- #1
zboomer
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Hi =) just signed up, and i really need help this term. I am really having a hard time this term absorbing what's going on, I am getting things mixed up, and I am falling behind.. I am finding myself quickly losing my confidence and becoming quite frustrated and stressed out ;( I am really sorry to open up that way but =) its good to be here, i'll probably be here quite frequently.. i appreciate all patience with me..
"A 2.0kg object's momentum at a certain time is 10kg.m/s 37° vertically upward from due west. What are the components and magnitude of its velocity at this time (in a frame in standard orientation)?"
p=mv
v=dr/dt
mag(v)=√((dx)^2 + (dy)^2 +(dz)^2)
a^2 + b^2 = c^2
im not getting very far but here's where I am at:
first i plug what i know into the formula for momentum.
10kg.m/s = 2kg.v
i divide 10kg.m/s by 2kg (canceling out my kg units) leaving me with a velocity of 5m.s
i understand that if i draw a picture, i basically am looking at an object that is rising at a rate of 5m.s 37° west of my position. what's really killin me is that i can't figure out how to go backwards to find any distances to go with to figure out my magnitude, or even the actual time to divide those distances from ;\
what i have done is since i havnt been given a time interval, i plugged 5 for my dr, and used 1 second for dt
since i had a distance of 5 meters:
5tan(37°) = -3.767770251m x-hat for my adj. (the ground), and i used the Pythagorean theorem to find a hypotenuse. i know I am just slaughtering the terminology..
a^2 + b^2 = c^2
√((5)^2 + (-3.767770251)^2) = 6.260678291m
i didnt know what to do from here to satisfy my dr portion so i used C as my dr so i could divide by my velocity (5m.s) to find a more accurate time interval to finish things off with
6.260678291m / 5m.s
t = 1.252135658seconds
canceling out the meters I am left with
i then used this to find my final magnitude
mag(v) = √(((5)^2 + (-3.767770251)^2 + (6.260678291)^2) / t)
mag(v) = 7.912444813
with my final components being
x - -3.767770251m x-hat
y - 6.260678291m y-hat
z - 5m z-hat
i know i skillfully messed this problem up early.. and I am sorry if the way i spewed this all out is not very conducive or easy to follow.. I am really struggling here ;\
i appreciate your time..
zboomer
Homework Statement
"A 2.0kg object's momentum at a certain time is 10kg.m/s 37° vertically upward from due west. What are the components and magnitude of its velocity at this time (in a frame in standard orientation)?"
Homework Equations
p=mv
v=dr/dt
mag(v)=√((dx)^2 + (dy)^2 +(dz)^2)
a^2 + b^2 = c^2
The Attempt at a Solution
im not getting very far but here's where I am at:
first i plug what i know into the formula for momentum.
10kg.m/s = 2kg.v
i divide 10kg.m/s by 2kg (canceling out my kg units) leaving me with a velocity of 5m.s
i understand that if i draw a picture, i basically am looking at an object that is rising at a rate of 5m.s 37° west of my position. what's really killin me is that i can't figure out how to go backwards to find any distances to go with to figure out my magnitude, or even the actual time to divide those distances from ;\
what i have done is since i havnt been given a time interval, i plugged 5 for my dr, and used 1 second for dt
since i had a distance of 5 meters:
5tan(37°) = -3.767770251m x-hat for my adj. (the ground), and i used the Pythagorean theorem to find a hypotenuse. i know I am just slaughtering the terminology..
a^2 + b^2 = c^2
√((5)^2 + (-3.767770251)^2) = 6.260678291m
i didnt know what to do from here to satisfy my dr portion so i used C as my dr so i could divide by my velocity (5m.s) to find a more accurate time interval to finish things off with
6.260678291m / 5m.s
t = 1.252135658seconds
canceling out the meters I am left with
i then used this to find my final magnitude
mag(v) = √(((5)^2 + (-3.767770251)^2 + (6.260678291)^2) / t)
mag(v) = 7.912444813
with my final components being
x - -3.767770251m x-hat
y - 6.260678291m y-hat
z - 5m z-hat
i know i skillfully messed this problem up early.. and I am sorry if the way i spewed this all out is not very conducive or easy to follow.. I am really struggling here ;\
i appreciate your time..
zboomer