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Animated Physics materials overview (XIth)

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Animated Physics materials for IIT-JEE, AIEEE, AIPMT, SAT & Boards

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Slide 3 : Q: A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is fixed point. Show that OP × v is independent of the position P. SOLUTION: Let us visualize the situation. Let us take point O (arbitrarily any point) & point P (on the given line) as shown. For the sake of convenience let us take a coordinate system with origin at O and x-axis parallel to given line. Let us assume the coordinate of P be (x, y). Hence, we can write vector OP in unit vector notation as: OP = x i + y j . As particle moves parallel to x axis, we can write v in vector form as v = v i . v P O x y

Slide 4 : www.vidyadrishti.com OP × v = (x i + y j) × v I ? OP × v = – v y k. (using cross product) Thus, OP × v depends on v and y only. It is independent of x. If we see on particle motion, it moves parallel to x axis. Therefore, y coordinate remain same, only x coordinate of particle position changes. Thus, wherever be the particle, OP × v is unaffected by its position as it does not depend on x. Hence, OP × v is independent of the position P.

Slide 7 : Q. A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take reaching the ground? Ground A C B Subscribe our Physics package for complete solution Call 933337772 Mail: vidyadrishti@gmail.com Website: www.vidyadrishti.com

Slide 8 : Q. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drop a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the ground in 1s. Calculate from these data the acceleration of the elevator. x ft 6 ft Subscribe our Physics package for complete solution Call 933337772 Mail: vidyadrishti@gmail.com Website: www.vidyadrishti.com

Slide 9 : Q. A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally? SOLUTION: Subscribe our Physics package for complete solution Call 933337772 Mail: vidyadrishti@gmail.com Website: www.vidyadrishti.com

Conversion of standard graphs : P. K. BHARTI, I.I.T. KHARAGPUR Conversion of standard graphs motion Time, t Position (upward parabola [one half]) Velocity (linear) Acceleration (constant) o

Slide 12 : A B NA x y Nx Ny

Slide 13 : 13 M2 M1 a a Fixed wedge

Slide 14 : 14 M1 P M2 2a a

Graph of friction vs. applied force [Revised] : P K BHARTI, IIT KHARAGPUR Graph of friction vs. applied force [Revised] . 16 APPLIED FORCE FRICTION 45o LIMITING FRICTION fs, max = msN O A KINETIC FRICTION fk = mkN B C D

Slide 17 : P K BHARTI, IIT KHARAGPUR Example 4: We will find expression for magnitude forces of example 2. [Assume blocks are moving] Solution: We draw FBD of A and B. NOTE: SAME COLOURED FORCES REPRESENT ACTION-REACTION PAIRS OF NEWTON’S 3rd LAW. 17 A f1 f2 B f2 F A B F mBg N1 mAg N2 N2

Banking of road analysis without friction : P. K. BHARTI, I.I.T. KHARAGPUR Banking of road analysis without friction 19 Here, we shall draw the FBD of blue person treating him to particle. Let the banking angle of the road be q. Forces are: mg [downward] N [perpendicular to road surface] N mg q mg N r

HOW TO FIND WORK DONE IN DIFFERENT SITUATIONS : HOW TO FIND WORK DONE IN DIFFERENT SITUATIONS 21 P. K. BHARTI, I.I.T. KHARAGPUR

HOW TO FIND WORK DONE IN DIFFERENT SITUATIONS : HOW TO FIND WORK DONE IN DIFFERENT SITUATIONS 22 P. K. BHARTI, I.I.T. KHARAGPUR

Example 14 : Example 14 A smooth sphere of radius R is made to translate in a straight line with a constant accln a. A particle kept on the top of the sphere is released from there at zero velocity wrt the sphere. Find the speed of the particle wrt the sphere as a function of the angle it slides. P. K. BHARTI, I.I.T. KHARAGPUR 23 mg N ma pseudo

Example 8 : Example 8 A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes several collisions on the floor and final lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L. Find the displacement of the cart during this process. SOLUTION: 25 P. K. BHARTI, I.I.T. KHARAGPUR

Example 7 [RECOIL OF A GUN] : Example 7 [RECOIL OF A GUN] 26 P. K. BHARTI, I.I.T. KHARAGPUR V v

OBLIQUE CENTRAL COLLISION : 27 P K BHARTI IIT KHARAGPUR OBLIQUE CENTRAL COLLISION OBLIQUE COLLISION (OBLIQUE CENTRAL COLLISION): It is that collision in which the colliding particles do not move along a straight line before the collision or after the collision. Generally, it is 2 dimensional. NOTE: You have to study only 2D motion. During collision Before collision After collision

Example 3 continued … : Example 3 continued … Angle made by final velocities with tangential direction: It is clear from animation, that mass 2 moves in normal direction after collision [because it starts moving when impulsive force from mass 1 acts on it along normal direction]. 28 P K BHARTI IIT KHARAGPUR 2 1 u1 1 1 v2 v1 n t 1 v1 ?1 t n v1,n v1,t

AXIS OF ROTATION : AXIS OF ROTATION Example 3: We take another example of a disc in combined motion as shown in the figure. Here motion is such that velocity of contact of the disc with respect to the surface is always zero. Actually, this type of motion is called rolling motion (We shall read in detail about rolling motion later on). Here as the point of contact of disc with the surface is zero. Therefore, axis rotation will pass through this point of contact (because, only point of contact has zero velocity here) and will be perpendicular to plane of rotation of disc. Therefore, AB is the axis of rotation. Clearly, axis of rotation (AB) is moving here. This kind of axis of rotation is known as “ INSTANTANEOUS AXIS OF ROTATION”. P. K. BHARTI, I.I.T. KHARAGPUR 30 A B

Slide 31 : 31 P. K. BHARTI, I.I.T. KHARAGPUR ?2 ?1

Slide 32 : 32 A at aR a v at aR r ? ? a O

Slide 33 : Gravitation © P K Bharti, IIT Kharagpur www.vidyadrishti.com N S A O P ø Equatorial plane Polar axis E M P r Height r = R At surface E= GM/R2

Slide 36 : STEP II: Assume x = 0 at the equilibrium position. Displace particle at a distance x from the equilibrium position. Let the speed of the particle at this position be v. If there is rotation, let angular speed be w. Now, suppose the centre of the pulley goes down further by a distance x from the equilibrium position as shown in the figure. As the pulley goes a distance x downward, spring further stretches by a distance 2x (net effect of x on left of the pulley and x on right of the pulley). Thus, total extension of the spring in this position is, y + 2x = (mg/2k) + 2x (using (4)) Therefore, spring potential energy at this position, Uspring = ½ k {(mg/2k) + 2x}2 …(5) x x = 0 Equilibrium position Example 6 Continued Gravitational PE = 0 2x

Slide 38 : Case 2: Reflection of a string wave from soft boundary (soft reflection) Slideshow to view this animation. Wait for two seconds. Waves Reflection & Refraction

Slide 39 : A whistle of frequency 540 Hz rotates in a circle of radius 2 m at a linear speed of 30 m/s. What is the lowest and highest frequency heard by an observer a long distance away at rest with respect to the centre of circle. Take speed of sound in air as 330 m/s. Can the apparent frequency be ever equal to actual ? Solution : Apparent frequency will be minimum when the source is at N and moving away from the observer Frequency will be maximum when source is at L and approaching the observer. L M O N K C

Slide 41 : THERMODYNAMICS Part 4 Molar specific heat of gases © P K Bharti, IIT Kharagpur www.vidyadrishti.com Cp & Cv