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Physics 103: Lecture 4 Vectors - Motion in Two Dimensions. Today’s lecture will be on Vectors Two dimensions projectile motion. One Dimension. }. Define origin Define sense of direction Position is a signed number (direction and magnitude)

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physics 103 lecture 4 vectors motion in two dimensions
Physics 103: Lecture 4Vectors - Motion in Two Dimensions
  • Today’s lecture will be on
    • Vectors
    • Two dimensions
      • projectile motion

Physics 103, Spring 2004, U.Wisconsin

one dimension
One Dimension

}

  • Define origin
  • Define sense of direction
  • Position is a signed number (direction and magnitude)
  • Displacement, velocity, acceleration are also vectors specified just by signed numbers

Reference Frame

…-4

-3

-2

-1

0

1

2

3

4…

Physics 103, Spring 2004, U.Wisconsin

two dimensions
Two Dimensions
  • Position can be anywhere in the plane
  • Again, select an origin
  • Draw two mutually perpendicular lines meeting at the origin
  • Select +/- directions for horizontal (x) and vertical (y) axes
  • Any position in the plane is given by two signed numbers
  • A vector points to this position
    • The square of its length is, R2= Rx2+ Ry2
    • The angle of that vector is,  = tan-1(Ry / Rx)

R

Ry

Rx = R cos 

Ry = R sin 

q

Rx

Physics 103, Spring 2004, U.Wisconsin

preflight 4 q 1 2
Preflight 4, Q 1 & 2
  • Can a vector have a component bigger than its magnitude?
    • Yes
    • No

The square of magnitude of a vector is given in terms of its components by R2= Rx2+ Ry2

Since the square is always positive the components cannot be larger than the magnitude

Physics 103, Spring 2004, U.Wisconsin

preflight 4 q 3 4
Preflight 4, Q 3 & 4
  • The sum of the two components of a non-zero 2-D vector is zero. Which of these directions is the vector pointing in?
    • 45o
    • 90o
    • 135o
    • 180o

135o

-45o

The sum of components is zero implies Rx = - Ry

The angle,  = tan-1(Ry / Rx) = tan-1 -1 = 135o = -45o (not unique, ± multiples of 2 )

Physics 103, Spring 2004, U.Wisconsin

vector algebra
Vector Algebra

R=R1+R2

D=R2-R1

R2

R1

  • Analytical method
    • Add the components separately to get the components of sum vector
      • Rx = R1x + R2x
      • Ry = R1y + R2y
    • Scalar multiplication of vector
      • Can change magnitude and sign
        • Multiply all components by scalar
          • Components of sR are sRx and sRy
    • Negation of vector (multiplying by -1)
      • Reverse signs of both components
      • Vector points in opposite direction

y

D

x

Physics 103, Spring 2004, U.Wisconsin

vectors
Vectors!

Physics 103, Spring 2004, U.Wisconsin

slide8

Summary of Lecture 3

  • Equations with constant acceleration
  • x = v0t + 1/2 at2
  • v = at
  • v2 = v02 + 2a x
  • Free-fall: ay = -g = -9.81 m/s2
  • y = y0 + v0yt - 1/2 gt2
  • vy = v0y - gt
  • vy2 = v0y2 - 2gy

Physics 103, Spring 2004, U.Wisconsin

two dimensional motion
Two Dimensional Motion

Physics 103, Spring 2004, U.Wisconsin

range of soccer ball
Range of Soccer Ball

Dimensional Analysis:

Strategy:

Physics 103, Spring 2004, U.Wisconsin

kinematics in two dimensions
Kinematics in Two Dimensions
  • x = x0 + v0xt + 1/2 axt2
  • vx = v0x +axt
  • vx2 = v0x2 + 2axx
  • y = y0 + v0yt + 1/2 ayt2
  • vy = v0y +ayt
  • vy2 = v0y2 + 2ayy

x andymotions areindependent!

They share a common time t

Physics 103, Spring 2004, U.Wisconsin

kinematics for projectile motion a x 0 a y g
Kinematics for Projectile Motionax = 0 ay = -g
  • y = y0 + v0yt - 1/2 gt2
  • vy = v0y -gt
  • vy2 = v0y2 - 2g y
  • x = x0 + vxt
  • vx = v0x

x andymotions areindependent!

They share a common time t

Physics 103, Spring 2004, U.Wisconsin

projectile motion maximum height reached time taken for getting there
Projectile Motion:Maximum height reachedTime taken for getting there

Physics 103, Spring 2004, U.Wisconsin

projectile motion maximum range
Projectile Motion: Maximum Range

Physics 103, Spring 2004, U.Wisconsin

soccer ball
Soccer Ball

Physics 103, Spring 2004, U.Wisconsin

slide16

A battleship simultaneously fires two shells at enemy ships from identical canons. If the shells follow the parabolic trajectories shown, which ship gets hit first?

1. Ship A.

2. Ship B.

3. Both at the same time

Higher the shell flies, the longer it takes.

What should the captain order if he wants to hit both ships at the same time?

A

B

Physics 103, Spring 2004, U.Wisconsin

lecture 4 pre flight 5 6
Lecture 4, Pre-Flight 5&6

correct

You and a friend are standing on level ground, each holding identical baseballs. At exactly the same time, and from the same height, you drop your baseball without throwing it while your friend throws her baseball horizontally as hard as she can. Which ball hits the ground first?

1. Your ball

2. Your friends ball

3. They both hit the ground at the same time

Physics 103, Spring 2004, U.Wisconsin

lecture 4 pre flight 5 6 great answers
Lecture 4, Pre-Flight 5&6(great answers)

They both have the same initial vertical component with the same acceleration due to gravity, therefore they hit the ground at the same time.

No matter how much horizontal velocity is put on an object it still falls at the same rate as any other dropped object.

  • y = y0 + voyt - gt2/2
  • v0y = 0 and y=0
  • Therefore, t=sqrt(2y0/g)
  • Result is independent of v0x

Physics 103, Spring 2004, U.Wisconsin

question
Question?

Without air resistance, an object dropped from a plane flying at constant speed in a straight line will

1. Quickly lag behind the plane.

2. Remain vertically under the plane.

3. Move ahead of the plane

Physics 103, Spring 2004, U.Wisconsin

slide20

correct

Lecture 4, Pre-Flight 7&8

A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. Neglecting air resistance, where will the ball land?

1. Forward of the center of the car

2. At the center of the car

3. Backward of the center of the car

Physics 103, Spring 2004, U.Wisconsin

slide21

Great Answers!

The train and the ball have the same horizontal velocity and by throwing the ball straight up, the horizontal component is not changed.

The ball has no acceleration in the horizontal direction. Therefore, the balls

remains directly above the center of the train at all times during the flight

and would fall directly back toward the center of the train.

Physics 103, Spring 2004, U.Wisconsin

summary
Summary

Projectile Motion

  • y = y0 + v0yt - 1/2 gt2
  • vy = v0y -gt
  • vy2 = v0y2 - 2g y
  • x = x0 + v0t
  • v = v0x

Physics 103, Spring 2004, U.Wisconsin