Tiêu đề MOTION IN A PLANE.pdf ✅

Kinematics (Motion in a Straight Line & Plane)  Digital www.allendigital.in [ 151 ] Ground to Ground Projectile Motion 2D Motion: • When the motion of an object is restricted within a plane, it is said to undergo a motion in 2D. • 2D motion can be studied as two independent 1D motions. (One along x-axis and the other along y-axis). Example: Motion of a carrom coin, Projectile Motion, Circular Motion. Projectile Motion When a body moves with constant acceleration such that its initial velocity and acceleration are non- collinear then its path is parabola and motion is known as projectile motion. Projectile Motion = Horizontal Motion+Vertical Motion u ucos x =  ; =  u usin y ax = 0 ay = –g v u ucos x x = =  v u gt y y = − x ucos t =  ( ) =  − ( ) 1 2 y usin t gt 2 Horizontal Motion • Initial velocity in horizontal direction = ucos u = x • Acceleration along horizontal direction = ax = 0. (Neglect air resistance) • Therefore, Horizontal velocity remains unchanged. • At any instant horizontal velocity v ucos x =  • In time t, displacement along horizontal direction or x co-ordinate is ,x = uxt or x ucos t =  ( ) Vertical Motion • It is motion under the effect of gravity so that as particle moves upwards the magnitude of its vertical velocity decreases. • Initial velocity in vertical direction = usin u = y • Acceleration along vertical direction = ay = – g (Neglect air resistance) • At any instant, vertical speed v u gt usin gt y y = − = − • In time t, displacement in vertical direction or "height" of the particle above the ground = − =  − 2 2 y 1 1 y u t gt usin t gt 2 2 Motion in a Plane Projectile Motion Oblique Horizontal Ground to Ground Height to Ground Height to Ground usin ucos R x u H   v= u cos usin–gt P(x,y) y
NEET : Physics [ 152 ] www.allendigital.in  Digital • Equations of motion should be applied separately for x and y directions. • Velocity and position can be dealt separately in x and y directions as shown in the table given below: x-axis y-axis Resultant v u a t x x x = + v u a t y y y = + = + = +2 2 x y x y ˆ ˆ v v i v j; v v v  = + 2 x x 1 x u t a t 2  = + 2 y y 1 y u t a t 2 = + = +2 2 ˆ ˆ r xi yj; r x y Illustration 1: A particle has initial velocity, u 2i 3j = + ˆ ˆ and acceleration, = + ˆ ˆ a 4i 2j then, Find (a) v at t = 2sec. (b) s from 0 to 2 sec. Solution: x-direction y-direction ux = 2 uy = 3 ax = 4 ay = 2 t = 2 t = 2 v u a t x x x = + v u a t y y y = + vx = 2 + 4 × 2 = 10 vy = 3 + 2 × 2 = 7   + =     x x x u v s t 2   + =     y y y u v s t 2   + = =     x 2 10 s 2 12 2   + = =     y 3 7 s 2 10 2 = + ˆ ˆ v 10i 7j and = + ˆ ˆ s 12i 10j Illustration 2: A roller coaster goes down along 45° incline with an acceleration of 5.0 ms–2. (Starts from rest) (a) How far will the roller coaster travel in 10 seconds horizontally? (b) How far will the roller coaster travel in 10 seconds vertically? Solution: The motion of the roller coaster can be considered as a 1D motion along the direction of acceleration. Taking the direction of acceleration as r-direction. = + + 2 0 1 x x ut at 2 In r-direction, = + + 2 0 1 r r ut at 2 = +  + ( )( ) 1 2 0 0 t 5 10 2 = 250 m Taking right direction as positive x axis and downwards as positive y axis, Displacement along y-direction  =  y rsin =250 sin 45° = 250 m 2 Displacement along x-direction  =  x rcos = 250 cos 45° = 250 m 2 45° +x r=250 m +y 45° 5 m/s2
