Collision Prob PPT from class
... 7. A stationary object explodes, breaking into three pieces of masses m, m, and 3m. The two pieces of mass m move off at right angles to each other with the same magnitude of momentum mV, as shown in the diagram above. What are the magnitude and direction of the velocity of the piece having mass 3m ...
... 7. A stationary object explodes, breaking into three pieces of masses m, m, and 3m. The two pieces of mass m move off at right angles to each other with the same magnitude of momentum mV, as shown in the diagram above. What are the magnitude and direction of the velocity of the piece having mass 3m ...
review – midterm 2017
... What is its acceleration.. A) when it is first hit after it leaves the club? B) when it’s at the top of its path? C) when it is about to hit the ground? Unit 4: Forces and Newton’s Laws ...
... What is its acceleration.. A) when it is first hit after it leaves the club? B) when it’s at the top of its path? C) when it is about to hit the ground? Unit 4: Forces and Newton’s Laws ...
Problem Set 6 - Cabrillo College
... Q9.5. Reason: The sum of the momenta of the three pieces must be the zero vector. Since the first piece is traveling east, its momentum will have the form ( p1,0), where p1 is a positive number. Since the second piece is traveling north, its momentum will have the form (0, p2 ), where p2 is a positi ...
... Q9.5. Reason: The sum of the momenta of the three pieces must be the zero vector. Since the first piece is traveling east, its momentum will have the form ( p1,0), where p1 is a positive number. Since the second piece is traveling north, its momentum will have the form (0, p2 ), where p2 is a positi ...
π π π λ ρ ρ ρ ρ ρ
... K = 0.5mv2 + 0.5Iω2 = 0.5mv2 + 0.5mr2ω2 = mv2 b. 5.46 m/s Energy conservation gives Uf + Kf = Ui + Ki c. 1.91 m/s Hence, (mg 3sin250) + (mv2) = 0 + m(6.5)2 d. 3.79 m/s v2 = 6.52 – 3gsin250 e. 1.27 m/s v = 5.46 m/s ...
... K = 0.5mv2 + 0.5Iω2 = 0.5mv2 + 0.5mr2ω2 = mv2 b. 5.46 m/s Energy conservation gives Uf + Kf = Ui + Ki c. 1.91 m/s Hence, (mg 3sin250) + (mv2) = 0 + m(6.5)2 d. 3.79 m/s v2 = 6.52 – 3gsin250 e. 1.27 m/s v = 5.46 m/s ...
Fall Semester Review
... Universal Gravitation: Know the concepts behind how two large bodies are attracted to each other and how to calculate the force of attraction, period and velocity. Statics: Forces in two dimensions that are not in motion. Dynamics: Force in two dimensions that are in motion. Torque: A force applied ...
... Universal Gravitation: Know the concepts behind how two large bodies are attracted to each other and how to calculate the force of attraction, period and velocity. Statics: Forces in two dimensions that are not in motion. Dynamics: Force in two dimensions that are in motion. Torque: A force applied ...
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