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?

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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.

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