motion in 2 dimensions practice

8/14/2019 Motion in 2 Dimensions Practice

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Motion in 2 Dimensions

1. Add vectors graphically:Big ideas: Vectors end at tip of arrow, showing direction and size. Vectors are added bythe 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 treatedindependently 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 cosrelationships 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 casesVector A 40 m/s @ 40above horizontal

Example: sin 40 = opp/hyp = vy/40 m/s vy = (40m/s)sin 40 = 26m/scos 40 = adj/hyp = vx/40 m/s vx = (40m/s)cos 40 = 31m/s

Vector B 10 m/s @ 70 to the right of vertical

Vector C 100 m/s @ 85 N of E

3. Add vectors using right angle trigBig idea: Use parallelogram or head to tail method to lay out a sketch. Then use thePythagorean theorem to find the magnitude of the resultant and the appropriate trigidentity (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 orhorizontal.


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

Practice:

A boat aims itself due east at 15 km/h across a northern current of 5 km/h. What is theboats resulting velocity?What is a balls 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 surfers velocity if the waves are moving him E at 1.5 m/s and the wind isblowing 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 tosolve problems. Usually you have to apply the equations to single component or to bothand then add the components together to get a final answer. Be very careful to make surethat 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 streams 4m/s current, giving it a total speedof 5m/s. How long does it take to cross the stream? How far downstream is it, when itfinally 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, partA. 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 atan angle of 37 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 25 from

the bank. How long does it take to cross the stream? How far downstream is it, when itfinally reaches the other side?

A ball is thrown at 30m/s at an angle of 30 from the horizontal. What is its height after 1second? 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 roomto keep falling)?

50 m, N

70 m E c2 = a2 + b2 c = a2 + b2c = 50m2 + 70m2=86m

tan = opp/adj = 70m/50 m = tan-1 (70/50) = 54

ans. The boy is 86 m from his starting point at a bearing of54 E of N


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5. Solve projectile motion problems.Big idea: This is a special case of the last topic in which solutions are usually found byfollowing variations on a three step approach. Step 1: break the velocity into horizontaland vertical components. Step 2: Use the vertical component to solve for the time in theair. 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 angleof 10 above the horizontal?

Practice:

What is the range of an arrow that is shot at 120 m/s at an angle of 40 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 of3 m/s?

Solutions1.

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 @ 18 N of E, 22.4 m/s @ 63 above horizontal, 4.3 m/s @ 20 E of N4. 64 m, 10 m, 26 m, 28 m @ 21 above the horizontal, 26 m/s along the horizontal, 36m/s at 44 below the horizontal5. 1416 m , 296 m, 2.12m

V= 8m/s @10 above ground

vy

vx

sin 10 = vy/8m/svy = (8m/s)sin 10 =1.4m/s

cos 10 = vy/8m/svy = (8m/s)cos 10 =7.9m/s

vy at top = 0m/s a = -10m/s2

a = v/t t = v/a =(0m/s 1.4 m/s)/ (-10 m/s2) = 0.14 st total = time up +time down = 2x time up = 0.28sOnly vertical components in the last step and only

horizontal in the next.d = vavet = (7.9 m/s)(0.28s) = 2.2 m


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motion in 2 dimensions practice

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