Newton’s Laws of Motion Chapter 5 PDF

216 Newton’s Laws of Motion 216 R sin R  60 R cos  E3 4.1 Point Mass. (1) An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own size. (2) Object with zero dimension considered as a point mass. ID (3) Point mass is a mathematic...

216 Newton’s Laws of Motion 216 R sin R  60 R cos  E3 4.1 Point Mass. (1) An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own size. (2) Object with zero dimension considered as a point mass. ID (3) Point mass is a mathematical concept to simplify the problems. 4.2 Inertia. U (1) Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called inertia. D YG (2) Inertia is not a physical quantity, it is only a property of the body which depends on mass of the body. (3) Inertia has no units and no dimensions (4) Two bodies of equal mass, one in motion and another is at rest, possess same inertia because it is a factor of mass only and does not depend upon the velocity. 4.3 Linear Momentum. U (1) Linear momentum of a body is the quantity of motion contained in the body. (2) It is measured in terms of the force required to stop the body in unit time. ST (3) It is measured as the product of the mass of the body and its velocity i.e., Momentum = mass × velocity. If a body of mass m is moving with velocity v then its linear momentum p is given by p  mv (4) It is a vector quantity and it’s direction is the same as the direction of velocity of the body. (5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.] (6) Dimension : [MLT 1 ] v p = constant m Newton’s Laws of Motion 217 217 (7) If two objects of different masses have same momentum, the lighter body possesses greater velocity. p  m 1 v1  m 2 v 2 = constant (8) For a given body p  v p 1 m [As p is constant] (9) For different bodies at same velocities p  m p m U 4.4 Newton’s First Law. v= constant ID m= constant v 60 i.e. v  E3 v1 m 2  v 2 m1  A body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon by some external force to change the state. D YG (1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot accelerate. (2) Newton’s first law defines inertia and is rightly called the law of inertia. Inertia are of three types : Inertia of rest, Inertia of motion, Inertia of direction U (3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This means a body at rest remains at rest and cannot start moving by its own. ST Example : (i) A person who is standing freely in bus, thrown backward, when bus starts suddenly. When a bus suddenly starts, the force responsible for bringing bus in motion is also transmitted to lower part of body, so this part of the body comes in motion along with the bus. While the upper half of body (say above the waist) receives no force to overcome inertia of rest and so it stays in its original position. Thus there is a relative displacement between the two parts of the body and it appears as if the upper part of the body has been thrown backward. Note :  If the motion of the bus is slow, the inertia of motion will be transmitted to the body of the person uniformly and so the entire body of the person will come in motion with the bus and the person will not experience any jerk. 218 Newton’s Laws of Motion 218 (ii) When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper part of the body as explained above. (iii) A bullet fired on a window pane makes a clean hole through it while a stone breaks the is small. So in case of bullet the motion is transmitted only to a small portion of the glass in that small time. During this time the motion is transmitted to the entire window, thus creating the cracks in the entire window. Hole by the bullet ID (iv) In the arrangement shown in the figure : Cracks by the ball E3 Hence a clear hole is created in the glass window, while in case of ball, the time and the area of contact is large. 60 whole window because the bullet has a speed much greater than the stone. So its time of contact with glass U (a) If the string B is pulled with a sudden jerk then it will experience tension while due to inertia of rest of mass M this force will not be transmitted to the string A and so the string B will break. A (b) If the string B is pulled steadily the force applied to it will be transmitted from string B to A through the mass M and as tension in A will be D YG greater than in B by Mg (weight of mass M) the string A will break. M B (v) If we place a coin on smooth piece of card board covering a glass and strike the card board piece suddenly with a finger. The cardboard slips away and the coin falls into the glass due to inertia of rest. (vi) The dust particles in a durree falls off when it is beaten with a stick. This is because U the beating sets the durree in motion whereas the dust particles tend to remain at rest and hence separate. (4) Inertia of motion : It is the inability of a body to change itself its state of uniform motion i.e., a body in uniform motion can neither accelerate nor retard by its own. ST Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall forward. This is because the lower part of his body comes to rest with the bus or train but the upper part tends to continue its motion due to inertia of motion. (ii) A person jumping out of a moving train may fall forward. (iii) An athlete runs a certain distance before taking a long jump. This is because velocity acquired by running is added to velocity of the athlete at the time of jump. Hence he can jump over a longer distance. (5) Inertia of direction : It is the inability of a body to change by itself direction of motion. Example : (i) When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string Newton’s Laws of Motion 219 219 was forcing the stone to move in a circle. As soon as the string breaks, the pull vanishes. The stone in a bid to move along the straight line flies off tangentially. (ii) The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia. 60 (iii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards. Sample problem based on Newton’s first law When a bus suddenly takes a turn, the passengers are thrown outwards because of E3 Problem 1. [AFMC 1999; CPMT 2000, 2001] (a) motion Inertia of motion (c) Speed of motion (d) of Both (b) and (c) A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball fall [EAMCET (Med.) 1995] Outside the car (b) In the car ahead of (c) In the car to the side of the person (d) Exactly in the hand which U (a) the person D YG Problem 2. ID Solution : (a) (b) Acceleration Solution : (d) Because the horizontal component of velocity are same for both car and ball so they cover equal horizontal distances in given time interval. 4.5 Newton’s Second Law. (1) The rate of change of linear momentum of a body is directly proportional to the external ST U force applied on the body and this change takes place always in the direction of the applied force.  (2) If a body of mass m, moves with velocity v then its linear momentum can be given   by p  m v and if force F is applied on a body, then   dp dp F FK dt dt  d p or (K = 1 in C.G.S. and S.I. units) F dt    d   dv dv or (As a  F  (m v )  m  ma  acceleration produced in the body) dt dt dt   F  ma  Force = mass  acceleration Sample problem based on Newton’s second law 220 Newton’s Laws of Motion 220 Problem 3. A train is moving with velocity 20 m/sec. on this, dust is falling at the rate of 50 kg/min. The extra force required to move this train with constant velocity will be (a) 16.66 N Problem 4. (c) 166.6 N (d) dm 50  20   16.66 N 60 dt 60 Solution : (a) Force F  v (b) 1000 N A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum is [MP PET 2002] (a) 5 kg m/s (b) 100 kg m/s Problem 5. E3 Solution : (b) Change in momentum  force  time  10  10  100 kg m / sec (c) 200 kg m/s A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in (a) 5000 N (b) 500 N ID force in opposite direction is (d) 1 sec, the required 10 (c) 50 N (d) Solution : (a) m  100 kg u  5 m / s, v  0 t = 0.1 sec mdv m(v  u) 100 (0  5)   dt t 0.1 U Force  4.6 Force. D YG F  5000 N (1) Force is an external effect in the form of a push or pulls which (i) Produces or tries to produce motion in a body at rest. (ii) Stops or tries to stop a moving body. (iii) Changes or tries to change the direction of motion of the body. U F ST u=0 v=0 Body remains at rest. Here force is trying to change the state of rest. F Body starts moving. Here force changes the state of rest. u=0 v>0 F u0 v>u In a small interval of time, force increases the magnitude of speed and direction of motion remains same. u In a small interval of time, force decreases the magnitude of F v