Collision and conservation of P

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SUMMARY

The discussion focuses on a physics problem involving the conservation of momentum during a collision between two balls. Ball A, with a mass of 1 kg, rolls west at 3 m/s and collides with stationary ball B, which has a mass of 2 kg. After the collision, ball A moves south at 2 m/s, resulting in ball B moving at a velocity of 1.8 m/s at an angle of 33.7 degrees north of west. The calculations confirm that the momentum is conserved, with ball B's final momentum calculated as 3.6 kg·m/s at the specified angle.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with vector components and vector diagrams
  • Knowledge of basic physics equations, specifically P = m x v
  • Ability to apply the Pythagorean theorem in physics contexts
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  • Study advanced momentum conservation problems in two dimensions
  • Learn about elastic and inelastic collisions and their differences
  • Explore vector addition and subtraction techniques in physics
  • Investigate real-world applications of momentum conservation in sports and vehicle collisions
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Students studying physics, educators teaching mechanics, and anyone interested in understanding collision dynamics and momentum conservation principles.

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Homework Statement


A ball (A) is rolling west at 3m/s and has a mass of 1kg. Another ball (B) has a mass of 2kg and is stationary. After a collision, ball A moves south at 2 m/s. Calculate the momentum and velocity of ball B after the collision.

Homework Equations


Pa(initial)=Pa(final)+Pb(final
P=m x v
a2+b2=c2


The Attempt at a Solution


Pa(initial)=1kg(3m/s W)= 3kgm/s W
Pa(final)=1kg(2m/s S)= 2kgm/s S
Pa(initial)-Pa(final)=Pb(final)
a vector diagram gives me a right triangle. Pythagoras gives me c=Pb(final) as 3.6kgm/s
33.7 degrees North of West.
Pb=m x v
3.6kgm/s 33.7degrees north of west /2 kg= vb
vb= 1.8 m/s 33.7 degrees north of west.
Is this correct?
 
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Absolutely.
 
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