Kinematics (Motion in a Straight Line & Plane)  Digital www.allendigital.in [ 153 ] Illustration 3: A body is projected with a velocity u at an angle  with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time t given by. Solution: v usin gt y = − v ucos x =    +    + − = ( ) ( ( ) ) ucos i usin j ucos i usin gt j 0 ˆ ˆ ˆ ˆ  + −  = 2 2 2 2 u cos u sin usin gt 0  =  2 u usin gt  =  u t gsin Illustration 4: A particle is projected with initial velocity u = 25m/s and  = 53° with horizontal? Find height of projectile when horizontal distance is 45 m? Solution: x-direction y-direction ux = 25 cos 53° uy = 25 sin 53° = 15 m/s 20 m/s ax = 0 ay = –10 sx = 45 t = 3 sec. 45 = 15 × t =    2 y 1 s 20 3– 10 3 2 = = 45 t 3s 15 sy = 60 – 45= 15 m Standard Results of Ground to Ground Projectile Motion Time of flight (T) Time of flight is the time for which projectile remains in air. At time T particle will be at ground again, i.e. displacement along Y-axis becomes zero. = − 1 2 y u t gt y 2  = − 1 2 0 u T gT y 2 or  = = 2uy 2usin T g g Time of ascent = Time of descent =  = = T usin uy 2 g g At time T 2 particle attains maximum height of its trajectory. Maximum height (H) At maximum height vertical component of velocity becomes zero. At this instant y coordinate is, its maximum height. = − 2 2 y y v u 2gy  = − = =   2 0 u 2gH v 0,y H y y so,  = = 2 2 2 uy u sin H 2g 2g Horizontal range or Range (R) It is the displacement of particle along X-direction during its complete flight. = = = =  y x y x x x 2u 2u u x u R u T u R g g t ; ( )  =   =    =  2 2 u sin2 R g (ucos )(usin ) R 2sin cos sin2 g +ve 53° +ve 25m/s Y u A B B X u H O
NEET : Physics [ 154 ] www.allendigital.in  Digital Maximum horizontal range (Rmax) If value of  is increased from  = 0° to 90°, then range increases from  = 0° to 45° but it decreases beyond 45°. Thus range is maximum at  = 45° For maximum range,  =  45 and  = = 2 2 max u sin2(45 ) u sin 90 R g g  = 2 max u R g Comparison of two projectiles of equal range When two projectiles are thrown with equal speeds at angles  and (90° –  ) then their ranges are equal but maximum heights attained are different and time of flights are also different. At angle  ,  = 2 u sin2 R g At angle ( )  −   −    −  = = = 2 2 2 u sin2(90 ) u sin(180 2 ) u sin2 90 ,R ' g g g Thus, R' R= Maximum height of projectiles   −   = = = 2 2 2 2 2 2 u sin u sin (90 ) u cos H and H' 2g 2g 2g •  = =   2 2 2 H sin tan H' cos •   = =  = 4 2 2 2 2 u sin cos R HH' R 4 HH' 4g 16 •   + = +  + = 2 2 2 2 2 u sin u cos u H H' H H' 2g 2g 2g Time of flight of projectiles  −   = = = 2usin 2usin(90 ) 2ucos T ;T' g g g •  = =   T sin tan T' cos •   = =   2 2 4u sin cos 2R TT' TT' R g g Equation of Trajectory Along horizontal direction x = uxt or x ucos t =  Along vertical direction = − 2 y 1 y u t gt 2 or =  − 1 2 y usin t gt 2 On eliminating t from these two equations 2 x 1 x y (usin ) g ucos 2 ucos =  −                2 2 2 1 x y x tan g 2 u cos =  −  This is an equation of a parabola so it can be stated that projectile follows a parabolic path. Again 2 gx x y x tan 1 x tan 1 2u sin cos R =  −  −       =         Kinetic Energy of a Projectile Kinetic energy =  1 2 Mass × (Speed)2 Let a body is projected with velocity u at an angle  . y T u u u H R H O x T