Physics Notes 2024 - 2025 - M.E.S Indian School PDF

M.E.S INDIAN SCHOOL, DOHA -QATAR Notes 2024 - 2025 Section: Boys / Girls Date: 06-05-2024 Class &Div.: XI (All divisions) Subject: Physics Lesson / Topic: MO...

M.E.S INDIAN SCHOOL, DOHA -QATAR Notes 2024 - 2025 Section: Boys / Girls Date: 06-05-2024 Class &Div.: XI (All divisions) Subject: Physics Lesson / Topic: MOTION IN A STRAIGHT LINE xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx CHAPTER 03 MOTION IN A STRAIGHT LINE Motion An object is said to be in motion, if its position changes with respect to time. This is related to the observer. If its position is not changing, the object is said to be at rest. Point object If the distance travelled by the object is very large compared to the size of the object, then the object can be considered as a point object or particle. For example, a train whose length is nearly a kilometer may be treated as a point objects if it travels very long distances. Frame of reference: It is a set of axes from which the state of body is observed with respect to time. Motion in one, two and three dimensions *If the motion of an object is restricted to a straight line it is said to execute one dimensional motion. e.g. A train running along a straight track *If the motion of an object is restricted to a plane it is said to be in two-dimensional Motion. e.g. The motion of a boat on a lake *An object moving in space in said to be in three-dimensional motion. e.g. A butterfly flying m air Vector: - A physical quantity having both magnitude and direction F 061, Rev 01, dtd10th March 2020 e.g. Force, velocity etc., Scalar- A physical quantity having only magnitude e.g. mass, distance etc., Differences between the terms ‘distance’ and ‘displacement’:- Distance Displacement 1. Distance is the length of actual 1. Displacement is the shortest path travelled by a body, distance between the initial irrespective of its direction of and final positions of a body motion. in a given direction. 2. Distance between two given 2. Displacement between two points may be the same or given points is always same. different for different paths chosen. 3. It is a vector quantity. 3. It is a scalar quantity. 4. Distance covered is always 4. Displacement covered may positive. be positive, negative or zero. Use the diagram to determine the resulting displacement and the distance travelled 1. Displacement = 2r Distance = πr 2 distance = h + h = 2h Displacement = 0 F 061, Rev 01, dtd10th March 2020  3. distance = h +2h’ Displacement = -h POSTION – TIME GRAPH The graph on which the instantaneous position of a particle is plotted on the y-axis and the time on the x-axis is known as the Position-Time graph. Speed- It is the distance travelled in unit time. Speed = distance travelled Time Uniform speed - when an object covers equal distances in equal intervals of time, however the small these time intervals may be, it is said to be in ‘uniform speed’. Non-uniform speed - when an object covers unequal distances in equal intervals of time then it is said to be in ‘non-uniform speed’. F 061, Rev 01, dtd10th March 2020  Average speed: When the speed of a body varies with time, we need to define its average speed. total distance Average speed = total time taken Note:-  If a particle travels at a speed v1 and v2 for intervals t1 and t2 respectively then total distance = v1 t1 + v2 t2 and total time = t1 + t2 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑐𝑒 𝑣1 𝑡1 +𝑣2 𝑡2  Average speed = = 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒 𝑡1 +𝑡2 v1 + v2 Case(i): - if t1 = t2 = t then, average speed (Vav ) = 2 2v1 v2  If S1 =S2 =S then  Vav = v1 + v 2 1. A car covers first half of the distance between two places at a speed of 40km/h andand the second half at 60 km/h. What is the average speed of the car? Let 2x be the distance covered then S1 =S2 = x ; Average speed ( Vav ) = 2𝑣1 𝑣2 2𝑥40𝑥60 = = 48 𝑘𝑚⁄ℎ 𝑣1 + 𝑣2 40 + 60 To convert km/h into m/s  multiply by 5 /18 5 i.e., 48km/h = 48 x = 13.33m/s 18 Velocity: - Velocity of a body is defined as the displacement produced per unit time. displacement Velocity = time *It’s S.I unit is m/s. F 061, Rev 01, dtd10th March 2020 Average velocity: -It is the ratio of displacement to the total time taken displacement Average Velocity = total time x(t') - x(t) V = av t' - t Distinguish between the terms ‘Speed’ and ‘Velocity’. Speed Velocity 1. It is the distance travelled by a body 1. It is the distance travelled by a per unit time in any direction. body per unit time in a given 2. It is a scalar quantity. direction. 