Motion In 2 Dimensions Practice

Motion in 2 Dimensions

1. Add vectors graphically: Big ideas: Vectors end at tip of arrow, showing direction and size. Vectors are added by the parallelogram method and/or the head to tail method. Practice: Draw the resultant for each case.

2. Break vectors into perpendicular components graphically and with trig. Big ideas: Perpendicular components are useful because they can be treated independently simplifying complicated motions into understandable, predictable pieces. The components must together add up to the total vector. The sin relationship will relate the opposite component to the vector and the cos relationships will relate the adjacent component to the vector. Practice Draw the components for each case. Label each component vx or vy. A

B

C

Determine the components using trig for each of the above cases Vector A 40 m/s @ 40ºabove horizontal Example: sin 40 = opp/hyp = vy/40 m/s  vy = (40m/s)sin 40 = 26m/s cos 40 = adj/hyp = vx/40 m/s  vx = (40m/s)cos 40 = 31m/s Vector B 10 m/s @ 70º to the right of vertical Vector C 100 m/s @ 85º N of E

3. Add vectors using right angle trig Big idea: Use parallelogram or head to tail method to lay out a sketch. Then use the Pythagorean theorem to find the magnitude of the resultant and the appropriate trig identity (usually inverse tangent) to find the angle. Label the angle in the proper format: the number of degrees in a direction from a direction of the compass or vertical or horizontal.

Example: A boy runs 50 m due North and then 70 m due East. What is his displacement? 70 m E

c2 = a2 + b2  c = √a2 + b2 c = √50m2 + 70m2=86m tanθ = opp/adj = 70m/50 m  θ = tan-1 (70/50) = 54º

θ 50 m,

N

ans. The boy is 86 m from his starting point at a bearing of 54º E of N

Practice: A boat aims itself due east at 15 km/h across a northern current of 5 km/h. What is the boat’s resulting velocity? What is a ball’s actual velocity if after 2 seconds in the air it has vy = 20 m/s up and vx = 10 m/s? What is a kite surfer’s velocity if the waves are moving him E at 1.5 m/s and the wind is blowing him North at 4 m/s?

4. Solve motion equations in 2 dimensions. Big idea: Apply motion equations from the first unit combined with vector addition to solve problems. Usually you have to apply the equations to single component or to both and then add the components together to get a final answer. Be very careful to make sure that you are treating the proper component with the proper equation. Example: A boat is going across a 30 m wide stream with a velocity of 3 m/s. At the same time it is going downstream with the stream’s 4m/s current, giving it a total speed of 5m/s. How long does it take to cross the stream? How far downstream is it, when it finally reaches the other side? Solution: Use only the across component to determine how long it takes to go across. Use only the downstream component to determine how far downstream it lands. So, part A. vave = d/t  t = d/vave = 30 m/3m/s = 10 s to cross. B. d = vavet = (4 m/s)(10 s) = 40 m. (This would have been harder if the question had read the boat had a velocity of 5 m/s at an angle of 37º from the bank, and you had been forced to solve for the two components.) Practice: A boat is going across a 30 m wide stream with a velocity 3m/s at an angle of 25º from the bank. How long does it take to cross the stream? How far downstream is it, when it finally reaches the other side? A ball is thrown at 30m/s at an angle of 30º from the horizontal. What is its height after 1 second? What is its total displacement after 1 second? What is its velocity after 1.5 s? What is its velocity after 4s (assuming it was thrown off the edge of a cliff and has room to keep falling)?

5. Solve projectile motion problems. Big idea: This is a special case of the last topic in which solutions are usually found by following variations on a three step approach. Step 1: break the velocity into horizontal and vertical components. Step 2: Use the vertical component to solve for the time in the air. Step 3 Use the horizontal component to solve for the horizontal displacement. Example: How far downfield does a soccer ball land if it is kicked at 8 m/s at an angle of 10º above the horizontal? V= 8m/s @10º above ground vy

vx

sin 10º = vy/8m/s vy = (8m/s)sin 10º =1.4m/s cos 10º = vy/8m/s vy = (8m/s)cos 10º =7.9m/s vy at top = 0m/s a = -10m/s2 a = Δv/t  t = Δv/a =(0m/s – 1.4 m/s)/ (-10 m/s2) = 0.14 s t total = time up +time down = 2x time up = 0.28s Only vertical components in the last step and only horizontal in the next. d = vavet = (7.9 m/s)(0.28s) = 2.2 m

Practice: What is the range of an arrow that is shot at 120 m/s at an angle of 40º above horizontal? What is the maximum height of the arrow in the last problem? How far does a ball land from the edge of 2.5 m high table if it had a horizontal speed of 3 m/s?

Solutions 1.

2. vx = 9.4 m/s, vy = 3.4 m/s

vN = 99.6 m/s, vE = 8.7 m/s

3. 15.3 km/h @ 18º N of E, 22.4 m/s @ 63º above horizontal, 4.3 m/s @ 20º E of N 4. 64 m, 10 m, 26 m, 28 m @ 21º above the horizontal, 26 m/s along the horizontal, 36 m/s at 44º below the horizontal 5. 1416 m , 296 m, 2.12m