Chapter 3: Forces and Motion - Physics Class 11 PDF

Here is the converted text from the images into markdown format: # Unit 3: Forces and Motion This unit covers forces and motion. **LEARNING OUTCOMES** After studying this unit, students will be able to: * Describe vector nature of displacement. * Describe average and instantaneous velocitie...

Here is the converted text from the images into markdown format: # Unit 3: Forces and Motion This unit covers forces and motion. **LEARNING OUTCOMES** After studying this unit, students will be able to: * Describe vector nature of displacement. * Describe average and instantaneous velocities of objects. * Compare average and instantaneous speeds with average and instantaneous velocities. * Interpret displacement-time and velocity-time graphs of objects moving along the same straight line. * Determine the instantaneous velocity of an object moving along the same straight line by measuring the slope of displacement-time graph. * Define average acceleration (as rate of change of velocity $a_{av}=\Delta v/\Delta t$) and instantaneous acceleration (as the limiting value of average acceleration when time interval $\Delta t$ approaches zero). * Distinguish between positive and negative acceleration, uniform and variable acceleration. * Determine the instantaneous acceleration of an object measuring the slope of velocity-time graph. * Manipulate equations of uniformly accelerated motion to solve problems. * Explain that projectile motion is two dimensional motion in a vertical plane. * Communicate the ideas of a projectile in the absence of air resistance that: * (i) Horizontal component ($v_x$) of velocity is constant. * (ii) Acceleration is in the vertical direction and is the same as that of a vertically free falling object. * (iii) The horizontal motion and vertical motion are independent of each other. * Evaluate, using equations of uniformly accelerated motion, that for a given initial velocity of frictionless projectile: 1. How higher does it go? 2. How far would it go along the level land? 3. Where would it be after a given time? 4. How long will it remain in air? * Determine for a projectile launched from ground height: 1. Launch angle that results in the maximum range. 2. Relation between the launch angles that result in the same range. * Describe how air resistance affects both the horizontal component and vertical component of velocity and hence the range of the projectile. * Apply Newton's laws to explain the motion of objects in a variety of contexts. * Define mass (as the property of a body which resists change in motion). * Describe and use the concept of weight as the effect of a gravitational field on a mass. * Describe Newton's second law of motion as rate of change of momentum. * Co-relate Newton's third law of motion and conservation of momentum. * Show awareness that Newton's Laws are not exact but provide a good approximation, unless an object is moving close to the speed of light or is small enough that quantum effects become significant. * Define Impulse (as a product of impulsive force and time). * Describe the effect of an impulsive force on the momentum of an object, and the effect of lengthening the time, stopping, or rebounding from the collision. * Describe that while momentum of a system is always conserved in interaction between bodies some change in K.E. usually takes place. * Solve different problems of elastic and inelastic collisions between two bodies in one dimension by using the law of conservation of momentum. * Describe that momentum is conserved in all situations. * Identify that for a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation. * Differentiate between explosion and collision (objects move apart instead of coming together). Motion is very important as nearly every physical process involve some kind of motion. Mechanics is the branch of science that deals with the study of motion of bodies, which is further sub-divided into kinematics and dynamics. In this chapter, we start our discussion from kinematics, which explains the motion without making any reference to the force (cause of motion). Later, this discussion is extended to dynamics, which deals with the study of motion under the action of force and its various types. ## 3.1 Rest and Motion A body is at rest with respect to an observer