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In this chapter, we are going to learn about the motion in a plane. It is also known as motion in two dimensions. To analyse this type of motion we will take a reference of an origin and two coordinates. Chapter 4 of the book, Physics Part-I, consists of 11 sub-sections with each section explaining the various aspects of the chapter in details.
Let us take a look at the subtopics of Class 11 Physics, Chapter 4, Motion in a Plane.
4.1 Introduction
4.2 Scalars and vectors
4.3 Multiplication of vectors by real numbers
4.4 Addition and subtraction of vectors – graphical method
4.5 Resolution of vectors
4.6 Vector addition — analytical method
4.7 Motion in a plane
4.8 Motion in a plane with constant acceleration
4.9 Relative velocity in two dimensions
4.10 Projectile motion
4.11 Uniform circular motion
Section 4.1 – In this section, you will learn that for a physical body, its position, displacement, velocity and acceleration along a straight line (1D) are defined by its magnitude and a + or – sign (for direction). In 2D (a plane) and 3D (space), this would be appropriately defined by vectors. Vectors enable us to define these physical parameters in planes and space. For more information, you may refer to NCERT Solutions For Class 11 Physics Chapter 4 here at UrbanPro.
Section 4.2 -In this section, you will learn that scalars mean simple numbers or magnitude such as temperature (say 300K), mass and speed of an object (5kg, 10m/s). It’s without a direction and can be added, subtracted, multiplied and divided as per normal arithmetic rules. Vectors mean both magnitude and direction, thus following triangle law of addition such as weight, velocity.
Section 4.3 – In this section, you will learn that multiplying a vector with a real number (R) results in a vector of magnitude that is equal to the multiplication of R and vector magnitude. If R is negative, then the direction of vector reverses (otherwise unchanged).
Section 4.4 – In this section, you will learn that the addition of two vectors say A and B, follow triangle law of addition. One vector’s (say B) tail is placed at the head of another (say A), in a continuous fashion. The resultant vector A+B will be the magnitude and direction of the line starting from the tail of A to the head of B, forming a triangle. You will also learn that subtraction is the sum of vectors A and -B (or vice-versa, as required), and follows the same procedure.
Section 4.5 – In this section, you will learn that resolution of vectors is a process of conversion into horizontal (i) and vertical (j) components (2D) and k component(3D), which are vectors themselves. It’s a measure of the vector’s effect on each axis of the plane or space.
Section 4.6 – In this section, you will learn the analytical addition of vectors – say A+B, employs scalar addition of their components. The resultant vector is initially in component form. Vector form can be regained through the triangle addition of three components. For a better understanding of this portion, you can refer to NCERT Solutions for Class 11 Physics Motion In A Plane, here at UrbanPro.
Section 4.7 – In this section, you will learn that the position vector r of a body at P is defined by its components that have been taken from the origin. Displacement from P to P’ is a vector defined as d=r-r’. The average velocity in a corresponding time interval t, from P to P’ is vector defined as v=dt. Instantaneous velocity would be defined by v=t0dt=drdt . Average acceleration here would be a=vt. Instantaneous acceleration is a=dvdt. These equations define the motion in a plane.
Section 4.8 – In this section, you will learn that for constant acceleration, a=v-ut-0. Thus, v=u+at. Integrating this, r=r0+ut+0.5at2, where r0is the initial position.
Assuming the initial position was, v2= u2+2ar.
Section 4.9 – In this section, you will learn that two objects, say A and B, moving with velocities vAand vBrespectively, will have a relative velocity of A wrt B vAB=vA-vB, and vice-versa vBA=vB-vA.Their directions are opposite to one another but their magnitudes are the same.
Section 4.10 – In this section, you will learn that the projectile motion is the motion experienced by an object thrown at an angle away from the earth, like in a goal-kick. The range of this kick is u2sin2 g, and the time of flight is 2usinθg. Neglecting air friction, the projectile path is a perfect parabola.
y=(tan)x – g2(u cosθ)2×2
Section 4.11 – In this section, you will learn that the Uniform Circular Motion occurs when an object moves in a circular path of radius r, along its path of inertia, at a uniform speed v and angular speed =vr. It is then balanced by the centripetal force of acceleration which is defined as =v2r=2r
All the above information is also available to you as an NCERT Solutions for Class 11 Physics Chapter 4 PDF Download, here at UrbanPro.
https://www.urbanpro.com/assets/new-ui/sharing_job.pngPublished on 2020-02-10 09:38:22 by arunima. Last Modified on 2020-02-10 09:38:22
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