{"id":46757,"date":"2024-02-17T05:03:53","date_gmt":"2024-02-16T23:33:53","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=46757"},"modified":"2024-02-17T14:42:01","modified_gmt":"2024-02-17T09:12:01","slug":"plus-one-physics-notes-chapter-4","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-physics-notes-chapter-4\/","title":{"rendered":"Plus One Physics Notes Chapter 4 Motion in a Plane"},"content":{"rendered":"
Summary<\/span> Scalars And Vectors<\/span><\/p>\n a. Scalars: b. Vectors: (ii) Vector: 1. Position and Displacement Vectors: The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\\(\\overrightarrow{\\mathrm{OP}}\\)|.<\/p>\n Displacement vector: 2. Equality of vectors: Question 1. Multiplication Of Vectors By Real Numbers<\/span> Addition And Subtraction Of Vectors – Graphical Method<\/span> 1. When two vectors are acting in the same direction: 2. When two vectors act in opposite direction: 3. When two vectors are inclined to each other: (i) Triangle method: Explanation<\/u> <\/p>\n (ii) Parallelogram law of vector addition: 4. Substraction of vectors: Question 2. Resolution Of Vectors Unit Vectors<\/span> Consider a vector \\(\\overrightarrow{\\mathrm{A}}\\) that lies in x-y plane as shown in figure. To resolve \\(\\overrightarrow{\\mathrm{A}}\\), draw lines from the head of \\(\\overrightarrow{\\mathrm{A}}\\) perpendicularto the coordinate axes as shown below. Question 3. Vector Addition – Analytical Method<\/span> Question 4. Motion In A Plane<\/span><\/p>\n 1. Position vector and displacement vector Position vector: Acceleration: <\/p>\n Motion In A Plane With Constant Acceleration<\/span> Relative Velocity In Two Dimensions<\/span> Projectile Motion<\/span> 1. Time of flight: 2. Vertical height: 3. Horizontal Range: Question 5. Uniform Circular Motion<\/span> a. Speed and angular speed in uniform circular motion: b. Acceleration in uniform circular motion: The direction of this acceleration should be in the direction of \\(\\overrightarrow{\\Delta V}\\). Fig (b) shows that \\(\\overrightarrow{\\Delta V}\\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration. Kerala Plus One Physics Notes Chapter 4 Motion in a Plane Summary Introduction In this chapter, we will study, about vector, its \u2019 addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail. Scalars And Vectors a. Scalars: A quantity which has […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\n
\nIntroduction<\/span>
\nIn this chapter, we will study, about vector, its \u2019 addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.<\/p>\n
\nA quantity which has only magnitude and no direction is called a scalar quantity.
\nEg: length; volume, mass, time, work etc.<\/p>\n
\n(i) The need for vectors:
\nIn one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.<\/p>\n
\nA physical quantity which has both magnitude and direction is called a vector quantity.
\nEg: Displacement, Velocity, Acceleration, Force, momentum.<\/p>\n
\nPosition vector:
\nConsider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.
\n
\nA vector representing the position of an object P with respect to an origin O is called position vector \\(\\overrightarrow{\\mathrm{OP}}\\) of the object. This position vector may be represented
\nby an arrow with tail at O and head at P.<\/p>\n
\n
\nConsider the motion of an object in a plane. Let P be the position of a moving object at a time t and p1<\/sup> that at a later time t1<\/sup>. \\(\\overrightarrow{\\mathrm{OP}}\\) and \\(\\overrightarrow{\\mathrm{OP}^{1}}\\) are the position vectors at time t and t1<\/sup> respectively. So the vector \\(\\overrightarrow{\\mathrm{PP}^{1}}\\) is called displacement vector corresponding to the motion in the time interval (t – t1<\/sup>).<\/p>\n
\nTwo vectors are said to be equal if they have the same magnitude and direction.
\n
\nThe above figure shows two vectors \\(\\vec{A}\\) and \\(\\vec{B}\\) having the same magnitude and direction.
\n\u2234 \\(\\vec{A}\\) = \\(\\vec{B}\\).<\/p>\n
\nObserve the following figures (a) and (b) and find which pair does represents equal vectors?
