A man standing on frictionless ice throws a 1.00-kg mass at 20.0 m/s at an angle

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  1. A man standing on frictionless ice throws a 1.00-kg mass at 20.0 m/s at an angle of elevation of 40.0°. What was the magnitude of the man’s momentum immediately after throwing the mass?

    a.
    b.
    c.
    d.
Jun 26th, 2015

Thank you for the opportunity to help you with your question!

Horizontal component of throw V = (cos 40) x 20, = 15.32m/sec.
Vertical component has no affect on the problem, because of lack of friction.
His momentum will be the momentum given the mass, which is 15.32kg/m/sec. (1kg x 15.32m/sec.).
Momentum is conserved.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 26th, 2015

Please let me know if you need any clarification.I request that you best my answer please I'm always happy to answer your questions.

Jun 26th, 2015

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