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?

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

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