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?

- anonymous

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- schrodinger

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- anonymous

I am using the trigonometry law to account for the final piece, well, as far as conservation of energy holds true .

- anonymous

The addition of all of these forces must equal zero

- 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

|dw:1440660915883:dw|

- 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

|dw:1440660968779:dw|

- anonymous

The triangle is imaginary, hypotenuse is pointing towards upper right.

- 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

Right, and final velocity obtained will be hence (17.5+37.5)N/3.0kg

- anonymous

Now I am only left to find the direction which satisfies the vector with cosine

- anonymous

or so I think

- 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

Now I get it

- anonymous

But I think this assumes one dimensional momentum though

- anonymous

Two dimensions where two are canceling each other out becomes a question of finding the hypotenuse.

- Jhannybean

This is a general physics question dealing with the momentum?

- anonymous

Look|dw:1440661555489:dw|

- anonymous

|dw:1440661454515:dw|

- anonymous

If you mean 30 degrees south of east, just switch the angles

- anonymous

|dw:1440662053313:dw|

- anonymous

Ops.

- anonymous

Now left with the sine law to get the vector version of the velocity.

- anonymous

This was a good question

- Astrophysics

You want a difficult physics question? :P

- Astrophysics

Try this |dw:1440662683161:dw|

- anonymous

Angle to the horizontal?

- Astrophysics

No numbers, but a hint: Both the incline and box are accelerating

- anonymous

I would think that it depends on the mass of each object

- anonymous

and gravity being 9.8m/s^2

- anonymous

and mass cancels out

- Astrophysics

Hmm nope, try setting it up, as you would a regular incline plane question, but remember the incline itself is accelerating.

- anonymous

|dw:1440663228557:dw|

- Astrophysics

Not bad but remember the incline also has weight and a normal force |dw:1440663465987:dw|

- anonymous

Center mass originates from the tip of force due to gravity huh?

- anonymous

Man gotta love physics

- imqwerty

The force that will generate acceleration in the wedge would be the horizontal component of the normal force by the smaller box :)

- anonymous

Yes, but how would you calculate that using gravity?

- anonymous

Angle would have to be known as well because normal force decreases as the angle increases.

- anonymous

Good question anyhow

- imqwerty

u can do this by 2 ways -
1) constraint relation
2)and the one that i told u :)

- imqwerty

|dw:1440664770880:dw|

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