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

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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?

Jun 18th, 2015

Hello friend...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.

Here is the complete answer! Please let me know if you need any clarification. Please best with a 5*. Thank you and Godbless!!!
Jun 18th, 2015

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