Conservation of Momentum of roller-skates

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SUMMARY

The discussion centers on the conservation of momentum in an inelastic collision involving roller skaters Anthony and Sissy. Anthony, weighing 75kg and moving backwards at 3m/s, collides with Sissy, who weighs 60kg and jumps into his arms at 5m/s. The momentum conservation equation is applied correctly, yielding a final velocity of approximately 3.9m/s after rounding. The calculation confirms the principles of inelastic collisions, where total momentum before the collision equals total momentum after.

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  • Concept of mass and velocity in physics
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  • Learn about the differences between elastic and inelastic collisions
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Homework Statement


Anthony and Sissy are participating in the "Roll-a-Rama" roller-skating dance championship. While 75kg Anthony roller-skates backwards at 3m/s, 60kg Sissy jumps into his arms with a velocity of 5m/s in the same direction.
m1= 75kg
v1= 3m/s
m2= 60kg
v2= 5m/s
v' = ?

Homework Equations



The inelastic collision equation possibly?

The Attempt at a Solution


(75kg)(3m/s) + (60kg)(5m/s)= (75kg+60kg)x
225kg*m/s + 300kg*m/s= 135kg(x)
525kg*m/s = 135kg(x)
(525kg*m/s)/135kg = x
3.8m/s= x

Is that correct?
 
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Yes, it is inelastic collision and your calculation is correct, but take care of rounding. x=3.888... round it to 3.9 m/s.

ehild
 
ehild said:
Yes, it is inelastic collision and your calculation is correct, but take care of rounding. x=3.888... round it to 3.9 m/s.

ehild

Thank you very much :)
 

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