## anonymous one year ago 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?

1. anonymous

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

2. anonymous

The addition of all of these forces must equal zero

3. 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

4. anonymous

|dw:1440660915883:dw|

5. anonymous

Use law of conservation of momentum $\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}$ momentum before=momentum after $M\vec u=m_{1}\vec v_{1}+m_{2} \vec v_{2} +m_{3} \vec v_{3}$ $\vec v_{3}=\frac{1}{m_{3}}.(M \vec u-m_{1} \vec v_{1}-m_{2} \vec v_{2})$

6. anonymous

|dw:1440660968779:dw|

7. anonymous

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

8. anonymous

We know that $\vec u=\vec 0$ Therefore our equation becomes $\vec v_{3}=-\frac{m_{1} \vec v_{1}+m_{2} \vec v_{2}}{m_{3}}$

9. anonymous

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

10. anonymous

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

11. anonymous

or so I think

12. anonymous

$v_{3}=|\vec v_{3}|=|-\frac{m_{1}\vec v_{1}+m_{2} \vec v_{2}}{m_{3}}|=\frac{1}{m_{3}}.(m_{1}v_{1}+m_{2}v_{2})$ $v_{1}=|\vec v_{1}|=\sqrt{v_{1x}^2+v_{1y}^2}, v_{2}=|\vec v_{2}|=\sqrt{v_{2x}^2+v_{2y}^2}$

13. anonymous

Now I get it

14. anonymous

But I think this assumes one dimensional momentum though

15. anonymous

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

16. Jhannybean

This is a general physics question dealing with the momentum?

17. anonymous

Look|dw:1440661555489:dw|

18. anonymous

|dw:1440661454515:dw|

19. anonymous

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

20. anonymous

|dw:1440662053313:dw|

21. anonymous

Ops.

22. anonymous

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

23. anonymous

This was a good question

24. Astrophysics

You want a difficult physics question? :P

25. Astrophysics

Try this |dw:1440662683161:dw|

26. anonymous

Angle to the horizontal?

27. Astrophysics

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

28. anonymous

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

29. anonymous

and gravity being 9.8m/s^2

30. anonymous

and mass cancels out

31. Astrophysics

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

32. anonymous

|dw:1440663228557:dw|

33. Astrophysics

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

34. anonymous

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

35. anonymous

Man gotta love physics

36. imqwerty

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

37. anonymous

Yes, but how would you calculate that using gravity?

38. anonymous

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

39. anonymous

Good question anyhow

40. imqwerty

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

41. imqwerty

|dw:1440664770880:dw|