VIEWS: 7 PAGES: 6 CATEGORY: High School POSTED ON: 9/18/2012
AIEEE and other Entrance Exams PHYSICS Notes & Key Point MOTION IN ONE AND TWO DIMENSIONS Chapter - 2 1. Motion in one dimension (Rectilinear motion) (a) To describe the motion of an object, we introduce four important quantities namely position, displacement, velocity and acceleration. (b) For a restricted motion confined to move only in a straight line, there are only two directions, one is designated as positive and the other negative. 2. Position (a) Since the motion is along a straight line, we choose an axis say X − axis so that it coincides with the path of the object. (b) Position : It is the location ( X − coordinate) of the particle with reference to a conveniently chosen origin, say O . (c) Position to the right of O is taken as positive and to the left of O is taken as negative. (d) In the figure shown, the position of the object at points A, B and C are respectively + 10m,+20m and − 10m C O A B X −10 0 +10 +20 ( in metres) 3. Distance and Displacement (a) Distance: It is the length of the actual path traversed by an object during motion in a given interval of time. Distance has only magnitude and hence is a scalar. (b) Displacement: It is the shortest distance between the starting and end positions of an object during motion in a given interval of time. Displacement has both magnitude and direction and hence is a vector. n (c) Suppose an object moves from position A(x = +10 m) to B(x = +20 m) and then to C(x = −10 m) .c o.i Then the distance travelled by the object from A to B is 10m and from B to C is 30m . e rs ach The total distance travelled from A to C is 40m . The displacement is given by e te ∆x = (Coordinate of end position, C ) – (Coordinate of starting position, A ) = (−10m) − (+10m) = −20 m in l where, ∆x denotes the change in position. /on : /Path A→B h t tpB → C A→C m DIstance Displacement 10m d f ro 30m − 30m 40m 10m e − 20m n l oad (d) If the particle is at x1 at time t 1 and at x 2 at time t 2 , then the displacement of the object is ∆x = x 2 − x1 D ow (e) If x 2 > x1 , ∆x is positive and hence displacement is positive. If x 2 < x1 , ∆x is negative and hence displacement is negative. 4. Speed and Velocity Distance travelled (a) Speed : It is defined as the rate of change of position with respect to time of the object. Speed = time taken Total distance travelled (b) Average speed : Average speed = total time interval x − x 1 ∆x (c) Average velocity : v av = 2 = t 2 − t1 ∆t ∆x dx (d) Instantaneous velocity : v = lim = ∆ t →0 ∆t dt 5. Acceleration (a) When the velocity of the object changes continuously as the motion proceeds, the body is said to have accelerated motion. v 2 − v1 ∆v (b) Average acceleration : a av = = t 2 − t1 ∆t ∆v dv (c) Instantaneous acceleration : a = lim = ∆t →0 ∆t dt Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in 6 Relation among Displacement, Velocity and Acceleration in Velocity-Time graph R Q (a) The average acceleration between two given points P and Q in time interval ∆t is equal to the slope of the chord connecting the points on a velocity-time graph. ∆v P (b) The instantaneous acceleration at a given point say R is the slope of the tangent ∆t drawn to the velocity-time graph. t (c) Given a velocity versus time graph, the displacement during an interval between times t 1 and t 2 is the area bounded by the velocity curve, the two vertical lines at t = t1 and t = t 2 and the X − axis. The shaded portion shown in the figure is the bounded area and hence displacement. (d) The area above X − axis is positive and below it is negative. t1 t2 t (e) If the area is summed up without taking signs into consideration, it gives distance covered in time interval t = 0 to t = t 1 A1 A3 Distance = A1 + A 2 + A 3 0 t1 A2 Displacement = A1 − A 2 + A 3 t in 7. Equations of motion for uniform accelerated motion . co. e rs (a) For a uniformly accelerated motion, the variables u , v, a , s and t are connected by the following relations v = u + at , 1 s = ut + at 2 , v 2 − u 2 = 2as te ach e 2 lin on (b) It should be noted that all of these equations contain exactly four variables