Angle from force applied / Momentum question

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SUMMARY

The discussion focuses on calculating the impulse delivered to a hockey puck with a mass of 0.250 kg and initial velocity of 3.00 i + 5.00 j m/s, resulting in a final velocity of -37.0 i + 25.0 j m/s. The magnitude of the impulse is determined to be 11.18 kg m/s. Additionally, the angle of the force applied to the puck is to be calculated with respect to the positive x-axis, requiring the use of vector components. The discussion also touches on the concept of total momentum being zero for multiple particles, even when individual momenta are not.

PREREQUISITES
  • Understanding of momentum (p = mv)
  • Familiarity with vector components and angles
  • Knowledge of impulse and its calculation
  • Basic physics concepts related to collisions
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  • Learn how to calculate angles using trigonometric functions in vector analysis
  • Study the principles of impulse-momentum theorem
  • Explore vector diagrams for visualizing forces and momentum
  • Investigate scenarios where total momentum can be zero despite non-zero individual momenta
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Kaln0s
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Homework Statement



Okay the first question is:

1. A hockey puck with mass 0.250 kg has initial velocity 3.00 i + 5.00 j m/s just before being hit by a hockey stick. The final velocity after the hit is -37.0 i + 25.0 j m/s. What was the magnitude of the impulse delivered by the stick to the puck, in kg m/s?

For this I just took the final velocity - initial and took the magnitude to get .250 sqrt[-40^2 + 20^2] = 11.18

Part two though I have no idea how to do.

2. What is the angle of the force applied to the hockey puck in the previous problem? The angle should be measured with respect to the positive x-axis in degrees.

below.

3. Is it possible for the total momentum of three particles to be zero, even though the momentum of the individual particles is not?

below.

Homework Equations



p= mv
Velocityf - Velocityi

The Attempt at a Solution



2. I'm not sure how to do this... I thought of just doing the y over x component 400/1600 = 1/4 = 45 degrees. I don't think that's right though because looking at the final velocity it looks like it will be in quadrant 2 maybe?

3. I said yes as long as they were in the same plane because they wouldn't be able to be expressed in terms of each other. (mostly need confirmation on this yes / no)

Thanks for your help!

:biggrin:
 
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2. you have the direction of the initial, final velocity, and you have also calulated the direction of the impulse. Draw a vector diagram and find any angle that is necessary.
 

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