Momentum and the Conservation Laws

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SUMMARY

The discussion centers on calculating the average force exerted on a 4.00 kg steel ball that strikes a wall at a speed of 9 m/s and an angle of 60.0°. The initial calculations incorrectly considered both components of velocity, leading to an erroneous average force of -360 N. The correct approach involves only the change in the x-component of velocity, as the y-component remains unchanged during the collision. The accurate calculation should focus solely on the change in the x-direction to determine the average force accurately.

PREREQUISITES
  • Understanding of momentum and impulse concepts
  • Familiarity with vector components of velocity
  • Knowledge of Newton's second law of motion
  • Basic proficiency in algebra for solving equations
NEXT STEPS
  • Review vector decomposition in physics
  • Study Newton's laws of motion in detail
  • Learn about impulse-momentum theorem applications
  • Practice problems involving collisions and force calculations
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Physics students, educators, and anyone interested in understanding momentum and force calculations in collision scenarios.

closer
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A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

p8-09.gif


F[tex]\Delta[/tex]T = [tex]\Delta[/tex]p
p = mv

[tex]\Delta[/tex]p = pf - pi = mvf - mvi = m(vf - vi)
[tex]\Delta[/tex]p = m(vf - vi)
[tex]\Delta[/tex]p = 4(-9 - 9)
[tex]\Delta[/tex]p = -72
Faverage[tex]\Delta[/tex]t = [tex]\Delta[/tex]p
Faverage(.2) = -72
Faverage = -360

Final answer is incorrect, any ideas?
 
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closer said:
A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

F[tex]\Delta[/tex]T = [tex]\Delta[/tex]p
p = mv

[tex]\Delta[/tex]p = pf - pi = mvf - mvi = m(vf - vi)
[tex]\Delta[/tex]p = m(vf - vi)
[tex]\Delta[/tex]p = 4(-9 - 9)
[tex]\Delta[/tex]p = -72
Faverage[tex]\Delta[/tex]t = [tex]\Delta[/tex]p
Faverage(.2) = -72
Faverage = -360

Final answer is incorrect, any ideas?


The only thing that reversed is the x-component of velocity.
The y-component remained the same and since you are taking differences in Vectors - Velocity is a vector after all - then the force should bbe determined off of the change in x-component of velocity.
 

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