Physics chap 3

TENSION IN THE STRING
The tension is defined as:   "The force exerted by a string when it is subjected to pull".
If a person is holding a block of weight W attached to the end of a string, a force is experienced by him .   This force is known as Tension. When the body is at rest, the magnitude of tension is equal to the weight   of the body suspended by the string. Tension and the weight acts in the opposite direction. Tension is   vector quantity, which has both magnitude and direction. Its magnitude remains constant at all points of   the string.
UNIT OF TENSION
Since tension is a force,therefore, it has same units as that of force.   In S.I. system : NEWTON   In C.G.S. system : DYNE   In F.P.S. system : POUND
MOTION OF BODIES CONNECTED BY A STRING
When the bodies move vertically
Consider two bodies of unequal masses m1 and m2 connected by the ends of a string, which passes over a   frictionless pulley as shown in the diagram.
If m1>m2, the body ‘A’ will move downward with acceleration ‘a’ and the body ‘B’ will move up with same   acceleration. Here we have to find the value of ‘a’ and tension ‘T’.
Forces acting on body A
There are two forces acting on A
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(i)Weight of body: w1 = m1g   (ii) Tension in the string = T
The net force acting on the body is F= m1g – T
Net force acting on body 'A' is given by Newton’s 2nd law as m1a. Thus we have the equation for the   motion of body "A" as:
m1g – T = m1a --------- (i)
Forces acting upon body B
There are also two forces acting on B
(i)Weight of body: w2 = m2g   (ii) Tension in the string = T
Since body "B" is moving up, the net force acting on body is
F= T – m2g
T – m2g = m2a---------- (ii)
Adding (i) & (ii)
(m1 – m2)g = (m1 + m2)a
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putting the value of 'a' in equation (ii) to find the magnitude of T
T – m2g = m2a T – m2g = m2

INCLINED PLANE     Any plane surface which makes an angle q with the horizontal surface is called inclined plane such that 0<q <90. . Inclined plane is an example of simple machine which is used to lift heavy bodies without applying very huge force. MOTION OF A BODY ON AN INCLINED PLANE   Consider a block of mass "m" placed on an inclined plane, which makes an angle q with the horizontal plane. The weight "W" of the block is acting vertically downward. The weight of the block can be resolved into two rectangular components: wcosq and wsinq other forces acting on the block are: (i) Surface reaction (R) which is perpendicular to the plane (ii) Force of friction (f) acting opposite to the direction of motion of block. Let us take x-axis perpendicular to the inclined plane. If the block is at rest, then wsinq acting down the plane balances the opposing frictional force. Applying 1st condition of equilibrium. For latest information , free computer courses and high impact notes visit : www.citycollegiate.com Fx = 0 f – wsinq = 0 -------(1) Fy = 0 R– wcosq = 0 -------(2) Since there is no motion in the direction perpendicular to the inclined plane, therefore wcosq is balanced by R i.e. R = wcosq If block slides down with an acceleration equal to 'a', then the resultant force is equal to 'ma' and the force on block will be: wsinq – f According to 2nd law of motion wsinq – f = ma If the force of friction is negligible, then wsinq = ma but w = mg This expression shows that if friction is negligible the acceleration of a body on an inclined plane is independent of mass but is directly proportional to sinq. WHEN FRICTION IS NOT NEGLIGIBLE   (i) If block moves upward f - wsinq = ma (ii) If block moves downward wsinq – f = ma MECHANICAL ADVANTAGE OF INCLINED PLANE   MECHANICAL ADVANTAGE OF INCLINED PLANE IS GIVEN BY M.A = 1/sinq PARTICULAR CASES     (i) When q = 0 a = g sin0 a = g x 0 a = o   (ii) When q = 30 a = g sin30 a = g x 0.5 a = g/2 a = 4.9 m/s2   (iii) When q = 60 a = g sin60 a = g x 0.866 a = 4.9 m/s2 For latest information , free computer courses and high impact notes visit : www.citycollegiate.com (iv) When q = 90 a = g sin90 a = g x 1 a = g a = 9.8 m/s2   Momentum     Physical quantity that describes the quantity of motion in a body is called momentum. The momentum of a moving body is defined as "the product of mass and velocity of a moving body is called linear momentum" Mathematically, Momentum is a vector quantity and its direction is the same as that of velocity. Explanation   Momentum is that property of a moving