Calculating Magnitude and Vector Componends given Momentum and Mass

In summary, the speaker is a new user who needs help with schoolwork and is feeling stressed and frustrated. They are specifically struggling with a problem involving an object's momentum and velocity. They have attempted to solve the problem but are unsure of their approach and have made several errors. They are seeking assistance and appreciate any patience and guidance.
  • #1
zboomer
2
0
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
 
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  • #2
sorry that final column vector probably should have read

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

Related to Calculating Magnitude and Vector Componends given Momentum and Mass

1. What is the equation for calculating magnitude given momentum and mass?

The equation for calculating magnitude is magnitude = momentum / mass.

2. How do you calculate the vector components given momentum and mass?

To calculate the vector components, you will need to use trigonometry. First, find the angle between the momentum vector and the x-axis. Then, use the formula for the x-component = magnitude * cos(angle) and the formula for the y-component = magnitude * sin(angle).

3. What units are used for momentum and mass?

Momentum is typically measured in kg*m/s, while mass is measured in kilograms.

4. Can you explain the concept of momentum and how it relates to mass?

Momentum is a property of a moving object that describes its tendency to keep moving in the same direction at the same speed. It is directly proportional to the mass of the object and its velocity. This means that the more mass an object has, the more momentum it will have, and the faster it is moving, the more momentum it will have as well.

5. How do you use the calculated magnitude and vector components in real-life scenarios?

Calculated magnitude and vector components can be used in various real-life scenarios, such as in physics experiments or in engineering designs. For example, they can help determine the force needed to move an object or to design a structure that can withstand a certain amount of momentum. They can also be used in sports to analyze and improve the performance of athletes, as well as in space exploration to plan trajectories and maneuvers of spacecraft.

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