Tiêu đề 4.Motion In a Plane-F.pdf ✅

1 | P a g e NEET-2022 Ultimate Crash Course PHYSICS Motion in a Plane
2 | P a g e Important Points to Remember 1. For maximum range, 0 = 45 . In this case, 2 0 Rmax g  = and ( ) 2 2 2 2 2 0 0 0 1/ 2 sin h 2g 2g 4g      = = = or R 4h max = Thus, the maximum horizontal range of a projectile is 4 times the maximum height attained when it is fired along the required oblique direction (i.e., 0 = 45 ) 2. For R to be n times h, i.e., for R = nh, 2 2 2 0 0 sin 2 sin g n 2g     =  or 2 sin sin 2 n 2     =     or 2 sin 2sin cos n 2      =     or 4 tan n  = 3. As 2 2 0 sin h , 2g   = it is maximum when 0 = 90 ; i.e., 2 0 max h 2g  = Clearly, 2 2 0 0 R 2 2h max max g 2g    = = =     Thus, for a given velocity (0 ) of projection, Rmax (when 0 = 45 )is twice the maximum height, max h (when 0 = 90 ) 4. At the highest point, Potential energy, 2 2 0 2 2 0 sin 1 U mgh mg m sin 2g 2   = =  =   and kinetic energy, 2 2 2 0 0 1 1 K m m cos 2 2 =  =   Thus, 2 2 2 2 2 0 0 0 1 1 1 U K m sin m cos m 2 2 2 + =   +   =  The sum of U and K, i.e., mechanical energy (E) 2 0 1 m 2 =  = constant. Thus in projectile motion, E is conserved. 5. All along the trajectory of the projectile a) horizontal component of velocity and mechanical energy remain constant b) speed, velocity, vertical component or velocity, K and U change 6. Further, ( ) ( ) 2 2 0 2 2 2 0 U 1/ 2 m sin tan K 1/ 2 m cos   = =    If case 0 U 45 , 1 K  = = or U = K ; Clearly, 1 U K E 2 = = Thus, for a body projected at an angle of 450 , at the highest point, half of its E is K and the other half is U. River Problem In Two Dimensions (Crossing River) Consider a man swimming in a river with a velocity of mR, v relative to river at an angle of  with the river flow. The velocity of river is R v and the width of the river is d. m m R R , v v v = + ( ) = + + v i v j ui cos sin   ˆ ˆ ˆ ( ) = + + v u i v j cos sin   ˆ ˆ Here vsin  is the component of the velocity of man in the direction perpendicular to the river flow. This component of velocity is responsible for the man crossing the river. Hence if the time to cross the river is t, then
3 | P a g e sin y d v t v v  = = It is defined as the displacement of man in the direction of river flow. It is simply the displacement along x-axis during the period the man crosses the river. (v u cos + ) is the component of the velocity of man in the direction of river flow, and this component of velocity is responsible for drift along the river flow. If the drift is x, then Drift x =  v t ( cos ) sin d x v u v   = +  Crossing the River in Shortest Time As we know that t d v = / sin . Clearly t will be minimum when 0  = 90 , i.e., time to cross the river will be minimum if the man swims perpendicular to the river flow. Which is equal to d v/ Crossing the River in Shortest Path, Minimum Drift The minimum possible drift is zero. In this case, the man swims in the direction perpendicular to the river flow as seen from the ground. This path is known as shortest path. Here, x v u min =  + = 0 cos 0 (  ) or cos u v  = − Since cos is –ve, therefore, 0   90 i.e., for minimum drift the man must swim at some angle  with the perpendicular in upstream direction, where , sin R m R v u v v  = = 1 , cos 1 R m R v u v v  −   −  =         i.e., u v  i.e., minimum drift is zero if and only if the velocity of man in still water is greater than or equal to the velocity of river. Time to cross the river along the shortest path, 2 2 sin d d t v  v u = = − Rain-Man Problems: If rain is falling vertically with a velocity r v and an observer is moving horizontally with velocity m v , the velocity of rain relative to the observer will be 2 2 r m r m r m r m , , v v v or v v v = − = + and direction 1 tan m r v v  −   =     with the vertical as shown in the figures
4 | P a g e Different situations in rain-man problem The man is stationary and the rain is falling at this back at an angle  with the vertical. The man starts moving forward. The relative velocity of rain w.r.t. man shifts towards vertical direction. As the man further increases his speed, then at a particular value, the rain appears to be falling vertically. If the man increases his speed further more, then the rain appears to be falling from the forward direction. Average Angular Velocity: It is defined as the total angular displacement divided by the total time taken. or av Angular displacement Total time taken  = 2 1 2 1 t t t  −   = = −  Angular Acceleration: The angular acceleration of P relative to O can be given as,  = d / dt where d dt  = ; Then, 2 2 d dt   = When  increases,  is directed along ; when  decreases,  is directed opposite to . Centripetal Acceleration: It is responsible for change in the direction of velocity. In circular motion, there is always a centripetal acceleration. Centripetal acceleration is always considered variable because it changes in direction. It is also called radial acceleration or normal acceleration. It is given by 2 r v a R = ; Here v is the speed of the particle and R is the radius of the circle. Tangential Acceleration: