Tiêu đề 2. Kinematics and Vector.pdf ✅

Kinematics and Vector 21 2 Kinematics and Vector QUICK LOOK Table 2.1: One, Two and Three Dimensional Motions One dimensional Two dimensional Three dimensional Motion of a body in a straight line is called onedimensional motion Motion of body in a plane is called two dimensional motion. Motion of body in space is called three dimensional motion When only one coordinate of the position of a body changes with time then it is said to be moving one dimensionally When two coordinates of the position f a body change with time then it is said to be moving two dimensionally When all three coordinates of the position of a body change with time then it is said to be moving three dimensionally Ex. (i) Motion car on a straight road. (ii) Motion of freely falling body. Ex. (i) Motion of car on a circular tum. (ii) Motion of billiards ball Ex. (i) Motion of flying kite. (ii) Motion of flying insect. Motion in One Dimension: Equations of Kinematics υ = + u at u = initial velocity 1 2 2 s ut at = + v = velocity after t se 2 2 v u as = + 2 s = displacement in t seconds 1 (2 1) 2 t s u a t = + − a = constant acceleration st = displacement in second Displacement-time Graphs and their Characteristics Figure: 2.1 If the graph is a straight line parallel to time-axis means that the body is at rest, i.e., v = 0.A straight line inclined to x-axis shows that body is moving with a constant velocity. Remember that a straight line inclined x-axis by an angle > 90° represent negative velocity. It is worth noting that no line can ever be ⊥ to the time axis because it implies infinite velocity. If the curve is of the type whose slope decreases with time, the velocity goes on decreasing i.e., motion is retarded. If the curve is of the type whose slope increases with time, the velocity goes on increasing i.e., motion is accelerated. The slope of s-t graph gives velocity. The speed is magnitude of velocity and it can never be negative. The distance is length of actual path and it can never be negative. Velocity-time Graphs and their Characteristics: If the graph is a straight line parallel to time axis means that the body is moving with a constant velocity or acceleration (1) is zero. If the graph is a straight line inclined to the x-axis with +ve slope means that the body is moving with constant acceleration. If the graph is a straight line inclined to x-axis with negative slope means that the body is under retardation. The slope of v t − graph gives acceleration Figure 2.2 The area enclosed by v–t curve and times axis gives displacement. Vertical Motion In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection. As well as the magnitude of velocity at any point on the path is same whether the body is moving in upward or downward direction. The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity Time in S 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Velocity in m/s Constant deceleration Constant Velocity Constant acceleration Time Distance Fast steady speed Getting faster Returning to start Stationary Steady speed
22 Quick Revision NCERT-PHYSICS i.e., 2h t g = and v hg = 2 In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance. Time of descent 2 ( ) t = time of ascent 1 ( ) u t g = ∴ Total time of light 1 2 2u T t t g = + = Let m be the mass of the body. Then in going from the ground to the highest point, following changes take place. Change in speed = u Change in velocity = u Change in momentum = mu Change in kinetic energy = Change in potential energy 1 2 2 = mu One return to the ground the changes in these quantities are as follows Change in speed = 0 change in velocity = 2u Change in momentum = 2mu Change in kinetic energy = Change in potential energy = 0 If, the friction of air be taken into account, then the motion of the object thrown upwards will have the following properties Time taken to go up (ascent) < time taken to come down (descent) The speed of the object on returning to the ground is less than the initial speed. Same is true for velocity (magnitude), momentum (magnitude) and kinetic energy. Maximum height attained is less than 2 2 u g A part of the kinetic energy is used up in overcoming the fraction. A ball is dropped form a building of height h and it reaches after t seconds on earth. From the same building if two balls are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after 1 t and 2 t seconds respectively then 1 2 t t t = A particle is dropped vertically form rest form a height. The time taken by it to fall through successive distance of 1m each will then be in the ratio of the difference in the square roots of the integers i.e., 1, ( 2 1), ( 3 2),....( 4 3),... − − − A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than that time of descent. 2 1 t t > Let u is the initial velocity of body then time of ascent 1 u t g a = + and 2 2( ) u g g a = + . Where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will work vertically downward. For downward motion a and g will work in opposite direction because a always work in direction opposite to motion and g always works vertically downward. So, ( ) 2 2 1 2 h g a t = − ⇒ ( ) ( ) 2 2 2 1 2 2 u g a t g a = − + ⇒ ( )( ) 2 - u t g a g a = + Comparing 1 t and 2 t we can say that 1 2 t t > ,since ( g a g a + > − ) ( ) Horizontal Motion If a particle is accelerated for a time t1 with acceleration a1 and for time t2 with acceleration a2 then average acceleration is 1 1 2 2 1 2 av a t a t a t t + = + If same force is applied on two bodies of different masses m1 and m2 separately then it produces acceleration 1 a and a2 respectively. Now these bodies are attached together and form a combined system and same force is applied on that system so that a be the acceleration of the combined system, then 1 2 1 2 a a a a a = + If a body start form rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to 2 2 t i e s t ( . . : ) ∝ So, we can say that the ratio of distance covered in 1 sec, 2 second and 3 sec is 2 2 2 1 : 2 : 3 or 1 : 4 : 9. A body moving with a velocity u is stopped by application of brakes after covering a distance s. If the same body moves with velocity nu and same braking force is applied on it then it will come to rest after covering a distance of n 2 s. As 2 2 2 v u u = − ⇒ = − 2as 0 2as ⇒ 2 2 , 2 u s s u a = ∝ [since a constant]
