Object Thrown Upward: Max Momentum & Equilibrium

[garva1]

an object is thrown upwards, at the top of the path it has
a) zero acceleration
b) maximum momentum
c) maximum kE
d) is at equillibrium
e)none of the above


i know accelarion is 9.8m/s^2
i also know is not KE since at the top is potential
im confused of the other two

 

[LowlyPion]

an object is thrown upwards, at the top of the path it has
a) zero acceleration
b) maximum momentum
c) maximum kE
d) is at equillibrium
e)none of the abovei know accelarion is 9.8m/s^2
i also know is not KE since at the top is potential
im confused of the other two

b) what is formula for momentum?
d) what is equilibrium?

 

[garva1]

those are the ones that i don't know i think momentum is = mv

 

[chislam]

Ok, so if you know that P = mv, and the object is at the max height of its trajectory, then what would be it's velocity and momentum?

 

[garva1]

v is zero

 

[garva1]

ok vo is zero but the final velocity =v + gt but I am confused

 

[gabbagabbahey]

yes, v=0 at the top, so what is the momentum at the top?

 

[garva1]

0 but what about equllibrium

 

[gabbagabbahey]

look up the definition of mechanical equilibrium

 

[garva1]

i don't get the definition is says something about rest but so does the free fall at top have equllibrium but then we only have one force gravity since air is neglected

 

[LowlyPion]

i don't get the definition is says something about rest but so does the free fall at top have equllibrium but then we only have one force gravity since air is neglected

OK. Is it at rest at the top?

 

[gabbagabbahey]

http://en.wikipedia.org/wiki/Mechanical_equilibrium"

"A rigid body is in mechanical equilibrium when the sum of all forces on all particles of the system is zero, and also the sum of all torques on all particles of the system is zero."

What forces are at play here? Is the net force on the particle zero at the top of its path?

 

[garva1]

there is only gravity

 

[garva1]

i don't get equllibrium sorry :(

 

[chislam]

Ok, try and think fundamentally. At the very top, the instantaneous velocity is zero. If the velocity is zero then the net acceleration must also be zero at that very instant. So if F = ma, then the net force must be zero too. If you are still confused, I frankly do not know how else to explain it.

 

[garva1]

so at the top there is equllibrium even though there is only one force the sum will be concidered zero

 

[garva1]

"If the velocity is zero then the net acceleration must also be zero at that very instant. So if F = ma,"
but how does velocity applied in the equation f=ma

 

[garva1]

but i thought that the acceleration was gravity and gravity was throughout the whole path

 

[gabbagabbahey]

Ok, try and think fundamentally. At the very top, the instantaneous velocity is zero. If the velocity is zero then the net acceleration must also be zero at that very instant. So if F = ma, then the net force must be zero too. If you are still confused, I frankly do not know how else to explain it.

That's incorrect, the velocity will be zero, but it will still be accelerating downwards; the net force will still just be the force due to gravity, which is nonzero.

 

[gabbagabbahey]

but i thought that the acceleration was gravity and gravity was throughout the whole path

This thought is correct; see above^^^.

 

[garva1]

but i thought the a was gravity and top gravity still 9.8

 

[garva1]

so there is no equillibrium nor momentum nor kinetic energy and the acceleration is 9.8

 

[gabbagabbahey]

so there is no equillibrium nor momentum nor kinetic energy and the acceleration is 9.8

Correct, so the answer is (e) none of the above.

 

[garva1]

yayy THANK YOU VERY MUCH I am really thankfull so no do i just leave the thread open or there is a way to close it
THANK YOU

 

[Kurdt]

Just leave the thread, so others may benefit from the discussion.

 

[garva1]

Thank you