Cue ball at rest on a frictionless table

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SUMMARY

A cue ball with a mass of 0.165 kg is initially at rest on a frictionless table and is struck by a pool stick applying an impulse of +1.50 Ns. According to the impulse-momentum theorem, the velocity of the cue ball after being struck is calculated using the formula Ft = mv, resulting in a velocity of approximately 9.09 m/s. This cue ball then undergoes an elastic collision with a second ball of equal mass, which is also at rest. The final velocity of the second ball after the collision is determined to be 9.09 m/s, demonstrating the conservation of momentum in elastic collisions.

PREREQUISITES
  • Understanding of the impulse-momentum theorem
  • Knowledge of elastic collision principles
  • Familiarity with basic physics concepts such as mass and velocity
  • Ability to perform calculations involving Newton's laws
NEXT STEPS
  • Study the impulse-momentum theorem in detail
  • Learn about conservation of momentum in elastic collisions
  • Explore the mathematical derivation of elastic collision equations
  • Investigate real-world applications of impulse and momentum in sports physics
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Physics students, educators, and anyone interested in understanding the dynamics of collisions and momentum transfer in sports scenarios.

rabid_baboon
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A cue ball (mass=0.165kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of +1.50Ns (Newton-seconds) to the ball. The ball the slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

I need help at least getting started.
 
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[tex]Ft=mv[/tex]
 

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