as shown in the table below. / Equation p: / tt v Contains uh a s t v = u + at ro m df yes yes yes no yes 1 s = ut + at 2 d e loa yes no yes yes yes 2 n v 2 − u 2 = 2as D ow yes yes yes yes no In most problems, in uniformly accelerated motion, three parameters are given and fourth or fifth or both are to be found. Depending upon convenience, one can choose any one or two of the three relations given in the table to calculate the unknown parameters. (c) Sometimes representing motion on a velocity versus time graph can help in solving problems faster than solving them by conventional methods. (d) It should also be noted that the variables u , v, a and s are vectors and t is a scalar. This aspect is used while solving problems. th (e) Distance covered in n second : (Distance covered in n second) =( Distance covered in n seconds) - ( Distance covered in n − 1 seconds) th 2 ∴ s n = un + an 2 − u (n − 1) + a (n − 1) = u + (2n − 1) 1 1 a 2 2 2 8. Approach to Solving Problems Step- I: Define initial and final points between which the equations of motion are to be written Step- II: With initial point as origin, define positive direction and call it X − axis. Step- III: Make a table with 5 variables u , v, a , s and t written in horizontal line and fill up the table with appropriate values. Step-IV: Count the no. of unknown variables and write as many no. of eqns. Solve the eqns to get the values of unknown vairables. Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in Example 1 : A stone is dropped from a tower of height h . Calculate the velocity and time after which the stone strikes the ground. Solution : Here, Initial point - A , Final point - B . Let origin be at A and X -axis be downwards. (A) Initial Final Accele Displac Time Velocity Velocity ration ement Interval (u ) ( v) (a ) (s) (t) h 0 v g h t Unknown variables are u and t (B) Equations: v 2 − u 2 = 2as and v = u + at X i.e, v − 0 = 2gh 2 2 and v = 0 + gt 2h Hence, v = 2gh and t= g Example 2 : A stone is thrown vertically upward with a speed 5 m / s from the top of a building 10m high. Find (a) the time after which it strikes the ground. (b) the velocity with which it strikes the ground. (Take g = 10 m/s 2 ) Solution: Here, Initial point - A , Final point - C (B) Let origin be at A and X − axis be downwards Initial Final Accele Displac Time n o.i (A) Velocity Velocity ration ement Interval rs ( t ) .c he t (u ) ( v) (a ) (s) ec t10am Path A to C via B − 5m / s v 2 10 m 10m / s e Note that s is displacement and hence a vector. . The displacement from point A to C via B is 10m n lin (C) / /o . Unkonwn variables are v and t tp: t and X Equations: v − u = 2as h 2 2 v = u + at v − (−5) = 2 × 10r× 10 2 2 o m and v = −5 + 10 × t d f and i.e., Hence, v = 15m / s e t = 2 sec o ad taken upwards. Accordingly, the signs of vectors u, v, a and s gets reversed. However, the Note: X − axis canlalso be w nthe same. Students should try to solve the same problem themselves by taking X − axis upwards. Do from the top of a building and 1 second later, another stone is thrown downwards with a velocity of final results remain Example 3 : A stone is dropped (Take g = 10 m/s ) 2 20m / s . How far below the top, will the second stone over take the first ? Solution: Here, Initial point - A Final point - B (where the two stones meet) 1 2 Let origin be at A and X − axis be downwards. (A) Initial Final Accele Displac Time h Velocity Velocity ration ement Interval (B) (u ) ( v) (a ) (s ) (t) Stone 1 0 – 10m / s 2 h t Stone 2 20m / s – 10m / s 2 h t −1 X Unknown variables are h and t 1 1 Eqn for stone 1: s = ut + at 2 i.e., h = 0 + × 10t 2 = 5t 2 ... (i) 2 2 1 1 Eqn for stone 2: s = ut + at 2 i.e., h = 20( t − 1) + × 10( t − 1) 2 = 20( t − 1) + 5( t − 1) 2 ... (ii) 2 2 From (i) and (ii) , 5t 2 = 20( t − 1) + 5( t − 1) 2 ⇒ 20( t − 1) + 5 (−2 t + 1) = 0 ⇒ 10t − 15 = 0 or t = 1.5 sec ⇒ h = 5t 2 = 5(1.5) 2 Hence, h = 11.25m Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in Example 4: A stone is dropped from the top of a tower of height h . Simultaneously, another stone is projected upwards from bottom. They meet at a height 2h / 3 from the ground level. If h = 60 m , find the initial velocity of the lower stone. (Take g = 10 m/s ) 2 Solution: Here, Initial point - A for stone 1 and B for stone 2 . Final point - C for both stones. Stone 1 Let A be origin and X − axis be downwards Stone 2 Let B be origin and X'− axis be upwards X' Note: For different particles, different axes and different directions may be considered. 