Lesson 2: Motion in Two Dimensions

Motion in Two Dimensions The equations for two dimensional motion are almost the same as those for motion in one dimension. The only difference is in notation. In one dimensional motion, the displacement is written as Δx = x2 − x1 . In two dimensional motion, we have both x and y directions and the notation is G G G Δr = r2 − r1 G G where r1 and r2 are vectors form the origin to the points A and B. Each of these vectors has x and y components. A y Δr B G r1 G r2 O x JJJG The average velocity is the displacement vector AB divided by the time it takes for the object to move from A to B. The acceleration is still the change in velocity over time: G G G v − v0 a= t Then: G G G v = v0 + a t These equations look like the corresponding equations for one dimensional motion. The only difference is the vector signs. Another important equation is: G G G G r − r0 = v0t + 12 a t 2 Again, it looks almost the same as the equation for displacement in one dimensional motion. It is very inconvenient to do calculations with vectors. To solve problems on two dimensional motion, we use the components of the vectors. Then the vector equations turn into algebraic equations, the same as the equations used in one dimensional motion, which are easy to solve using elementary mathematical operations. In component form the two equations above become: vx = v0 x + ax t v y = v0 y + a y t and: x − x0 = v0 x t + 12 ax t 2 y − y0 = v0 y t + 12 a y t 2 “Breaking into components” turns a two dimensional motion problem into a system of two equations with two unknowns. Note that the time ‘t’ in the equations above is the same for the two components. Projectile Motion An example of an object moving in 2-dimensions is throwing the object (a ‘projectile’) near to the surface of the Earth. We will be using the following terms with regard to projectile motion: • • • Range is the distance covered in the x direction. Maximum height is the maximum value of y during the motion. Time of flight is the length of time the projectile is in the air. The path of a projectile is influenced by gravity. G G The acceleration a = g y G v0 2-D φ x R G G ⎧⎪ ax = 0 ⎪⎩ a y = g The components of a = g are: ⎨ ax = 0 because there are no horizontal forces on the projectile if we ignore air resistance. G G G Let us break the vector equation for velocity v = v0 + a t into components. v vy φ vx a. x-direction: vx = v0 x + ax t , because ax = 0 , v0 x = v0 cos φ + 0 t v0 x = v0 cos φ The velocity in this direction will not change. If the projectile has an initial horizontal velocity of v0x it will always have this horizontal velocity. b. y-direction: v y = v0 y + a y t , because a y = − g v0 y = v0 sin φ v y = v0 sin φ − g t At any time t, the speed is given by s = v = vx2 + v y2 = v02 − 2v0 gt sin φ + g 2t 2 (c) The trajectory G G G G r = r0 + v0t + 12 a t 2 ⎧⎪ x = ⎨ ⎪⎩ y = From (A), t = x0 + v0 x t + 12 ax t 2 y0 + v0 y t + 12 a y t 2 = (v0 cos φ )t = (v0 sin φ )t − 12 g t 2 ( A) ( B) x , substitute it into (B) v0 cos φ y = x tan φ − g 2v cos 2 φ Trajectory equation 2 0 y G v0 G vx P x R Finding the Range At the moment the projectile hits the ground its height y = 0. To calculate the range, we plug in y = 0, the solutions are (a) (b) x=0 xmax 2v02 v02 2sin φ cos φ =R= sin φ cos φ = g g Using the identity sin 2φ = 2 sin φ cos φ R= v02 sin 2φ g Find the Maximum Height v y = v0 sin φ − gt At t = 0, v y = v0 sin φ > 0 as t increases, vy becomes smaller, but v y > 0 At some moment v y = 0 and when that happens we are at the top of the trajectory. We can find the time ttop when the projectile is at the top of the trajectory by plugging 0 in the equation for vy v y = v0 sin φ − g ttop 0 = v0 sin φ − g ttop ttop = v0 sin φ g when t > t top , v y < 0 , so at t top = v0 sin φ the projectile reaches the highest point. Then the g maximum height: ymax 1 = (v0 sin φ ) ttop − g (ttop ) 2 2 2 2 v sin φ 1 v02 sin 2 φ = 0 − g g 2 g2 v02 sin 2 φ 2g = Finding the Time of Flight The time of flight can be found by determining the time when the projectiles height returns to 0. The projectiles height is given by: y = y0 + v0 y t + 12 a y t 2 Since the initial height y0 = 0 and a y = − g y = v0 y t − 12 g t 2 And the projectile returns to it’s starting height when y = 0 , the time in flight t f will be given by: 0 = v0 y t f − 12 g t 2f 0 = v0 sin φ t f − 12 g t 2f 0 = v0 sin φ − 12 g t f tf = 2v0 sin φ g Notice that this exaclty double the time it took to get to the top of the trajectory ttop = v0 sin φ . g