\n
\nAnswer:
\nFigure a represent that A and B are equal vectors. Two vectors A1<\/sup> and B1<\/sup> are unequal, because they were in different directions.<\/p>\n
\nMultiplying a vector \\(\\vec{A}\\) with a positive number I gives a vector whose magnitude is changed by the factor \u03bb.
\n
\nThe direction \u03bb\\(\\vec{A}\\) is the same as that of \\(\\vec{A}\\).
\nExamples:
\n
\nA vector \\(\\vec{A}\\) and the resultant vector after multiplying \\(\\vec{A}\\) by a positive number 2.
\n
\nA vector A and resultant vector after multiplying it by a negative number-1 and -1.5.<\/p>\n
\nVectors representing physical quantities of the same dimensions<\/u> can be added or subtracted. The sum of two or more vectors is known as their resultant.<\/p>\n
\n<\/p>\n
\nIn this case, the angle between the vectors is 180\u00b0.
\n
\nThe resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.<\/p>\n
\nThe sum of two vectors inclined at an angle q can be obtained either by<\/p>\n\n
\nThis law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.<\/p>\n
\nConsider two vectors \\(\\vec{A}\\) and \\(\\vec{B}\\) as shown in figure.
\n<\/p>\n
\nThis law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
\nExplanation<\/u>
\nConsider two vectors \\(\\vec{A}\\) and \\(\\vec{B}\\) as shown in figure.
\n
\nTo find \\(\\vec{A}\\) + \\(\\vec{B}\\), we bring theirtails to a common origin Q as shown below.
\n
\n
\nThe diagonal of parallelogram OQSP, gives the resultantof (\\(\\vec{R}\\) = \\(\\vec{A}\\) + \\(\\vec{B}\\)) of two vectors \\(\\vec{A}\\) and \\(\\vec{B}\\).
\nNote: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.<\/p>\n
\n
\nTo substract \\(\\vec{B}\\) from \\(\\vec{A}\\), reverse the direction of \\(\\vec{B}\\).
\n
\nThen add –\\(\\vec{B}\\) with \\(\\vec{A}\\) using parallelogram law or tri\u00acangle law.
\n
\nThe resultant of \\(\\vec{A}\\) and \\(\\vec{B}\\) is given by \\(\\vec{R}\\).
\nNull vector or zero vector:
\nA vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \\(\\vec{O}\\). Since the magnitude is zero, we don\u2019t have to specify its direction.
\nProperties of null vector:
\n<\/p>\n
\nExplain a zero vector using an example.
\nAnswer:
\nSuppose that an object which is at P at time t, moves to p1<\/sup> and then comes back to P. In this case displacement is a null vector.<\/p>\n
\nA vector divided by its magnitude is called a unit vector along the direction of that vector. A unit vector in the direction of \\(\\vec{A}\\) is written as \\(\\hat{A}\\).
\n
\nOrthogonal unit vectors:
\n
\nIn the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) respectively and are known as orthogonal unit vectors.
\nFor unit vectors
\n
\nResolution of vector into rectangular components:
\nThe components of a vector in two mutually perpendicular directions are called its rectangular components.
\nExplanation<\/u>
\n<\/p>\n
\n
\nThe quantities Ax<\/sub> and Ay<\/sub> are called x and y components of the vector \\(\\overrightarrow{\\mathrm{A}}\\). Hence the vector \\(\\overrightarrow{\\mathrm{A}}\\) can be written in terms of rectangular components as
\n
\nMagnitude of \\(\\overrightarrow{\\mathrm{A}}\\):
\n
\nFrom the figure, the magnitude of \\(\\overrightarrow{\\mathrm{A}}\\) can be written as,
\n<\/p>\n
\nA vector \\(\\overrightarrow{\\mathrm{A}}\\) in xyz plane is given below. Ax<\/sub>, Ay<\/sub> and Az<\/sub> are the perpendicular components in x,y and z directions respectively.<\/p>\n\n
\nAnswer:
\n
\nThe magnitude of vector \\(\\overrightarrow{\\mathrm{A}}\\) is
\n<\/p>\n
\nThe graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use the analytical method to add vectors.
