    ## Physics

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01_Unit_and_Dimension.pdf 02_Motion_in_1-D (1).pdf 04_Motion_in_2-D1.pdf 05_Laws_of_Motion_and_Frictions.pdf 06_Work_power_and_Energy.pdf 09_Magnetism.pdf 10__Electromagnetic_Induction.pdf

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MOTION IN 2-DIMENSION (Projectile & Circular motion And Vectors) INTRODUCTION The motion of an object is called two dimensional, if two of the three co-ordinates required to specify the position of the object in space, change w.r.t time. In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc. Two special cases of motion in Two Dimension are 1. Projectile motion 2. Circular motion Projectile Any object that is given an initial velocity obliquely, and that subsequently follows a path determined by the gravitational force (and no other force) acting on it, is called a Projectile. Examples of projectile motion : A cricket ball hit by the batsman for a six A bullet fired from a gun. A packet dropped from a plane; but the motion of the aeroplane itself is not projectile motion because there are forces other than gravity acting on it due to the thrust of its engine. Assumptions of Projectile Motion We shall consider only trajectories that are of sufficiently short range so that the gravitational force can be considered constant in both magnitude and direction. All effects of air resistance will be ignored. Earth is assumed to be flat. Projectile Motion : The motion of projectile is known as projectile motion. It is an example of two dimensional motion with constant acceleration. Projectile motion is considered as combination of two simultaneous motions in mutually perpendicular directions which are completely independent from each other i.e. horizontal motion and vertical v cos 0 = vo cos00 motion. u o u cos x ax = 0 u cos 0 Horizontal motion u sin 0 Vertical motion Parabolic path = vertical motion + horizontal motion. Types of Projectile Motion (1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion on an inclined plane Oblique Projectile Motion Oblique Projection on a Horizontal Surface (a) (b) Change in position vector r 2 gt sin 00 If vot (P = tan 2vo 2vo sinoo —gt 2vo cos 00 00 x Average Velocity
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2 gt sin 00 av 2vo 1 2vo sinoo —gt = tan 2vo cos 00 Instantaneous Velocity (c) 2 0 = tan i {tanoo —(gt / vo)sec00} (d) (g) (h) (k) v 0 0 tan tan vo sinoo —gt v 0 cos 00 v: sin2 00 —2gt tan tan = tan 1 v: sin2 00 —2gy vo cos00 Equation of Trajectory - sec Y — (x tan00) Time of Flight 2vo sinoo Maximum height sin2 00 vo cos00 00 H max Range 2g v: sin 200 R is maximum when sin200 is maximum 00 = 450. Angle of Projection of Given Ratio of Range and Maximum Height Attained tanoo = 4 / 11 41—1 760 (when H = R orn = 1) 00 tan Projectile Passing Through Two Different Points of same height at Time ti and t2 1 = —gtltl Y 2 Speed and Angle of Projection so that Projectile Passes Through Two Given Points Angle of projection (0) = tan The speed of projection 0 ( — +tan2 0 2 -XlY2) where tan 00 YlX2 - Y2Xl 2 Minimum Velocity of Projection Required to Pass Through a Given Point
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(m) (n) min Critical angle of projection to pass through a given point. 2 (Oc) = tan x Position, Time and Speed at Any Angular Elevation 9 vo cos00 cos0 vo sin(00 —0) g cos0 v: sin(00 0)cos00 g cos0 Radius of Curvature at any Point on the Path of a Projectile 0 2 gcos0 tan v: sin2 00 —2gy vo cos00 2. HORIZONTAL PROJECTION FROM A GIVEN HEIGHT (a) (b) (c) (d) Displacement 2 2y 2 m +1 and - tan 2y / g Velocity vo x x - tan —l tan 1 v 2 v voi-gtj +2gy and (P Range 2H Equation of trajectory 2v20 Example 1 : A bomb is fired from a cannon with a velocity of 1000 m/s making an angle of 300 with the horizontal (g = 9.8 m/s2). (i) What is the time taken by the bomb to reach the highest point? (ii) what is the total time of its motion ?