Kinematics and Vector 23 So, we can say that if u becomes n times then s becomes n 2 times that of previous value. A particle moving with uniform acceleration from A to B along a straight line has velocities v1 and v2 at A and B respectively. If C is the mid-point between A and B then velocity of the particle at C is equal to 2 2 1 2 2 v v v + = Motion with Variable Acceleration If acceleration is a function of time a f t = ( )then 0 ( ) t v u f t dt = + ∫ and ( ) 0 ( ) t s ut f t dt dt = + ∫ ∫ If acceleration is a function of distance a f x = ( ) then 2 2 2 ( ) v u v u f x dx = + ∫ If acceleration is a function of velocity a f v = ( ) then ( ) v u dv t dx f v = ∫ and 0 ( ) v u vdv x x f v = + ∫ Projectile Motion Figure: 2.3 A projectile is any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown. (This is an informal definition.) The path of a projectile is called its trajectory. If g can be taken as constant in magnitude and direction, i.e., if velocity of projection is small, the path is a parabola. If g cannot be taken as constant in magnitude and direction, i.e., if velocity of projection is large, the path is a conic section. Let the particle acquire a position P having the coordinates ( , ) x y just after time t from the instant of projection. The corresponding position vector of the particle at time t is r as shown in the figure. ˆ ˆ r xi yj = + . . .(i) The horizontal distance covered during time t is given as cos x x u t x u t = ⇒ = θ . . .(ii) The vertical velocity of the particle at time t is given as y y v u gt = − . . .(iii) Now the vertical displacement y is given as 1 2 sin 2 y u t gt = − θ . . .(iv) Putting the values of x and y form equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as ( ) ( ) 1 2 ˆ ˆ cos sin 2 r u ti u t gt j θ θ   + −     ⇒ ( ) ( ) 2 2 1 2 cos sin 2 r u u gt θ θ   = + −     2 sin 1 2 gt gt r ut u u   θ = + −     and 1 tan y x φ −   =     ( ) 2 1 1 sin 2 tan cos ut gt ut θ φ θ −   −   =       or 2 1 2 sin tan 2 cos u gt u θ φ θ −   − =     Note The angle of elevation φ of the highest point of the projectile and the angle of projection θ are related to each other as 1 tan tan 2 θ θ = Range on an Inclined Plane Figure: 2.4 tan tan y y x x α α = ⇒ = . . . (i) Equation of trajectory 2 2 2 1 tan tan 2 cos x x x x α θ θ   = −     or ( ) 2 2 2 tan tan cos u x g θ α θ − = . . . (ii) ∴ ( ) 2 2 2 tan tan cos tan u y g θ α θ α − = . . . (iii) 2 2 R x y = + . . . (iv) Using (ii) + (iii) equation (iv) gives ( ) 2 2 2 tan tan cos cos u R g θ α θ α − = Resulted velocity of a projectile at a give instant 2 2 dy dx v dx dy     = +         y X Y Q R α x Y O R X H Vx Vy u θ P (x, y)
24 Quick Revision NCERT-PHYSICS ( ) ( ) 2 2 v u gt u = − + sin cos θ θ ( ) 2 2 v u u g t g t = − ⋅ + 2 sinθ Resulted velocity angle two horizontal say β sin 1 tan tan tan cos cos u gt gt u u θ β β θ θ θ − −   = ⇒ = −     Relative Motion: A particle P is moving and is observed from two frames S and S'. The frame S is stationary and the frame S' is in motion. Let at any time position vector of the particle P with respect to S is, OP r = p s, and with respect to S′ O P rp s, ′ ′ = Position vector of the origin of S' with respect to S is, O P rp s, ′ ′ = Figure: 2.5 From vector triangle OO'P, we get, O P OP OP ′ ′ = − ⇒ p s p s s s , , , r r r ′ ′ = − ⇒ ( ) p s p s s s , , , ( ) ( ) d d d r r r dt dt dt ′ ′ = − ⇒ p s p s s s , , , v v v ′ ′ = − ⇒ p s, p absolute s absolute ( ) ( ) v v v ′ = − ′ Relative Motion between Rain and Man: Let r v = velocity of rain w.r.t. ground, m v velocity of man w.r.t. ground so the velocity of rain w.r.t. man. r rm m v v v = + That means the vector addition of the velocity of rain with respect of man ( ) rm v and the velocity of man (vehicle) ( ) rm v yield the actual velocity of rain . r v The magnitude and direction of r v can be given as, {( ) ( ) } 2 2 2 cos r rm m rm m v v v v v = + + θ 1 sin tan cos rm rm m v v v θ φ θ −   =     + with horizontal m v River Cases: Boat–River problem: Let 1 v = velocity of boat in still water, 2 v = velocity of flow of water in river, d = width or river. To cross the river in the shortest path: Here it is required that the boat starting from A must reach the opposite point B along the shortest path AB. For the shortest path, the boat should be rowed up stream making an angle θ with AB such that AB gives the direction of resultant velocity. So 2 1 sin v v θ = and 2 2 2 1 2 v v v = − . Also; 2 2 1 2 s s t v v v = = − Figure: 2.6 To cross the river in the shortest time: For the boat to cross the river in shortest time, the boat should be directed along AB. Let ν be the resultant velocity making an angle θ with AB. Then 2 1 tan v v θ = and 2 2 2 1 2 v v v = + ∴ Time of crossing, 1 . d t v = Now the boat reaches the point C rather than reaching point B. If BC x = , then 2 1 tan v x v d θ = = or 2 1 v x d v   = ×     Figure: 2.7 If a man travels downstream in a river, then the time taken by the man to cover a distance d is 1 1 2 . d t v v = + If a man swims upstream in a river, then the time taken by him to cover a distance d is 2 1 2 . d t v v = − So 1 1 2 2 1 2 . t v v t v v − = + 1 v B C A d 2 v 2 v v θ x 1 v C B A d 2 v 2 v v θ r v m v m −v r v m v m v θ θ Rain and Man