Calculating Magnitude and Vector Componends given Momentum and Mass

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SUMMARY

The discussion focuses on calculating the components and magnitude of velocity from momentum and mass, specifically for a 2.0 kg object with a momentum of 10 kg·m/s at a 37° angle vertically upward from due west. The user correctly applies the momentum formula (p=mv) to find a velocity of 5 m/s. They utilize trigonometric functions and the Pythagorean theorem to derive the components of the velocity, ultimately calculating the magnitude as approximately 7.91 m. The final components are identified as x = -3.77 m, y = 5 m, and z = 6.26 m.

PREREQUISITES
  • Understanding of momentum (p=mv)
  • Basic knowledge of trigonometry (e.g., tangent function)
  • Familiarity with the Pythagorean theorem
  • Concept of vector components in physics
NEXT STEPS
  • Study vector decomposition techniques in physics
  • Learn about momentum conservation principles
  • Explore the relationship between velocity, distance, and time
  • Investigate advanced applications of trigonometry in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to clarify concepts related to momentum and velocity calculations.

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..

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
 
Physics news on Phys.org
sorry that final column vector probably should have read

x - -3.767770251m x-hat
y - 5m y-hat
z - 6.260678291m z-hat
 

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