\nExplanation
\n
\nThe vectors obey commutative and associative laws. Hence
\n<\/p>\n
\nFind the magnitude and direction of the resultant of two vectors \\(\\overrightarrow{\\mathrm{A}}\\) and \\(\\overrightarrow{\\mathrm{A}}\\) in terms of their magnitudes and angle between them.
\nAnswer:
\n
\nConsider two vectors \\(\\vec{A}(=\\overrightarrow{O P}) \\text { and } \\vec{B}(=\\overrightarrow{O Q})\\) making an angle q. Using the parallelogram method of
\nvectors, the resultant vector \\(\\overrightarrow{\\mathrm{R}}\\) can be written as,
\n
\nSN is normal to OP and PM is normal to OS. From the geometry of the figure
\nOS2<\/sup> = ON2<\/sup> + SN2<\/sup>
\nbut ON = OP + PN
\nie. OS2<\/sup> = (OP+PN)2<\/sup> + SN2<\/sup> ______(1)
\nFrom the triangle SPN, we get
\nPN = Bcosq and SN = Bsinq
\nSubstituting these values in eq.(1), we get
\nOS2<\/sup> = (OP + Bcosq)2<\/sup> + (Bsinq)2<\/sup>
\nBut OS = R and OP = A
\nR2<\/sup> = (A + Bcosq)2<\/sup> + B2<\/sup>sin2<\/sup>q
\n= A2<\/sup> + 2ABcosq + B2<\/sup>cos2<\/sup>q + B2<\/sup>sin2<\/sup>q
\nR2<\/sup> = A2<\/sup> + 2 ABcosq + B2<\/sup>
\n
\nThe resultant vector \\(\\overrightarrow{\\mathrm{R}}\\) make an angle a with \\overrightarrow{\\mathrm{A}}. From the right angled triangle OSN,
\n
\nBut SN = Bsinq PN = Bcosq
\n<\/p>\n
\n
\nConsider a small body located at P with reference to the origin O. The position vector of the point \u2018P\u2019
\n
\nDisplacement vector
\n<\/p>\n
\nwhere Dx = x1<\/sup> – x1<\/sup>, Dy = y1<\/sup> – y
\nVelocity:
\nIf Dt is the time taken to reach from P to P1<\/sup>
\nThe average velocity, \\(\\overrightarrow{\\mathrm{v}}_{\\mathrm{av}}=\\frac{\\overrightarrow{\\Delta r}}{\\Delta \\mathrm{t}}\\) ____(3)
\nSubstitute eq.(2) in eq.(3), we get
\n
\nThe direction of average velocity is the same as that of \\(\\overrightarrow{\\Delta r}\\).
\nThe instantaneous velocity can be written as
\n
\n<\/p>\n
\nIf the velocity of an object changes from \\(\\overrightarrow{\\mathrm{v}} \\text { to } \\overrightarrow{\\mathrm{v}^{1}}\\) in time Dt, then its average acceleration is given by
\n
\nInstantaneous acceleration:
\nThe acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
\nie. Instantaneous acceleration
\n<\/p>\n
\nConsider an object moving in XY plane with a constant acceleration \u2018a\u2019. Let \\(\\vec{u}\\) be the initial velocity at t=0 and \\(\\vec{v}\\) be the final velocity at time t.
\nThen by definition acceleration
\n
\nIn terms of components
\nvx<\/sub> = ux<\/sub> + ax<\/sub>t
\nvy<\/sub> = uy<\/sub> + ay<\/sub>t
\nDisplacement in a plane<\/u>
\nIf \\(\\overrightarrow{\\mathrm{r}_{0}}\\) and \\(\\vec{r}\\) be position vectors of particle at t = 0 and time t respectively, then
\ndisplacement = \\(\\vec{r}-\\vec{r}_{0}\\) _______(1)
\nFor uniformly accelerated motion, displacement,
\n
\nIn terms of components
\nx = x0<\/sub> + ux<\/sub>t + 1\/2 ax<\/sub>t2<\/sup>
\ny = y0<\/sub> + uy<\/sub>t + 1\/2 ay<\/sub>t2<\/sup>
\nThe eq.(2) shows that, the above motion in xy plane can be treated as two separate one-dimensional motions along two perpendicular directions.<\/p>\n
\nConsider two bodies A and B moving along a plane with velocities \\(\\overrightarrow{\\mathrm{V}}_{\\mathrm{A}}\\) and \\(\\overrightarrow{\\mathrm{V}}_{\\mathrm{B}}\\). Then the velocity of A relative to that of B is,
\n
\nSimilarly velocity of B relative to that of A
\n<\/p>\n
\nProjectile:
\nA body is projected into air and is allowed to move under the influence of gravity is called a projectile.