3. It is always positive or zero but 2. A vector quantity. never negative. 3. It may be positive or negative or zero. Instantaneous velocity: - The velocity of particle at any instant of its motion is called instantaneous velocity. Acceleration (a):- It is the rate of change of velocity. Its S.I unit is m/s2. 𝑣−𝑢 𝑎= 𝑡  (where v- final velocity & u- initial velocity) F 061, Rev 01, dtd10th March 2020 Instantaneous acceleration:- The acceleration of an object at a given instant of time. Uniform acceleration:- If the velocity of the particle changes equal amounts in equal interval of time, then particle is said to move with uniform acceleration. Graphical representation of motion:- *Time is always plotted on the X-axis. Distance-time graph:- It is a graph obtained by plotting distance travelled along Y-axis and time along X-axis. The uses of distance -time graph are as follows: (a) It tells the position of the body at any instant of time. (b) The distance covered by the body during a particular time interval can be seen from this graph. (c) The velocity of the body at any instant of time can be determined. 1. The steeper the graph, the faster the motion 2. A horizontal line means the object is not changing its position - it is not moving, it is at rest F 061, Rev 01, dtd10th March 2020 3. A downward sloping line means the object is returning to the start. Velocity –time graph:- velocity along Y-axis. Note: - NOTE: The steeper the graph, the greater the acceleration. Area under velocity –time graph gives acceleration e.g. Here the shaded area = 30 x 6 m2 = 180 m2. i.e., The distance travelled in 6 second = 180m. Q1.Shows the velocity –time graph of a particle travelling along a straight line. Find the distance and displacement covered in 6s. F 061, Rev 01, dtd10th March 2020 Ans: Distance covered = area 1+area 2+ area 3+ area 4 = ½ x2x2 + 2x2 + ½ x 1x2 + ½ x 1 x 2 = 8 m Displacement= = area 1+area 2+ area 3- area 4 = ½ x2x2 + 2x2 + ½ x 1x2 - ½ x 1 x 2 = 6 m Position –time Velocity –time Acceleration –time F 061, Rev 01, dtd10th March 2020 Equation of Motion by Graphical Method Consider an object moving along a straight line with an initial velocity u and uniform acceleration a. suppose, it travels distance s in time t. As shown in figure its velocity-time graph is a straight line. Here OA = ED = u, OC = EB = v, OE = AD = t. 1. Equation for velocity-time relation: Velocity We know that, Acceleration = time 𝐵𝐸 − 𝐸𝐷 𝑣 − 𝑢 𝑎= = 𝑂𝐸 𝑡 or, v - u = at  v = u + at. This proves the first equation of motion. 2. Equation for position-time relation: Distance travelled by an object in time t is s, Or, s = Area of the trapezium OABE = Area of OADE + Area of ADB Or, s = (OA x OE) + (½ x DB x AD) ------------ Eq. (1) Now, DB = BE - DE = v - u = at Putting this value for DB in Eq. (1), we get s = ut + ½at2  this proves the second equation of motion. 3. Equation for position-velocity relation: The distance travelled by an object in time t is (Sum of parallel sides) s = Area of the trapezium OABE = x height 2 ½ (EB + OA) x OE = ½ (EB + ED) x OE F 061, Rev 01, dtd10th March 2020  or, substituting EB, ED and OE with v, u and t we get, s = ½ (v + u) t ---------- Eq. (2) But from the first equation of motion we know that v = u + at (v-u) Or, t= a Substituting t in Eq. 2 with this value we get, 𝑣 2 − 𝑢2 𝑠= 2𝑎 v2 - u2 = 2as This proves the third equation of motion. Free Falling bodies The motion of a body falling under the effect of gravity alone with a constant acceleration is known as free fall. This acceleration is due to the gravitational attraction of the earth and is called acceleration due to gravity. It is represented by g. Its value is the same for all bodies at the same place. The value of g is nearly 9.8 m/s2 at our place. Note:- (i): When a body is projected vertically up , ‘g’ is taken as negative. eqns. Of motion becomes a=-g v = u - gt s = ut - ½ gt2 v2 - u2 = - 2gs (ii). When a body is projected vertically down , ‘g’ is taken as positive.  a= + g v = u + gt s = ut + ½ gt2 v2 - u2 = + 2gs F 061, Rev 01, dtd10th March 2020