if it does not change its position with respect to an observer. A body is in state of motion with respect to an observer if it changes its position with that observer. Rest and motion are relative. Rest and motion depends upon the state of the observer. Two observers can have disagreeing observation about the state of motion or rest. **Point to Ponder** When sitting on a chair, your speed is zero relative to Earth but 30 km/s relative to the Sun. For example a body in moving train is in motion with respect to an observer on ground. Whereas the same object is at rest with respect to another observer in train. Thus the motion and rest are not absolute. This means that specification of the observer is important while inferring about the state of rest or motion of the body. ## 3.2 Displacement Displacement is the shortest directed distance between two positions. Displacement is a vector quantity and has SI unit meter. Displacement is usually denoted by $\Delta x$, $\Delta r$, $\Delta s$, $\Delta l$ or $\Delta d$. The magnitude of the displacement vector is the shortest distance between the initial and final position of the object. However, this does not mean that displacement and distance are the same physical quantities. Figure shows the motion of an object at two different positions $A$ and $B$. These positions are identified by the vectors $r_i$ and $r_f$ which are drown from an arbitrary coordinate origin $O$. The displacement $\Delta r$ of the object is the vector drown from the initial to the final position $B$, such that: $\Delta r = r_f - r_i$ ## 3.3 Velocity Measure of displacement covered ($\Delta s$) with passage of time ($\Delta t$) is called velocity (denoted by $v$). Mathematically $$velocity = \frac{displacement}{elapsed time}$$, or $$v = \frac{s_f - s_i}{t_f - t_i}$$ or $v = \frac{\Delta s}{\Delta t}$ Velocity is a vector quantity having the same direction as displacement vector. The SI unit of velocity is meter per second $(\text{m/s})$. **A. Average Velocity** Average velocity is the net (total) displacement (s) divided by the total time (t). Mathematically: $$<v> = \frac{\text{Total displacement}}{\text{Total time}}$$, or $$<v> = \frac{s}{t}$$ **B. Instantaneous Velocity** Velocity at a particular instant of time is known an instantaneous velocity. The instantaneous velocity is the change in displacement ($\Delta s$) is measured in short interval of time ($\Delta t$) such that the time interval is so small that take the limit to approach zero. Mathematically: $v = Lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$ If a body covers equal displacements in equal interval of time a body is said to be moving with uniform velocity. At uniform velocity the average and instantaneous velocity become equal. In all other cases body moves with nonuniform velocity. Speed is a scalar quantity and is obtained by dividing distance covered by time. As distance remains the same or increase, with time. Therefore, both the average speed and instantaneous speed can not be negative. Velocity on the other hand is a vector quantity can be negative. ## 3.4 Acceleration The measure of change in velocity ($\Delta v$) with the passage of time ($\Delta t$) is called acceleration. Or 'Time rate of change in velocity is called acceleration'. Mathematically: change in velocity elapsed time $a=\frac{\text{change in velocity}}{\text{elapsed time}}$ or $a = \frac{v_f - v_i}{t_f - t_i}$ $a= \frac{\Delta v}{\Delta t}.$ Acceleration is also a vector quantity having the same direction as change in velocity. SI Unit of acceleration is meter per second squared $(m/s^2)$. Acceleration is a measure of how rapidly the velocity is changing. **A. Average Acceleration** Average acceleration is the net (total) velocity (v) divided by the total time. <a> : Total change in velocity / Total time. **B. Instantaneous Acceleration** Acceleration at a particular instant of time is known as instantaneous acceleration. The value of instantaneous acceleration is obtained if the change in velocity ($\Delta v$) is measured in small time Interval ($\Delta t$), such the time is so small that it approaches to zero. Mathematically: $a = Lim_{\Delta t \to 0}\frac{\Delta v}{\Delta t}$ **C. Uniform and Variable Acceleration** A body is said to have uniform acceleration if its velocity changes by equal amount in equal intervals