body which determines how much effort is required to accelerate or stop a body. Hence it may also be termed as quantity of motion of a body. From various observations it is concluded that greater effort is required to stop a body if it possess either greater mass or greater velocity or both. Units Of Momentum   In S.I. system : NS [1 NS = 1 kg m/s] In C.G.S. system : Dyne.S In F.P.S. system : Lb.S Dimensions Of Momentum   The dimension of momentum is [MLT-1] Law Of Conservation Of Momentum The law of conservation of momentum states that: "When some bodies constituting an isolated system act upon one another, the total momentum of the system remains constant." In other words "Total momentum of an isolated system before and after collision is constant." Proof   Consider an isolated system of two bodies "A" & "B" having masses m1 & m2 moving initially with velocities u1 & u2 respectively. They collide with each other and after the impact their velocities become v1 & v2. Total momentum of system before collision = m1u1 + m2u2 Total momentum of system after collision = m1v1 + m2v2 When the two bodies collide with each other, they come in contact for a short time "t". During this interval, let the average force exerted one of the bodies is F. We know that the rate of change of linear momentum is equal to applied force, therefore FA = ( m1v1- m1u1)/t -------------(1) FB = ( m2v2- m2u2 )/t--------------(2) According to the third law of motion : FA = -FB Putting the values of FA and FB ( m1v1- m1u1)/t = - ( m2v2- m2u2 )/t   m1v1 - m1u1 = - ( m2v2- m2u2 ) m1v1 - m1u1 = - m2v2 + m2u2 This is known as the Law of conservation of momentum. This expression shows that the total momentum of an isolated system before and after collision remains constant i.e. the total momentum of the system is conserved. Isolated System   An isolated system is one in which constituents of the system interact with one another and no external force is applied on any of them. Actually a perfect isolated system is not possible in the physical world, but a group of objects whose mutual interaction with other objects can frequently be treated as if they are isolated. For example molecules of gas enclosed in a vessel at constant temperature.   ELASTIC AND INELASTIC COLLISION       ELASTIC COLLISION www.citycollegiate.com    An elastic collision is that in which the momentum of the system as well as kinetic energy of the system    before and after collision is conserved. INELASTIC COLLISION    An inelastic collision is that in which the momentum of the system before and after collision is conserved    but the kinetic energy before and after collision is not conserved. ELASTIC COLLISION IN ONE DIMENSION www.citycollegiate.com    Consider two non-rotating spheres of mass m1 and m2 moving initially along the line joining their centers    with velocities u1 and u2 in the same direction. Let u1 is greater than u2. They collide with one another    and after having an elastic collision start moving with velocities v1 and v2 in the same directions on the    same line. Momentum of the system before collision = m1u1 + m2u2 Momentum of the system after collision = m1v1 + m2v2    According to the law of conservation of momentum: m1u1 + m2u2 = m1v1 + m2v2 m1v1 – m1u1 = m2u2 – m2v2 m1(v1 – u1) = m2(u2 – v2) -------(1)    Similarly                                              www.citycollegiate.com K.E of the system before collision = ½ m1u12 + ½ m2u22 K.E of the system after collision = ½ m1v12 + ½ m2v22    Since the collision is elastic, so the K.E of the system before and after collision is conserved . For latest information , free computer courses and high impact notes visit : www.citycollegiate.com    Thus ½ m1v12 + ½ m2v22 = ½ m1u12 + ½ m2u22 ½ (m1v12 + m2v22) = ½ (m1u12 + ½ m2u22 m1v12-m1u12=m2u22-m2v22 m1(v12-u12) = m2(u22-v22) m1(v1+u1) (v1-u1) = m2(u2+v2) (u2-v2) ------- (2)    Dividing equation (2) by equation (1) V1+U1 = U2+V2    From the above equation V1=U2 +V2 -U1_________(a) V2=V1+U1 -U2_________(b)    Putting the value of V2 in equation (1) m1 (v1-u1) =m2 (u2-v2) m1 (v1-u1) =m2{u2-(v1+u1-u2)} m1(v1-u1)=m2{u2-v1-u1+u2} m1(v1-u1)=m2{2u2-v1-u1} m1v1-m1u1=2m2u2-m2v1-m2u1 m1v1+m2v1=m1u1-m2u1+2m2u2 v1(m1+m2)=(m1-m2)u1-2m2u2    In order to obtain V2 putting the value of V1 from equation (a) in equation (i) m1 (v1-u1) = m2(u2-v2) m1(u2+v2-u1-u1)=m2(u2-v2) m1(u2+v2-2u1)=m2(u2-v2) m1u2+m1v2-2m1u1=m2u2-m2v2 m1v2+m2v2=2m1u1+m2u2-m1u2 v2(m1+m2)=2m1u1+(m2-m1)u2