1 (A) Initial Final Accele Displac Time Velocity Velocity ration ement Interval (u ) ( v) (a ) (s ) (t) h (C) Stone 1 0 – g h /3 t 2h/3 Stone 2 u – −g 2h / 3 t (B) Unknown variables are u and t 2 X 1 h 1 Equation for stone 1: s = ut + at 2 i.e., = 0 + gt 2 ............. (i) 2 3 2 1 2 1 s = ut + at 2 h = ut − gt 2 Equation for stone 2: i.e., in ............ (ii) 2 3 2 . co. 2h 2 × 60 e rs ach From (i) , t= = = 2 sec 3g 3 × 10 e te u= 2h 1 + gt = 2 × 60 1 lin + × 10 × 2 = 30 m / s on From (ii) , 3× 2 2 : // 3t 2 p 9. ht tin a plane (two dimension) with constant acceleration . Motion in two dimensions with constant acceleration. (a) We shall now consider the special case of motion rm (b) As the particle moves, the accelerationo does not vary either in magnitude or direction. fr a d Consequently, the componentse acceleration a and a remain constant. o ad can be described as the sum of two component motion occurring simultaneously and each motion with (c) We then have a situationl which of x y wn Do constant acceleration along each of the two perpendicular directions. (d) The equations of motion, i.e., 1 v = u + at , s = ut + at 2 , v 2 − u 2 = 2as 2 can be applied separately for X − axis and Y − axis. 10. Projectile Motion (a) The most common example of a particle moving with uniform acceleration is the Y motion of particle near the surface of earth . Such a motion is called projectile motion . v v r ˆ (b) Consider a particle projected with initial velocity u which is directed in direction j a = −gˆ j v which makes a certain angle θ with the horizontal. Such a particle has two ˆ a = −gˆ θ a = −gj j dimensional motion and will move in a curve as we know from experience X (c) If we choose X − axis along the horizontal and Y − axis along the vertical with the ˆ i origin at the initial position of the particle, the motion of the particle will be along the a = −gˆ j curve as shown. 11. Important points in projectile motion (a) The acceleration of the particle is constant. Its magnitude is g = 9.8 m s - 2 and direction of acceleration is directed along the vertical r downward direction. In the chosen coordinate system, shown in the figure, a = −g ˆ = −9.8 ˆ m s −2 j j (b) The velocity vector changes with time both in magnitude and direction . The direction of velocity is always along the tangent to the curve. (c) The acceleration being in the vertical direction, the horizontal component of velocity remains constant. (d) When the particle is at the highest point, velocity is directed towards the horizontal. This means that at the highest point, v y = 0 (e) When the particle again hits the ground, its y − coordinate is zero. Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in 12. Approach to solving problems Step-I: Define initial and final points between which the equations of motion are to be written Step-II: With initial point as origin, define two mutually perpendicular axes. Call them X − axis and Y − axis. Step-III: Make a table with 5 variables u , v, a , s and t written in horizontal row and axes X and Y in vertical column. Fill up the table with appropriate values. Step-IV: Count the no. of unknown variables and write as many no. of eqns. Solve the eqns to get the values of unknown variables. 13. Particle projected at some angle to horizontal (a) Consider a particle projected with velocity u at an angle θ to horizontal. Y We shall derive an expression for (B) Horizontal range ( R ) Time of flight (T ) ( x, y ) (D) Maximum height attained (H) H (b) Let origin be point A and X and Y axes as shown. θ R X (c) Derivation for Range and Time of flight (A) (C) Here, Inital point is A and Final point is C Initial Final Accele Displac Time Axis Velocity Velocity ration ement Interval (u ) ( v) (a ) (s) (t) X u cos θ − 0 R T Path A to C Y u sin θ − −g 0 T n Unknown variables are R and T .c o.i Equation along X − axis 1 s = ut + at 2 R = u cos θ T e rs ach i.e., ............ (i) 2 Equation along Y − axis s = ut + at 1 2 i.e., e te1 0 = u sin θ T − g T 2 in ............ (ii) l /on 2 2 2u sin θ :/ From eqn (ii), we obtain T = g h t tp m 2u sin θ u 2 sin 2θ Putting the value of T in eqn (i) , we get, d f ro R = u cos θ × g ∴ R= g (d) Derivation for Maximum heighta de Inital point is A n l oand attained Here, Final point is B D ow Initial Final Accele Displac Time Axis Velocity Velocity ration ement Interval (u ) ( v) (a ) (s) (t) Path A to B Y u sin θ 0 −g H − u 2 sin 2 θ Equation is v 2 − u 2 = 2as i.e., 0 2 − (u sin θ) 2 = 2(−g )H ∴ H= 2g 14. Particle projected horizontally from some height (a) Consider a particle projected horizontally with velocity u from a point at height h above the ground. We shall derive an expression for A X • Horizontal distance (s) at which the particle will strike the ground • Time taken ( t ) to strike the ground. h (b) Here, Initial point is A and Final point is B Let origin be at point A with X and Y axes as shown. Initial Final Accele Displac Time S B Axis Velocity Velocity ration ement Interval Y (u ) ( v) (a ) (s) (t) X u − 0 s t Y 0 − g h t Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in Unknown variables are s and t 1 2 Equations along X − axis, s = ut + at i.e., s = ut ............(i) 2 1 2 1 Equations along Y − axis, s = ut + at i.e., h = gt 2 ............(ii) 2 2 2h 2h From eqn (ii), we get, t= From eqn (i), we get, s=u g g 15. Relative velocity in two dimensions r r (a) Consider two objects A and B moving with velocities v A and v B (with respect to some common frame of reference, say ground) r r r (b) Then velocity of object B relative to object A , v BA = v B − v A r r r Similarly, the velocity of object A relative to object B , v AB = v A − v B r r r r (c) Therefore, v BA = − v AB and | v AB | =| v BA | Q 16. Uniform Circular Motion s (a) Consider a particle moving in a circle of radius r with its centre at O . Since, the distance of θ the particle from origin O remains constant, its position can be uniquely specified by angle θ . o.in r P O PQ s . c (b) Angular displacement : θ= = e rs and is expressed in radians. ach r r ∆θ d θ Fort e (c) Angular velocity: ω = Lt ∆t →0 = ∆t dt li ne a uniform circular motion, ω is a constant. / on (d) Angular acceleration: α = Lt ∆ω dω = t p: / For a uniform circular motion, α is zero. ∆t →0 ∆t dt ht m (e) The relation between particle speed ( v ) and angular velocity ( ω ) is given by v = rω d f ro e 17. Relationship between linear and angular variables in Circular Motion d Linear Motion n loa Circular Motion Relation Distance - s Velocity - v D ow Angular Displacement - θ Angular velocity - ω s = rθ v = rω Tangential Acceleration - a t Angular Acceleration - α a t = rα Equations for linear motion with constant acceleration and initial velocity u v = u + at s = ut + (1 / 2)at 2 v 2 − u 2 = 2as Equations for circular motion with constant angular acceleration and initial angular velocity ωi ω = ωi + αt θ = ωi t + (1 / 2)α t ω2 − ωi2 = 2αθ 18. Acceleration of particle in Uniform Circular Motion (a) Consider a particle moving in a circle of radius r with a constant speed v . Since, its velocity (which is in the direction of tangent to the circle) changes its direction continuously, the particle is said to be undergoing an acceleration. (b) This acceleration is directed towards the centre of the circle known as centripetal acceleration. v2 Its magnitude is given by a r = = ω2 r r 19. Acceleration of particle in non-uniform circular motion (a) Consider a particle moving in a circle of radius r whose speed v changes at a uniform rate. (b) The acceleration in this case is the vector sum of v2 dv centripetal acceleration, a r = and tangential acceleration, at = r dt 2 v 2 dv 2 a = a2 + a2 = + r t r dt (c) The total acceleration is given by Downloaded from http://onlineteachers.co.in, portal for CBSE, ICSE, State Boards and Entrance Exams. Prepared by Alpha Education , Sec-4, Gurgaon, Haryana. sales@onlineteachers.co.in