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SACE Stage 2 Physics

SACE Stage 2 Physics. Motion in 2 Dimensions. Motion in 2 - Dimensions. Errors in Measurement. Suppose we want to find the area of a piece of paper (A4) Length = 297 ± 0.5 mm Width = 210 ± 0.5 mm Area max = 62623.75 mm 2 Area min = 62116.75 mm 2 Area = 62370 ± 253.5 mm 2.

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SACE Stage 2 Physics

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  1. SACE Stage 2 Physics Motion in 2 Dimensions

  2. Motion in 2 - Dimensions • Errors in Measurement Suppose we want to find the area of a piece of paper (A4) Length = 297 ± 0.5 mm Width = 210 ± 0.5 mm Areamax = 62623.75 mm2 Areamin = 62116.75 mm2 Area = 62370 ± 253.5 mm2

  3. Motion in 2 - Dimensions • Significant Figures When calculating data, the accuracy of the answer is only as accurate as the information that is least accurate. 307.63 – 5 significant figures 0.00673 – 3 significant figures 12000 – can be 2,3,4, or 5 significant figures depending on whether the zeros are just place holders for the decimal point. 12.45 x 1012 – has 4 significant figures

  4. Motion in 2 - Dimensions • Scientific Notation The diameter of the solar system is 5 946 000 000 000 metres. Can write this as 5.946 x 1012m. The decimal place has moved 12 places to the left. Calculations

  5. Motion in 2 - Dimensions • Scientific Notation Example Evaluate where, k = 9.00 x 109, q1 = 1.60 x 10-19, q2 = 3.20 x 10-19, r = 6.273 x 10-11

  6. Motion in 2 - Dimensions • Scientific Notation Example Evaluate where, k = 9.00 x 109, q1 = 1.60 x 10-19, q2 = 3.20 x 10-19, r = 6.273 x 10-11 Answer given to three significant figures as the least accurate piece of data was given to three sig. figs.

  7. Motion in 2 - Dimensions • Equations of Motion Average Velocity Average Acceleration

  8. Motion in 2 - Dimensions • Equations of Motion Using average velocity and average acceleration to derive two other equations. (a) Assuming velocity and acceleration remain constant, Become,

  9. Motion in 2 - Dimensions • Equations of Motion Combining,

  10. Motion in 2 - Dimensions • Equations of Motion (b) equation (1) = equation (2)

  11. Motion in 2 - Dimensions • Equations of Motion Hence, Ie, Note: (1) the acceleration is constant, (2)the directions for velocity and acceleration are used correctly

  12. Motion in 2 - Dimensions • Uniform Gravitational Field • Gravity acts vertically downwards. • A mass can only accelerate in the direction of gravity in the absence of all other forces (including air resistance). • 3. Gravity g = 9.8 ms-2 vertically down.

  13. vH vv vv vH vH a = g = 9.8 m.s-2 vH vv vv a = g = 9.8 m s-2 vv vH vv vH vH v2 v1 vH Motion in 2 - Dimensions • Uniform Gravitational Field – vector diagram

  14. Motion in 2 - Dimensions • Uniform Gravitational Field – multi-image photograph • Vertical separation the same for both balls at the same time interval. • Horizontal separation constant. • 3. Vertical and horizontal components are independent of each other.

  15. v q vh = vcos q Motion in 2 - Dimensions • Vector Resolution A vector can be resolved into components at right angles to each other. vv = vsin q

  16. v = 40 m s-1 vvertical 30o vhorizontal Motion in 2 - Dimensions • Example 1 – Known vector Trigonometric ratios, vvertical = 40 sin 30o = 20 m s-1 vhorizontal = 40 cos 30o = 34.6 m s-1

  17. v = ? vv= 20m s-1 q vh= 50m s-1 Motion in 2 - Dimensions • Example 2 – Unknown vector Pythagoras’ Theory,

  18. Motion in 2 - Dimensions • Time of Flight • Note: • Acceleration present is from gravity and remains constant. • Horizontal velocity remains constant (Ignore air resistance) • Vertical motion is independent of horizontal motion. • The launch height is the same as the impact height. • We can now determine the time of flight by only considering the vertical motion of the projectile.

  19. Motion in 2 - Dimensions • Time of Flight Can use the following equations for the vertical motion, (a = -g = 9.8ms-2) Can use the following equation for the horizontal velocity,

  20. Motion in 2 - Dimensions • Time of Flight We assume the launch point has position s1 = 0. The projectile is launched with some initial horizontal velocity (vh1) and some initial vertical velocity (vv1). The only acceleration is due to gravity acting vertically downwards. It reaches a maximum height at the time Dtmax, when, a = 9.8ms-2 down (take a =-g assuming acceleration down & vv1 up - ie. up is a positive direction) vv1 vh1

  21. Motion in 2 - Dimensions • Time of Flight At the time the maximum height is reached, gives,

  22. Motion in 2 - Dimensions • Time of Flight Time of impact occurs when DS = 0. ie, This equation has two solutions, at Dt = 0 and equation for the time of flight Comparing the two equations, and The time of flight is exactly twice the time taken to reach the maximum height.

  23. Motion in 2 - Dimensions • Range The range is simply the horizontal distance attained at the time Dt = Dtflight.

  24. Motion in 2 - Dimensions • Example A rugby player kicks a football from ground level with a speed of 35 ms-1 at an angle of elevation of 250 to the horizontal ground surface. Ignoring air resistance determine; (a) the time the ball is in the air, (b) the horizontal distance travelled by the ball before hitting the ground (c) the maximum height reached by the ball.

  25. Motion in 2 - Dimensions • Example (a) the time the ball is in the air, • vH = v cos • = 35cos(25) = 31.72 m s-1 • vv = v sin(25) • = 35(sin25) = 14.79 m s-1 • Using vertical components to determine time to reach maximum height • vv = vo + at • t = 14.79/9.8 = 1.509 = 1.5 s • Hence time in the air = 2(1.509) = 3.02 s 35 m s-1 vv m s-1 25o vH m s-1

  26. Motion in 2 - Dimensions • Example (b) the horizontal distance travelled by the ball before hitting the ground sH = vHt = (31.72)(2(1.5)) = 2(47.8766) = 2(47.9) = 96 m (c) the maximum height reached by the ball. s = (14.79)(1.5) + (0.5)(-9.8)(1.5)2 = 11.16 = 11.2 m

  27. Motion in 2 - Dimensions • Launch Angle and Range The following diagram shows the trajectories of projectiles as a function of elevation angle. Note that the range is maximum for q = 45o and that angles that are equal amounts above or below 45o yield the same range, eg, 30o and 60o. Ignoring air resistance

  28. Motion in 2 - Dimensions • Air Resistance • Affects all moving through air. • The force due to air resistance always acts in the opposite direction to the velocity of the object. • Air resistance is proportional to the speed of the object squared. • As speed changes, the air resistance must also change.

  29. Motion in 2 - Dimensions • Air Resistance • Horizontal velocity always decreasing. • No vertical air resistance at max height as vv = 0. • Time of Flight is reduced. • Range also reduced.

  30. Motion in 2 - Dimensions • Application: Projectiles in Sport • Launch height affects the range of the football. • Maximum distance achieved for elevation angle of 45o. • Air resistance will depend on the type of projectile, ie, basketball, football, ball of paper.

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