1. Dynamics + 2D Motion!
Created by Educational Technology Network

2. Proportions
Vectors
Gravity
Graphs and Diagrams 10 20 30 40

3. Question
If the mass of an object is tripled, then the inertia associated with the object is…

4. Answer 1 – 10
Also tripled
Mass = inertia

5. Question
If the normal force of an object is reduced to half of its original size, then the force of friction associated with the object is..

6. Answer 1 – 20 Also half of its original size Ff = M*Fn

7. Question
If the average acceleration of an object is doubled and the mass is reduced to one third of its original size, then the net force on the object is now..

8. Answer 1 – 30
a = Fnet/m
-> Fnet = ma
Fnet = (1/3)*(2) = (2/3)

9. Question
If the gravitational attraction between two objects is originally 2*10^7N and the distance between the two objects is halved, then the gravitational attraction between the two objects becomes …

10. Answer 1 – 40 Fg = (Gm1m2)/r^2 Fg = (G(1)(1))/((.5)^2) Fg = G/.25 = 4G
The force will be 4x as great
It was originally 2*10^7N, so it will be come 4 * (2*10^7N) = 8*10^7N

11. Question
Find the resultant of adding vectors A and B:

12. Answer 2 – 10 Step 1: Set up tip-to-tail
Step 2: Draw the resultant from the tail of the first vector to the tip of the last vector

13. Question
Find the resultant for the following set of vectors:

14. Answer 2 – 20
The vectors have equal magnitude and opposite direction. They would cancel out and the resultant would have a length of zero.

15. Question

16. Answer 2 – 30
The vector in black is the given vector (B); the vector in red is the resultant. The vector in yellow is the missing vector (tip to tail).

17. Question
Add the following vectors:

18. Answer 2 – 40

19. Question

20. Answer 3 – 10
4. Gravitational field lines always point in to show attraction

21. Question

22. Answer 3 – 20
3. 3m/s/s

23. Question

24. Answer 3 – 30
Answer: 8.17*10^-10

25. Question

26. Answer 3 – 40 Answer: 25N Proportion Problem:
The person is already standing one earth radius from the center. If he increases his distance by one earth radii, then he is doubling his distance. If he is doubling his distance, then his force gets ¼ as great (Fg = (G(1)(1))/(2^2) = G/4)

27. Question

28. Answer 4 – 10
Answer: 2 (Net force is to the left, so acceleration is to the left and velocity is to the right (opposite vectors means slowing down)).

29. Question

30. Answer 4 – 20
Answer: 1 (force is upward and it’s measured in Newton’s. The magnitude of the normal force is equivalent to calculating the weight)

31. Question

32. Answer 4 – 30 Answer: 2 Centripetal acceleration = (v^2)/r
So it proportional to V squared, which means the graph should show a quadratic

33. Question

34. Answer 4 – 40
Answer: 2
Magnitude stays the same. Direction changes, so if you are graphing the magnitude, then you should see a constant slope of zero to indicate no change.