A community for students.
Here's the question you clicked on:
 0 viewing
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?
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am using the trigonometry law to account for the final piece, well, as far as conservation of energy holds true .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The addition of all of these forces must equal zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00.5kg/s*35m/s=17.5kgm/s 1.5kg/s*25m/s=37.5kgm/s Now use the vector arrows to indicate for the hypotenuse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440660915883:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use law of conservation of momentum \[\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}\] momentum before=momentum after \[M\vec u=m_{1}\vec v_{1}+m_{2} \vec v_{2} +m_{3} \vec v_{3}\] \[\vec v_{3}=\frac{1}{m_{3}}.(M \vec um_{1} \vec v_{1}m_{2} \vec v_{2})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440660968779:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The triangle is imaginary, hypotenuse is pointing towards upper right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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
 one year ago
Best ResponseYou've already chosen the best response.0Right, and final velocity obtained will be hence (17.5+37.5)N/3.0kg

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now I am only left to find the direction which satisfies the vector with cosine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[v_{3}=\vec v_{3}=\frac{m_{1}\vec v_{1}+m_{2} \vec v_{2}}{m_{3}}=\frac{1}{m_{3}}.(m_{1}v_{1}+m_{2}v_{2})\] \[v_{1}=\vec v_{1}=\sqrt{v_{1x}^2+v_{1y}^2}, v_{2}=\vec v_{2}=\sqrt{v_{2x}^2+v_{2y}^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I think this assumes one dimensional momentum though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Two dimensions where two are canceling each other out becomes a question of finding the hypotenuse.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0This is a general physics question dealing with the momentum?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lookdw:1440661555489:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440661454515:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you mean 30 degrees south of east, just switch the angles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440662053313:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now left with the sine law to get the vector version of the velocity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This was a good question

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You want a difficult physics question? :P

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Try this dw:1440662683161:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Angle to the horizontal?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0No numbers, but a hint: Both the incline and box are accelerating

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would think that it depends on the mass of each object

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and gravity being 9.8m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and mass cancels out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Hmm nope, try setting it up, as you would a regular incline plane question, but remember the incline itself is accelerating.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440663228557:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Not bad but remember the incline also has weight and a normal force dw:1440663465987:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Center mass originates from the tip of force due to gravity huh?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Man gotta love physics

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0The force that will generate acceleration in the wedge would be the horizontal component of the normal force by the smaller box :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, but how would you calculate that using gravity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Angle would have to be known as well because normal force decreases as the angle increases.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good question anyhow

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0u can do this by 2 ways  1) constraint relation 2)and the one that i told u :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440664770880:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.