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?
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.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?