
You should be able to calculate the momentum at the maximum height knowing this.
 
  • #13
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so horizontal components will be the same as initial for max height.
based on the values I have for initial x=5.08, y=9.96 , I dont know what to do with them. I tried multiplying them by mass to get momentum.
 
  • #14
Doc Al
Mentor
44,945
1,206
I ended up with x=5.08, y=9.96 ,
The angle given is 27°. Although it doesn't specify, I suspect that the angle is with respect to the horizontal. (If so, you have the initial momentum components reversed.)

so horizontal components will be the same as initial for max height.
based on the values I have for initial x=5.08, y=9.96 , I dont know what to do with them. I tried multiplying them by mass to get momentum.
Those are already the momentum components. (You already multiplied by the mass in an earlier step.)

At the highest point, one component goes to zero while the other remains what it was at the start.
 
  • #15
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you are correct.
a)x=9.96 y=5.08
b)x=0.96 y=0
c) x=9.96 y=?
I still dont know the y component for y when the ball hits the ground.
 
  • #16
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you are correct.
a)x=9.96 y=5.08
b)x=0.96 y=0
c) x=9.96 y=?
I still dont know the y component for y when the ball hits the ground.
I presume you mean x=9.96 for (b).

From 2D kinematics, what is the y-component of the velocity when the ball hits the ground compared with its value to start with.
 
  • #17
Doc Al
Mentor
44,945
1,206
I still dont know the y component for y when the ball hits the ground.
If you throw something straight up with a speed V, what speed (and direction) will it have when it falls back down to your hand?
 
  • #18
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yes sorry 9.96=x for b

its negative from the downward force of gravity
 
  • #19
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@doc al the speed falling back down would be the same(speed V)
 
  • #20
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I got it now it was -5.08=y ,
Thank you all for your help!
 

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