of time, however these interval may be small. In uniform acceleration its average and instantaneous acceleration become equal. A body is said to be moving with variable acceleration if its velocity changes by unequal amount in equal intervals of time, however small these intervals may be. **D. Deceleration/Retardation** When an object is slowing down, we can say that there is Declaration on retardation. We have a deceleration or retardation whenever the magnitude of the velocity is decreasing; thus the velocity and acceleration point in opposite directions when there is deceleration. **E. Positive And Negative Acceleration** Figure 3.2 shows the motion of a care along x-axis. The velocity of a object moving to the right along the positive x-axis is positive; if the object in speeding up, the acceleration is positive as shown in figure 3.2 (a); and when the object in slowing down, the acceleration in negative as shown in figure 3.2 (b). However, the same object moving to the left (Decreasing x), and slowing down, a positive acceleration that points to the right positive as shown in figure 3.2 (c); and when the object is moving to the left ( Decreasing x), and speeding up, has negative acceleration that points to the left as shown in figure 3.2 (d). Thus negative acceleration in not simply retardation or deceleration; the acceleration being positive or negative depends upon the positive and negative direction as defined for displacement. ## 3.5 Graphical Analysis of Motion Graph is an effective way for showing relationship between physical quantities by using coordinate system **A. Displacement-Time Graph** The slope of distance-time curves only gives speed, as the distance always increase the slope can never be negative. The slope of displacement-time graph gives velocity, since displacement can be negative which indicate the reverse motion. The slope of displacement-time graph can also be negative. The displacement time graph is an easy way to understand the velocity of the object as show in the following graph *Table 3.1: Graphical Interpretation* | | x (m) | x(m) | x(m) | x(m) | | :-------------- | :-------------------------------------------------------------------------------------------------------------------------- | :-------------------------------------------------------------------------------------------------------------------------- | :-------------------------------------------------------------------------------------------------------------------------- | :-------------------------------------------------------------------------------------------------------------------------------- | | | *Image of graph with horizontal line representing zero velocity* | Image of linearly increasing line representing velocity with uniform velocity Image of linearly decreasing line representing velocity with uniform velocity | *Image of non-linear line representing variable velocity * | | **Time (s)** | *t (s) under graph with horizontal line* | *t (s) under graph with increasing line* | *t (s) under graph with decreasing line* | *t(s) under non-linear line * | | **Description** | Time is passing and no change in displacement since there is not slope so the velcoity is zero It mens the body is at rest | The displacment is increasing linearly with time. The slope is constant therefore object if moving with uniform velocity | The displacment is decreasing linearly with time. The slope is constant therefore object if moving with uniform velocity | The displacement is changing non linearly with time therefore object is moving with variable velocity and also moved pass it | If we know the position of particle at all times, we can complexly specifies its motin. Consider a car moving back and forth along the straight line as shown in figure 3.3 and we take data on the position of the car every 10 s, as depicted in table 3.2. **Table 3.2: Position of Car Various Times** | Position | Time (s) | Displacement (m) | | :------- | :-------- | :----------------- | | A | 10 | 30 | | B | 20 | 52 | | C | 30 | 38 | | D | 40 | 0 | | E | 50 | -37 | | F | 60 | -53 | Here is a description of an image: A car moves in relation to a speed limit sign labeled A. Point B occurs after the car passes the sign. A police car is located at points C, D, E and F but the first four points it moves to the left and then moves back to the right. The six data points we have recorded are represented by letter A through F. Figure 3.4 shows the graphical representation of one -dimensional motion for the positions x (m) of the car at regular interval (s) is represented by position time graph. *Figure 3.4 shows a line graph of displacement vs time with the car moving forward and returning to it's original starting position.