Chapter 5 Newton's Third Law of Motion
Exercises Unit II
31.
 | magnitude/no direction: age, speed, temp. |
| magnitude/direction: velocity, acceleration. |
32. Equal magnitude, but opposite direction.
40. Draw 1m/s vector down and add 1m/s vector horizontally. Measure the angle of the resultant vector with a protractor. Both magnitudes are equal and directions are in 90o relative to each other the result will be a 45 o streak on the window.
Problems Unit II
5. Hint: the ground velocity is the resultant vector sum. Use the Pythagorean Theorem (or Unit Triangles) to determine the magnitude and a diagram to determine the direction. 141 km/h Northeast
6.
| (a) 5 km/h Southeast |
| (b) Northeast |
Exercises Unit III
13. 100 N on the scale the same as tying to the wall.
14.
| (a) Upward the Force of athlete up > Force of barbell down. |
| (b) Downward the Force of barbell down > Force of athlete up. |
15. Fg = normal force. Fnet = 0 to be in static equilibrium.
21. The cars are equal mass, so he will push both with equal force. He can not give one greater speed.
23. Constant force 200N push must equal the force of friction for the crate to have a = 0. Fnet = 0.
25. They will both move towards each other.
29. Two have equal mass and the force will be equal and opposite. They accelerate the same magnitude but different directions. Will meet in the middle at 6 m.
47. Same answer as throwing a ball up and it reaches the top of its path. Acceleration = g = -10 m/s2 at the top and anywhere during its path.
Problems Unit III
2.
| (a) Wall pushes back with 40 N. |
| (b) a = F/m = 40 N / 80 kg = 0.5 m/s2. |
