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Alika [10]
3 years ago
9

A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a velocity of 8 m/s. What is the velocity of t

he skater immediately after?
Physics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

The velocity will be v1 = 0.58[m/s]

Explanation:

This problem can be solved by the law of conservation of the moment, which explains that the moment of a system remains constant because there are no external forces acting on it.

We have the following initial data:

m1 = mass of the skater = 55 [kg]

m2 = mass of the ball = 3 [kg]

v2 = velocity of the ball = 8 [m/s]

Therefore:

m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}+m_{2}*v_{2}\\(50*0)+(3*0)=(50*v_{1})+(3*8)\\50+3-24=50*v_{1}\\v_{1}= 0.58[m/s]

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