* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

# Download Chapter 3 Review Questions

Survey

Document related concepts

Specific impulse wikipedia , lookup

Coriolis force wikipedia , lookup

Jerk (physics) wikipedia , lookup

Electromagnetic mass wikipedia , lookup

Classical mechanics wikipedia , lookup

Relativistic angular momentum wikipedia , lookup

Fictitious force wikipedia , lookup

Newton's theorem of revolving orbits wikipedia , lookup

Modified Newtonian dynamics wikipedia , lookup

Center of mass wikipedia , lookup

Centrifugal force wikipedia , lookup

Rigid body dynamics wikipedia , lookup

Equations of motion wikipedia , lookup

Seismometer wikipedia , lookup

Relativistic mechanics wikipedia , lookup

Classical central-force problem wikipedia , lookup

Transcript

Laws of Motion Review Questions Name _________________________ Round all calculations. Given, formula, setup & solution is required. 1. State Newton’s 1st Law of Motion – An object maintains a constant velocity unless a net force acts on it 2. The tendency of an object to resist any change in motion is called Inertia. 3. What is required to accelerate an object? Net Force 4. Write the equation that represents Newton’s 2nd Law of Motion F = ma or a = F/m or m = F/a 5. T/F If equal forces are applied to two objects, the object with greater mass will accelerate faster than the object with less mass. 6. How much force is required to accelerate a bowling ball with a mass of a 7 kg at 11m/s2? F = m∙a m = 7 kg a = 11 m/s/s F = 7kg ∙ 11 m/s/s F = 77 N 7. Calculate the mass of a bowling ball that weighs 75 N. Wt (or F) = 75 N g (or a) = 9.8 m/s/s m = F/a m = 75 N / 9.8 m/s/s m = 7.65 kg Your little brother (if you don’t have one just play along) asks you to pass the milk at dinner. You decide to slide it across the table. The milk jug has a mass of 2 kg and you apply a force of 4 N. 8. What is the rate of acceleration for the milk jug? m = 2 kg F=4N a = F/m a = 4 N / 2 kg a = 2 m/s/s 9. If it takes you .85 seconds to accelerate the jug, how fast is it going when it leaves your hand? (Hint: use your answer from the previous question in your calculation) a = 2 m/s/s t = .85 s v = a∙t v = 2 m/s/s ∙ .85s v = 1.7 m/s 10. State Newton’s 3rd Law of Motion – For every action there is an equal and opposite reaction 11. According to Newton’s third law of motion, if you apply a force of 3 N to a wall, the wall must apply a force of 3 N on you. 12. State the Law of Conservation of Momentum – in a collision total momentum is conserved or Total momentum before a collision = total momentum after a collision 13. What is the momentum formula? P = m∙v or m = P/v or v = P/m 14. Which would have more momentum moving at the same speed down the grocery aisle, a full shopping cart or an empty shopping cart? Explain your answer. Full cart – since v is the same the mass makes the difference and the full cart has more mass so more P 15. What is measured in m/s2? acceleration 18. What is the metric standard unit for force? N 16. What is measured in m/s? velocity 19. What is the metric standard unit for weight? N 17. What is the metric standard unit for mass? kg 20. What is the metric standard unit for inertia? kg Laws of Motion Review Questions Name _________________________ A grocery cart with a mass of 50 kg is moving at a velocity of 4.0 m/s West. It bumps into another cart with a mass of 40 kg moving in the same direction at 2.0 m/s. The two get locked and move off together. Assume that the Law of Conservation of Momentum is in effect. 21. Calculate the momentum of each cart separately before they collide. m1 = 50 kg P1 = m1∙v1 v1 = 4 m/s W P1 = 50 kg∙4 m/s W m2 = 40 kg v2 = 2 m/s W P1 = 200 kg∙m/s P2 = m2∙v2 P2 = 40 kg ∙ 2 m/s W P2 = 80 kg∙m/s 22. Calculate the total momentum and the total mass of the two carts right after the crash? PT = P1 + P2 = 200 kg∙m/s + 80 kg∙m/s = 280 kg∙m/s mT = m1 + m2 = 50 kg + 40 kg = 90 kg 23. Moments after the crash, what is the velocity of the two carts together? mT = 90 kg v = P/m v = 280 kg∙m/s / 90 kg PT = 280 kg∙m/s v = 3.1 m/s 24. Draw and Label at least 2 (total of 4 arrows) action-reaction pair forces involving the fish on the diagram to the right. 1 1 = Force of hook on fish 2 2 = Force of fish on hook 3 = Force of fish on Earth These 2 (3 & 4) make up the force of gravity 4 = Force of Earth on fish 25. Calculate the Net force on the person from the diagram below. Eliminate the forces not on the person & you end up with 1000 N up & 900 N down Since they are opposite you subtract so NF =1000 N up – 900 N down NF = 100 N up 3 Force of board on person (1000 N) Force of person on earth (900 N) 4 Force of person on board (1000 N) Force of earth on person (900 N) 26. Calculate the mass, net force and acceleration of a 1000 N skydiver experiencing 250 N of air resistance AND draw & label the unbalanced forces on the skydiver. mass = 102 kg Net Force = 750 N down Acceleration = 7.35 m/s/s Mass: Wt = 1000 N, g = 9.8 m/s/s; m = Wt/g (F/a); m = 1000N/9.8 m/s/s; m = 102 kg Net Force: (opposites subtract) 1000 N down – 250 N up = 750 N down Acceleration: m = 102 kg, F = 750 N; a = F/m; a = 750 N/102 kg; a = 7.35 m/s/s