\n
\nConsider a body which is projected into air with a velocity u at an angle q. The initial velocity \u2018u\u2019 can be divided into two components ucosq along horizontal direction and using along vertical direction.<\/p>\n
\nThe time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
\ns = ut + 1\/2 at2<\/sup>
\nTaking vertical displacement s = 0, a = -g and initial vertical velocity = u sin q, we get
\n0 = usinqt – 1\/2gt2<\/sup>
\n1\/2 gt2<\/sup> = usinqt
\n<\/p>\n
\nVertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v2<\/sup> = u2<\/sup> + 2as
\nWhen we substitute v=0, a = -g, s = H and u = usinq, we get
\n0 = (usinq)2<\/sup> + 2 – g \u00d7 H
\n2gH = u2<\/sup>sin2<\/sup>q
\n<\/p>\n
\nIf we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be constant.
\nHence the horizontal distance (R) can be found as
\nR = horizontal velocity \u00d7 time of flight
\n
\nThe eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q0<\/sub> = 45\u00b0.
\nThe maximum horizontal range
\n
\nEquation for path of projectile<\/u><\/p>\n
\nWhat is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
\nAnswer:
\nS = ut+ 1\/2at2<\/sup>
\ny = usinqt – 1\/2gt2<\/sup>
\nBut we know horizontal displacement, x = ucosq \u00d7 t
\n
\nIn this equation g, q and u are constants. Hence eq.(4) can be written in the form
\ny = ax + bx2<\/sup>
\nwhere a and b are constants. This is the equation of the parabola, ie. the path of the projectile is a parabola.<\/p>\n
\nThe motion of an object along the circumference of a circle is called circular motion.
\nUniform circular motion:
\nWhen an object follows a circular path at a constant speed, the motion is called uniform circular motion.
\nPeriod:
\nThe time taken by the object to complete one full revolution is called the period.
\nFrequency:
\nThe number of revolutions completed per second is called the frequency u of the circular motion.
\nIf the period of a circular motion isT, its frequency
\n
\nAngular Displacement (Dq):
\nThe angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
\nAngular velocity:
\nThe rate of change of angular displacement is called the angular velocity.
\n
\nIf T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
\nTherefore in one second it sweeps out an angle \\(\\frac{2 \\pi}{T}\\).
\n\u2234 Angular velocity of the object
\n
\nExpression for velocity and acceleration in uniform circular motion:
\n
\nThe direction of velocity is in the direction of tangent at that point. The change in velocity vectors \\((\\overrightarrow{\\Delta v})\\) is obtained by triangle law of vector as shown in figure (b).<\/p>\n
\nLet the Dq be the angle constructed by the body during the time interval \u2206t. The angular velocity can be written as
\n
\nIf the distance travelled by the object during the time Dt is Dr (ie. PP1<\/sup> = Ds) then speed
\n
\nBut Ds = RDq
\nwhere R = \\(|\\vec{r}|=|\\overrightarrow{r^{\\prime}}|\\)
\nSubstituting Dr = RDq in eq.(1)
\nwe get
\n<\/p>\n
\n
\n<\/p>\n
\n
\nThe force which produces this centripetal acceleration is called centripetal force.
\nCentripetal force can be written as
\nF = mac<\/sub>
\n
\nBut \u03c9 = \\(\\frac{V}{R}\\). Hence we get K
\n<\/p>\nPlus One Physics Notes<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"