* Scale 1 SD= 5m 1 SD=2s Let us consider a car already in motion as shown in Figure 3.3 which cover distance in equal interval of time. We calculate its velocity between $A$ and $B$. The average velocity during this period is: $$v = \frac{x_B - x_A}{t_A - t_B}$$, or $$v= \frac{52m-30m}{20s-10s}$$, $$v = \frac {22 m}{10 s}$$, therefore $$v= 2.2 m/s$$ On the graph, this is represented by the gradient of stright line joining $A$ and $B$ as shown in figure 3.5 (a). At B, for a moment the car is at rest and after B it has reversed its direction and is heading back towards the reference 'O'. Between B and C the average velocity is $$ v = \frac{x_C-x_B}{t_C-t_B}$$, or $$ v =\frac {38 m - 52 m}{30s - 20s}$$, $$ v = \frac {-14 m}{10 s}$$, therefore $$v= -1.4 m/s$$ Since $x_i$ is greater than $x_f$, it gives a negative quantity indicating reverse direction calculating the average velocity of the can over relatively long tome Intervals will not gives us the complete the motion as shown in figure 3.5 ( b), since the can was not moving all the way throught with the speed.To describe the motion exactly,we need to know the car velocity at every instant of time. *Figure 3.5 is composed of 3 curved lines and 3 graphs showing data values representing a car traveling across specific points.* The displacement time graph through car through points $A,B,C,D,E$ and $F$ is showin in the figure. The average velocity in shroter interval of time is dissimilar both in magnitude and direction at different points.(b) the averange velocity over lnger interval of time reamain same at all points.(c) The instantaneous valocity tangent to the cruved path. The Instantaneous Velocity Is Obtained By Making The Time Intervals Smaller (Mathematically We Say That The Limit In Which Time Approcah To Zero ) in Displacement-time Graph. This Gives Us a Series Of Shorter Straigt-line Segments which have the same direct as the tangent to the curved, as show in figure 3.5 (c). **B. Velocity –Time Graph** The graph ploted between velotity (v )and time (t) is Velocity .Time Garph The Slope of the line one a velocity -Time Graph Reveals Useful information about the acceleartion of the Object: slope = $\frac{\Delta V}{\Delta t}= \frac {V_f - V_i}{ t_f - t_i}$ Graph in figure 3.6 show a details analysis of an Object In Mortion From point A TO B the Objects speed increase over the timethe line on the graph plotting tis motion slope up the accleaertin can the Otained By Calculating Slope as. *Figure 3.6 is graph representing car speed vs time and the scale used to measuse these attributes.* as $$a = \frac{V_f - V_i}{ t_f - t_i} \quad \text{or} \quad a = \frac {20 m/s - 0 \frac{m}{s}} {10 s -0 s}$$ or a = 20m/s / 10s = 2 m/s^2 \quad \text{therefore} \quad * a=2 m/s^2* From point $B$ to $C$ the object has maintained its speed of 20$\, m/s$ and there is zero action. From point $C$ to $D$ its velocity decreases over time, represented by a graph segment sloping down the above method can be used to calculate the negative deceleration of -2m/s2. The segment of of the graph points $E$ to $F$, the object accelerates the oppisite direction. The acceleration can be calculated by measuing the lope as $$a = \frac{V_f - V_i}{ t_f - t_i} \quad \text{or} \quad a = \frac {-6 m/s -0 \frac{m}{s}} {50 s -40 s}$$ It is equal to : $$a = -\frac {6 m/s -0 \frac{m}{s}} {10 s}$$ *therefore a = -0.6 $a= \frac{m}{s^2}$* This show that even object has gained but still acceleration is negative, The segment of the graph From points $F$ to $G$, Represents The Speed is Steading in Opposite Direction **Example 3.1** The velcoty time graph Show the motion of bicyclist In at Straight Lion. (a)From The Slope of The Grah Calculating The Acceleration The bicyclitist and The Segment A andB, Bande C , C and Dane. The Velcoty- Time graph shows the motion of bicycle. *Scale* 1 SD- Im/s ISD = 15s. (A) B ,B and C Candy C , D and Dare. Solution (a) The Acceleration from points A tob C