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PHYSICS 231 Lecture 6: Motion in 2 dimensions

PHYSICS 231 Lecture 6: Motion in 2 dimensions. Remco Zegers Walk-in hour: Thursday 11:30-13:30 HELPROOM. Often, we replace motion in 2D into horizontal and vertical components. Displacement in 2D. In vector notation: r= x + y. Velocity and acceleration.

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PHYSICS 231 Lecture 6: Motion in 2 dimensions

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  1. PHYSICS 231Lecture 6: Motion in 2 dimensions Remco Zegers Walk-in hour: Thursday 11:30-13:30 HELPROOM PHY 231

  2. Often, we replace motion in 2D into horizontal and vertical components. Displacement in 2D In vector notation: r= x+ y PHY 231

  3. Velocity and acceleration The definitions made in 1D remain the same in 2D: PHY 231

  4. While studying motion in 2D one almost always makes a decomposition into horizontal and vertical components of the motion, which are both described in 1D • Remember that the object can accelerate in one direction, but remain at the same speed in the other direction. • Remember that after decomposition of 2D motion into horizontal and vertical components, you should should investigate both components to understand the complete motion of a particle. • After decomposition into horizontal and vertical directions, treat the two directions independently. PHY 231

  5. vx=v0cos vy=v0sin-2g=0 vx=v0cos vy=v0sin-1g vx=v0cos vy=v0sin-3g V=v0 vx=v0cos vy=v0sin vx=v0cos vy=v0sin-4g  vx=v0cos vy=v0sin-5g Parabolic motion: a catapult Vt=v0+at t=0 t=1 t=2 t=3 t=4 t=5 PHY 231

  6. Question • A hunter aims at a bird that is some distance away and flying very high (i.e. consider the vertical position of the hunter to be 0), but he misses. If the bullet leaves the gun with a speed of v0 and friction by air is negligible, with what speed vf does the bullet hit the ground after completing its parabolic path? v0 vf PHY 231

  7. Answer • First consider the horizontal direction: V0x=V0cos() Since there is no friction, there is no change in the horizontal component: Vfx=V0cos()=V0x • Next the vertical direction: V0y=V0sin() Vy(t)=V0y-gt xy(t)=Voyt-0.5gt2 (g=9.81 m/s2) Boundary condition: bullet hits the ground: 0= Voyt-0.5gt2 t=0 or t=2V0y/g So, Vfy(t)=V0y-(2V0y/g)g=-V0y • Total velocity=(V0x2+(-V0y)2)=V0!!!! • The speed of the bullet has not changed, but the vertical component of the velocity has changed sign. PHY 231

  8. Pop and Drop B A For A: Vy=-0.5gt2 Vx=0 For B: Vy=-0.5gt2 Vx=V0 For A: Xy=X0-0.5gt2 Xx=0 For B: Xy=X0-0.5gt2 Xx=V0t PHY 231

  9. Shoot the monkey The hunter aims his arrow exactly at the monkey At the moment he fires, the monkey drops off the branch. What happens? PHY 231

  10. Always hit!! The hor. Position of the arrow is: x(t)=d-v0cos()t x(t)=0 at t=d/v0cos()=tx=0 Its vertical position is: y(t)=v0sin()t-0.5gt2 y(tx=0)=dtan()-0.5gt2=h-0.5gt2 h The vertical position of the monkey is: y(t)=h-0.5gt2 The horizontal position is 0  d PHY 231

  11. Another example • A football player throws a ball with initial velocity of 30 m/s at an angle of 30o degrees w.r.t. the ground. How far will the ball fly before hitting the ground? And what about 60o? And at what angle is the distance thrown maximum? X(t)=30cos()t Y(t)=30sin()t-0.5gt2 =0 if t(30sin()-0.5gt)=0 t=0 or t=30sin()/(0.5g) X(t=30sin()/(0.5g))=900cos()sin()/(0.5g) =900sin(2)/g if =30o X=79.5 m if =60o X=79.5 m !! Maximum if sin(2) is maximum, so =45o X(=45o)=91.7 m PHY 231

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