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Solution : (iii) With what speed the bomb will hit the ground and what will be its direction of motion while hitting? (iv) What is the maximum height attained by the bomb? (v) At what distance from the cannon the bomb will hit the ground? Let u be the initial velocity of the bomb and its angle of projection. The time taken by the bomb to reach the highest point is given by 1 1000 usin0 1000 9.8 2 9.8 The total time of its motion is, T = 2t = 2 x 51 = 102 s. (iii) The bomb will hit the ground with the same speed with which it was fired. Hence its speed of hitting = 1000 m/s. Also, the angle of hitting with respect to the horizontal is 300. (iv) The maximum height attained by the bomb is u2 sin2 0 2g (1 000)2 sin2 300 (v) Horizontal range is (1000)2 (0.5)2 =1.27 x 104 m 3. (1 000)2 x sin 600 u2 sin 20 9.8 PROJECTILE ON AN INCLINED PLANE (a) (b) (c) Time of flight 2vo sin(0 — B) g cosß Range — B) — sin B} gcos2 ß = 8.83 x 104 m gsin vo g cos 0 x Range is maximum when sin(20 —B) is maximum, that is equal to 1. max R' max 2 g(l + sinß) 2 g(l —sinß) (up the plane) (down the plane) Condition for retracing the path of a projectile on an inclined plane Example 2 : Solution : vo cosu , where u = gsinß 2tanu = cot ß cot 0 = ß+tan 2 0 = tan I(3cotß) From a point high enough on an inclined plane, whose rise is 7 in 25, a shot is fired with a velocity of 19.6m/s at an angle of 300 with the horizontal (a) up the plane, (b) down the plane. Find the range in each case. (g = 10 m/s2). Let be the inclination of the plane. Then 7 24 sinß - and cos 13 - 25 25
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21.12 cos(Yv sin(u — B) (a) Let RI be the range up the plane. Then RI = Here u = 19.6 m/s and = 300 Putting the values we get RI = 17.15 m (b) For motion down the plane, u cos (30 + P) and ax = g sin uy = u sin (30 + P) and ay = -g cos 1 — ayt2 Then using sy and uyt + noting gcos2 ß that u 300 x 2 sy = 0 we get time of flight T as 2usin(300 + p) gcosß Let R2 be the range down the plane. = uxt + —axt2 we get 2112 cos300 sin(300 + p) = 52.16m gcos2 ß CIRCULAR MOTION Then using When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant, then its motion is known as circular motion with respect to that fixed (or moving) point. The fixed point is called centre, and the distance of particle from it is called radius. Uniform Circular Motion 4 Ar (a) (b) (c) (d) S = 2r sin— 2 Average velocity IAfl 2rsin(0 / 2) av At x Angular and linear speeds v=r0/t and v = ro Change in velocity 0 CD 2v2 (1 — cos0) = 2v sin— Av 2 Centripetal acceleration 2v(0 /2) vo t t 2 Putting 0/ t —CD —— we obtain ar Example 3 : A body of mass 10 kg revolves in a circle of diameter 0.4 m, making 1000 revolutions per minute. Calculate its linear velocity and centripetal acceleration. Solution : If the body makes n revolution per second, then its angular velocity is 1000 = 1 0071/3 rad/s (D = 27tn = 2 It x 60 If the radius of the circle is r, then the linear velocity of the body is v = = 0.20 x (100W3) = 20n/3 m/s The centripetal acceleration is = = 0.20 x (100W3)2 2000TT2 2 m/s 9
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Non-uniform circular motion Radial acceleration, aR = Tangential acceleration, aT Resultant acceleration, a = roc an Non-Uniform Circular Motion with constant Angular Acceleration CD = CDO + Ott O = Ct)0t 4-—01t2 2 2 — 0302 -k 2010 . CD Analogy between translatory and angular motions in terns of equations CD = CDO + Olt S = ut + —at2 2 v2 —u2 = 2as CD CDC) t -k — Olt2 2 2 —CD{ = 2010 Example 4 : Solution : A car is moving with a speed of 30 m/s on a circular track of radius 500 m. Its speed is increasing at the rate of 2 m/s2. Determine the magnitude of its acceleration. The speed of the car moving on a circular track is increasing. Therefore, besides the centripetal acceleration ac, the car has a tangential acceleration at also. ac and at are mutually at right angles, 2 = 1.8 m/s2 Here ac = and 500 at = 2 m/s2 (given) . resultant acceleration a - VECTORS Introduction (1.8)2 = 2.7 m/s2 The physical quantities specified completely by their magnitude as well as direction are called vector quantities. The magnitude and direction alone cannot decide whether a physical quantity is a vector. In addition to the above characteristics, a physical quantity, which is a vector, should follow laws of vector addition. For example, electric current has magnitude as well as direction, but does not follow laws of vector addition. Hence, it is not a vector. A vector is represented by putting an arrow over it. The length of the line drawn in a convenient scale represents the magnitude of the vector. The direction of the vector quantity is depicted by placing an arrow at the end of the line. For example : If 1 cm length is equal to 20 km/hr, then vector km/hr due east. The point A is called initial point or tail and terminal point or head. SCALARS AND VECTORS Scalars AB represents 60 w point B is called 3 cm B A s Scalars are physical quantities which are completely described by their magnitude only. For example: mass, length, time, temperature energy etc.