can be Calculate Buy Measuring The Slope has:- $a = \frac{V_f - V_i}{ t_f - t_i}$ or $\frac{ (10 \frac{m}{s} -0 \frac{m}{s} )} {4s-05}$ $$ a = {10 \frac{m}}/{4s} =2.5 m/s $$ The acceleration from Points B B tot C calculating buy measuring the slope has:- $a = \frac{V_f - V_i}{ t_f - t_i}$ or $\frac{ {15 \frac{m}{s^- 10 \frac{m}{s}}} } {8s-4s}$ $a= 2.5 m/s$$ Or $a= (\frac{5 m}{s} ){4s}- {1.2 5 m/s Ans we r The the Acceleration from Point C tob D can be calculated By :- $a = \frac{V_f - V_i}{ t_f - t_i}$ or $\frac{ {0 \frac{m}{s} - 15 \frac{m}{s}}}{12s-as}$ Or $$ a = {15 \frac{m}{s^}}{4s}= -3.75 m/s $$ Therefore(a) The Acceleration From Point A tob B can be measuing. The Slope HAS. $ a = {4\frac{m}{s}} {4s-0s}$ Or * Therefore A: 2 $2m/s $ ANswer. a The Accelation form point B tob C Buy measuyng The SLOE HAS. $ = {1 5 \farc{M}}{S)-} 7 = 14/{ S\sqrt{5}}/{ AS}-= {075) 2}7{M}$ANS WER 14: The The Acceteration from point C Tob O can be calculated By "Them $ = {C\farc(M}}{5)24) = -8 (ANS WER = ($ * a12-37(220, $=-417 (a=- Q0e00 S07 $==\vert Similarly the acceleration from point D to E can be calculated as $a={-5\ f (5)/(-0 (12/125)=1) and & ==0 7:The The acceberatien from point C Tob $ *Figure shows 2 graphs plotting results obtained using calculations to measure acceleration .* An swerve = 1125/(The B the average acceleration dan The calculated 11252 "the slop 17 tom poine a tor Eas =-1.587 +31" "A(x=(T3-100/7)x(4x)=0-3 The velocity time graph shows the montion to reading the seal cartially Calculate The are acceleiation of the cat between Dand A and D and E rom T slope of the grapphe The Complete journey 8.0m/.0m/: - toom/, 1 67m/ and $50+614% The Acceleration from 14. D "4--41144= The three famous equation of motion are only apply cable action A frist CQUATION of Motion v-ve af. v =ve tat Be SECOND "Sorte 482 = 166 "S Vate # "03 1148 "4441 *7 25° 2 "7 ""774903. The albsence C alr resistin Earth moves earth. with a corm alceletia tion this This as acceleretou valwe near. 71 4th earth & ## Unit 3 Forces and Motion **3.6 EQUATIONS TO UNIFORMLY ACCELERATED MOTION** Equations are A. FIST EQUATION MOTION $\quad\nu = - _l\quad$(3.8) B.Second EQUATION MOTION $\quad S = _1t + \hat{w}*(3,9)$ (871, 1490" "C.THTRD EQUATION MOTION * as 2 7a $2a^2$ + 10910 to be apple of the Objects # 79 & The Eardh 1414 4 the abjeds Thes A (Jaint Fighter.9) 70m/ # "54252" * 2. " "4290 to be = The After The 1, 404 Toa 7.148+Calculate and The takeall "41485778 "4648419 *7115 774041 "27 4443 "10473 71 4173 = 771 541 381 To # Unit Forces" 4:34 *1008" 41 + 101= "3.15 ""71T2 The V.for phaset 4-27 7 4+99-44th=5+ 843=5 "A. FIRST LAY (409 A Objects Remains at Pest or in a Uniformly "74, **3.7 NEWTOVATION # 49 ## Unit B * [ A = to (The *81 "3.1 111 't-f 0 The # "7 -241 714 *D 40- "7 **3.8 LINEAR MOUT** "7368 (2) 2 A-Newtons Second and Lineal MOMENTUM Bay Newton see ond Law =ma (1 Dy Delination af to. \(2) Putting *N° -P) [ *95874 **1** # **98"1** "7#71 # "1#11 # A *47 the The 201 *D4 : # "7th" *43 3&A' *606 4 2. [A] *117 "4t [4]= a 631-42)5 (a ""Th"A. 43 9-041-0"71" # A41a43 *100 ##Unit3 FORCES AND MOTION "1 = 20U To the for "To compare These $20 + + 70 \times [ \to + 1a4 04-974 77 += #414857 +he e A the ""477" ""18+7#19""B++"" # "75T2" # A"" 75T2A *26 * 2 *3744 *191 71#04-83 #47-91 & # "& # 404871-43. 3. [ 4/3/ ""The "3648" "27 A B = "7 *Unit3* ==&110. FORCES AND MOTION = *KEY Displacement Velocity Acceleration * = "47380377=727 To The a "1 a- -The 2 There at 7447]7270 "74 *541 "471A T # TO" "To" * 7 "542" "*2 The The 4 The # "TO 607 The The ""3744 **18*1** *4947 ""THE "540 "407#24 * 63"" "37" the #5 ""941 1 The (247 T [ ""14 74 ""247#24 #4144 145 *The A7-6407 to The """1125 *4347 "7403774## "2 3+""23" *+43704-37 1 1# "24-04 **86' "#13##66" *""&#110901 # "746" ""77 114=755.59112145 10 ""[A4" *195 8 88 7 TO 12 # "7146 #A 49651 B"7 = TO *27 the the TO # "B447" The A 1A399 to a *7 A-7 (4)4 a # "7541 A To" * "3 7 #4 ""45-5 30% (470-7 [A **"""* # ""The "71+ T [TO 0 A "A the to"8 # 00301 *Units* *FORCES AND MOTION* &""781: What is tinear momentum? "the" a To of the 3 &state and explain Lay of Cansavatian of Linear Momentum 04 an 8 "" "the elastic of The the final velocities ot The 1277 2 & 8-951"4) = 85 the the The # "97"8/ """ # a #### # 499" "The I # 9 1446671 &1441 *309 *13399: &TO "" *#51##: 41.3 (194+84 7-0 "Tho 10470 """#""a""