anonymous
  • anonymous
A 5.0 kg stationary grenade explodes into three pieces. If a 0.5 kg piece moves at 35 m/s [E] and a 1.5 kg piece moves at 25 m/s [S 30° E], determine the velocity of the third piece?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I am using the trigonometry law to account for the final piece, well, as far as conservation of energy holds true .
anonymous
  • anonymous
The addition of all of these forces must equal zero
anonymous
  • anonymous
0.5kg/s*35m/s=17.5kgm/s 1.5kg/s*25m/s=37.5kgm/s Now use the vector arrows to indicate for the hypotenuse

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More answers

anonymous
  • anonymous
|dw:1440660915883:dw|
anonymous
  • anonymous
Use law of conservation of momentum \[\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}\] momentum before=momentum after \[M\vec u=m_{1}\vec v_{1}+m_{2} \vec v_{2} +m_{3} \vec v_{3}\] \[\vec v_{3}=\frac{1}{m_{3}}.(M \vec u-m_{1} \vec v_{1}-m_{2} \vec v_{2})\]
anonymous
  • anonymous
|dw:1440660968779:dw|
anonymous
  • anonymous
The triangle is imaginary, hypotenuse is pointing towards upper right.
anonymous
  • anonymous
We know that \[\vec u=\vec 0\] Therefore our equation becomes \[\vec v_{3}=-\frac{m_{1} \vec v_{1}+m_{2} \vec v_{2}}{m_{3}}\]
anonymous
  • anonymous
Right, and final velocity obtained will be hence (17.5+37.5)N/3.0kg
anonymous
  • anonymous
Now I am only left to find the direction which satisfies the vector with cosine
anonymous
  • anonymous
or so I think
anonymous
  • anonymous
\[v_{3}=|\vec v_{3}|=|-\frac{m_{1}\vec v_{1}+m_{2} \vec v_{2}}{m_{3}}|=\frac{1}{m_{3}}.(m_{1}v_{1}+m_{2}v_{2})\] \[v_{1}=|\vec v_{1}|=\sqrt{v_{1x}^2+v_{1y}^2}, v_{2}=|\vec v_{2}|=\sqrt{v_{2x}^2+v_{2y}^2}\]
anonymous
  • anonymous
Now I get it
anonymous
  • anonymous
But I think this assumes one dimensional momentum though
anonymous
  • anonymous
Two dimensions where two are canceling each other out becomes a question of finding the hypotenuse.
Jhannybean
  • Jhannybean
This is a general physics question dealing with the momentum?
anonymous
  • anonymous
Look|dw:1440661555489:dw|
anonymous
  • anonymous
|dw:1440661454515:dw|
anonymous
  • anonymous
If you mean 30 degrees south of east, just switch the angles
anonymous
  • anonymous
|dw:1440662053313:dw|
anonymous
  • anonymous
Ops.
anonymous
  • anonymous
Now left with the sine law to get the vector version of the velocity.
anonymous
  • anonymous
This was a good question
Astrophysics
  • Astrophysics
You want a difficult physics question? :P
Astrophysics
  • Astrophysics
Try this |dw:1440662683161:dw|
anonymous
  • anonymous
Angle to the horizontal?
Astrophysics
  • Astrophysics
No numbers, but a hint: Both the incline and box are accelerating
anonymous
  • anonymous
I would think that it depends on the mass of each object
anonymous
  • anonymous
and gravity being 9.8m/s^2
anonymous
  • anonymous
and mass cancels out
Astrophysics
  • Astrophysics
Hmm nope, try setting it up, as you would a regular incline plane question, but remember the incline itself is accelerating.
anonymous
  • anonymous
|dw:1440663228557:dw|
Astrophysics
  • Astrophysics
Not bad but remember the incline also has weight and a normal force |dw:1440663465987:dw|
anonymous
  • anonymous
Center mass originates from the tip of force due to gravity huh?
anonymous
  • anonymous
Man gotta love physics
imqwerty
  • imqwerty
The force that will generate acceleration in the wedge would be the horizontal component of the normal force by the smaller box :)
anonymous
  • anonymous
Yes, but how would you calculate that using gravity?
anonymous
  • anonymous
Angle would have to be known as well because normal force decreases as the angle increases.
anonymous
  • anonymous
Good question anyhow
imqwerty
  • imqwerty
u can do this by 2 ways - 1) constraint relation 2)and the one that i told u :)
imqwerty
  • imqwerty
|dw:1440664770880:dw|

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