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Vectors Vectors are those physical quantities having both magnitude as well as direction and they obeys vector algebra (eg. parallelogram law or triangle law of vector addition). For example: displacement, velocity, acceleration, force, momentum, impulse, electric field intensity etc. Types of Vector (1) Equal vectors : Two vectors A and B are said to be equal when they have equal magnitudes and same direction. (2) Parallel vector : Two vectors A and B are said to be parallel when (i) Both have same direction. (ii) One vector is scalar (positive) non-zero multiple of another vector. (3) Anti-parallel vectors : Two vectors A and B are said to be anti-parallel when (i) Both have opposite direction. (ii) One vector is scalar non-zero negative multiple of another vector. (4) Collinear vectors : When the vectors under consideration can share the same support or have a common support then the considered vectors are collinear. (5) Zero vector (0) : A vector having zero magnitude and arbitrary direction (not known to us) is a zero vector. (6) Unit vector : A vector divided by its magnitude is a unit vector. Unit vector for A is A (read as A cap or A hat). Since, A Thus, we can say that unit vector gives us the direction. (7) Orthogonal unit vectors i, j and k are called orthogonal unit vectors. These vectors must form a Right Handed Triad (It is a coordinate system such that when we Curl the fingers of right hand from x to y then we must get the direction of z along thumb). The x x x Y Y z z z x=xi, zk (8) Polar vectors : These have starting point or point of application . Example displacement and force etc. (9) Axial Vectors : These represent rotational effects and are always along the axis of rotation in accordance with right hand screw rule. Angular velocity, torque and angular momentum, etc., are example of physical quantities of this type. Axial veeor Anticbcg wtse rotat Axis 0t Axjs Chck '*tse Axial vector (10) Coplanar vector : Three (or more) vectors are called coplanar vector if they lie in the same plane. Two (free) vectors are always coplanar.
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Solution : Two forces are drawn from a common origin O, making an angle of 600. OA and OC the forces 60N and 80N represent respectively. The diagonal 0B represents the resultant R. . R2 = 602 + 802 + 2.60.80 cos 600 = 3600 + 6400 + 4800 = 14800 . R = 121.7N = 15 N o c Angle is given, tan+ Which gives, = 34.70 Subtraction of vectors : 80sin600 60 +80 COS 600 Example 3 : If the sum of two unit vectors A and the vector A— B. Solution: Given that IA +BI = 1 Hence the angle between A and B is 1200 2 2 2 Now Ipsl =IAI +1-B + 21AII-BIcos1200 1 3 2 Relative velocity (i) VAB VA-VB (ii) I VAB + VS —2VAVB COS0 B is also equal to a unit vector, find the magnitude of A 1200 -B (iii) If relative velocity makes an angle u with VA then, VB sino tanu VA - VB COS0 Resolution of vectors The component of Fin a direction making an angle 0 is Fcos 0. The other component of Fat right angles to F cos 0 is F sin 0. F _ + Fy2 F cos 0 Example 4 : Solution : A force of 30 N is acting at an angle of 600 with the y-axis. Determine the components of the forces along x and y-axes. F x = F sin600 30 x 2 Fy = F COS60 F: 30 N 600 2
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Direction cosines (d.c's) of a vector The components are ax = acosu ay = acosß az = acosy cos u, cos p, cos Y are cosines of a vector A The vectors will be parallel, cosines are same. Example 5 : called direction if their direction + 24 j , 01 x find the vector having the same magnitude as b If a - 3 i + 4 j andb and parallel to a . Solution: Magnitude of a = la 32 +4 And magnitude of b = b 7 Now a unit vector parallel to a = ä . The vector having the same magnitude as b and parallel to a 2+242 = 25 37 + 4 j 5 1 +20j = 25 Scalar product or Dot product A.B A B pos O if A = ax -k ay j + Note: i. i Vector product or Cross product Ax B Al I Bl sine h where fi = unit vector perpendicular to plane containing A and B. j a b [i (aybz azby ) j(axbz —azbx)+k (axby aybx a Note: xj=k ixk= —j j Example 6 . Find c = ä.b - (2i = + = 14 Solution : c-
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Example 7 : Solution : Find d = ax b Since i x .ä 1212- i x

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