Momentum and Impulse questions

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SUMMARY

The discussion centers on calculating the total normal force exerted by the ground on a skateboard during the impact of a 45-kg boy landing on it at a velocity of 5 m/s, with an impact duration of 0.05 seconds and an angle of 30 degrees below the horizontal. The calculations presented indicate a normal force of approximately -2009.5 N, which does not match any of the provided multiple-choice answers. Participants emphasize the importance of correctly applying the impulse-momentum theorem and ensuring accurate sign usage in momentum calculations.

PREREQUISITES
  • Understanding of impulse-momentum theorem
  • Basic knowledge of vector components in physics
  • Familiarity with Newton's second law of motion
  • Ability to perform calculations involving forces and mass
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  • Review the impulse-momentum theorem and its applications in collision problems
  • Learn how to resolve vectors into their components, particularly in projectile motion
  • Study Newton's laws of motion, focusing on applications in real-world scenarios
  • Practice solving problems involving normal forces and impacts in physics
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This discussion is beneficial for physics students, educators, and anyone involved in mechanics, particularly those studying dynamics and impulse-related problems in collisions.

sugz
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Homework Statement


The 45-kg boy has taken a running jump from the upper surface and lands on his 5-kg skateboard with a velocity of 5 m/s in the place of the figure as shown. If his impact with the skateboard has a time duration of 0.05 s, determine the total normal force N exerted by the ground on the skateboard wheels during the impact. The speed is 30 degrees below the horizontal.
a. 1.73 kN b. 2.22 kN c. 2.74 kN d. 2.44 kN e.None of these

Homework Equations



(Fnet)(delta t) = (delta p) = mvf-mvi, where vi=0
vfx=(-1)*(5sin30)
vfy=5cos30

The Attempt at a Solution


Fx = mvf/(delta t)
Fx = [(50)(5)*cos(30)]/0.05 = 4330.12 N
Fy = mvf/(delta t)
Fy = [(50)(5)*(-1)*sin(30)]/0.05 = -2500

The force equations for the boy on the skateboard:

(Fnet)y=-2500=Fn-mg
=-2500+(50)(9.81)
=-2009.5N

Is my answer correct? This does not seem to be any of the multiple choices give. If incorrect, please help me find where I made the mistake.
 
Last edited:
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sugz said:

Homework Statement


The 45-kg boy has taken a running jump from the upper surface and lands on his 5-kg skateboard with a velocity of 5 m/s in the place of the figure as shown. If his impact with the skateboard has a time duration of 0.05 s, determine the total normal force N exerted by the ground on the skateboard wheels during the impact. The speed is 30 degrees below the horizontal.
a. 1.73 kN b. 2.22 kN c. 2.74 kN d. 2.44 kN e.None of these

Homework Equations



(Fnet)(delta t) = (delta p) = mvf-mvi, where vi=0
vfx=(-1)*(5sin30)
vfy=5cos30

The Attempt at a Solution


Fx = mvf/(delta t)
Fx = [(50)(5)*cos(30)]/0.05 = 4330.12 N
Fy = mvf/(delta t)
Fy = [(50)(5)*(-1)*sin(30)]/0.05 = -2500
Does the skateboard move with a vertical component, along with the boy, when the boy is in the air? The problem statement sounds to me as if the boy lands on the skateboard, when the skateboard was already, firmly on the ground.

If only the boy was moving in the air with a vertical component, what is the mass of what was moving vertically? (50 kg or 45 kg?)

[Edit: In other words, make sure you are using the correct mass when calculating the initial momentum components before the impact.]

The force equations for the boy on the skateboard:

(Fnet)y=-2500=Fn-mg
=-2500+(50)(9.81)
=-2009.5N

Is my answer correct? This does not seem to be any of the multiple choices give. If incorrect, please help me find where I made the mistake.

I think you might have subtracted in there somewhere when you meant to add. [Edit: remember, impulse is equal to the body's change in momentum. You must subtract off the initial momentum component from the final momentum component to find the change in the vertical momentum. I think you got your signs mixed up in there somewhere.]
 
Last edited:

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