PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1Z_ETAlDh6zE4ZWqdRgq9eeASLK0jjFKv/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Second Law of motion; Impulse; Newton’s Third Law of motion. Law of conservation of linear momentum and its applications, Equilibrium of concurrent forces. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion: Centripetal force and its applications. FORCE Force is the action of one body upon another. It is a vector quantity with units newton (N). Types of Forces Weight : Weight of a body is the force with which earth attracts it. It is also defined as force of gravity or the gravitational force. Contact forces : Whenever two bodies come in contact they exert forces on each other, that are called contact forces. Normal reaction (N or R) : It is the component of contact force normal to the surface. It measures how strongly the surfaces in contact are pressed together. Frictional force (f ) : It is the component of contact force parallel to the surface. It opposes the relative motion (or attempted motion) of the two surfaces in contact. Tension : The force exerted by the end of a taut string, rope or chain is called the tension. The direction of tension is always pulling in nature. Spring force : Every spring resists any attempt to change its length, the more you change its length the harder it resists. The force exerted by a spring is given by F = –kx, where x is the change in length and k is spring constant or stiffness constant (units N/m). MOMENTUM Momentum of a body is defined to be the product of its mass m and velocity v . It is a vector quantity and is represented by P , ρ P = mv NEWTON’S LAWS OF MOTION First law (Law of inertia) It states that a body continues to be in its state of rest or of uniform motion until and unless it is acted upon by some external force If F = 0, ⇒ v = Constant, ⇒ a = 0 i.e. The body opposes any change in its state of rest or of uniform motion, it is also known as the law of inertia. CHAPTER COVERS : Force Momentum Newton’s laws of Motion Frame of Reference Conservation of linear momentum Rocket propulsion Applications of Newton’s laws of Motion Friction Motion in a horizontal circle Equilibrium of Contact Forces If three forces P , Q and R action on an object such that forces are concurrent and object is in equilibrium then P P sinα = sinβ = R R sinγ Second Law of Motion It states that the net force acting on a particle or a system of particles is directly proportional to the rate of change of its linear momentum. ρ dp Mathematically F ∝ dt ρ kdp F = dt ρ d where k = 1 ρ F = (mv ) dt If the mass of the system is constant, then ρ dv F = m dt ⇒ F = ma ⇒ Finst ρ ρ mainst Fav = maav ⎛ → →⎞ ⎜vf − vi ⎟ = m ⎝ ⎠ t Third Law of Motion To every action there is equal and opposite reaction → → Mathematically ⇒ F 12 = − F 21 Here F12 is force acting on body 1 due to 2 and F21 is force acting on body 2 due to 1. IMPULSE Impulse of a force is defined as the change of linear momentum imparted to the body ρ ρ J = P2 − P1 = m(v 2 − v1) If the body is acted upon by a variable force as shown in figure, then Impulse acting on the body in the interval t1 to t2 ρ ρ ρ t2 J = P2 − P1 = ∫F dt t1 = Area below the F ~ t graph. FRAME OF REFERENCE It is a set of co-ordinate axes with respect to which the position, velocity and acceleration of a particle or a system of particles can be measured. There are two types of reference frames: Inertial frames of reference If two frames of reference are either at rest or moving with uniform relative velocity, then they are said to be inertial frames of reference. All Newton's laws are valid in inertial frames of reference. Non-inertial frames of reference Accelerated frames are known as non-inertial frames of reference. Newton's laws are not valid in non-inertial frames. Pseudo forces (fictitious forces): It is a slight adjustment which enables us to apply Newton's laws in Non- inertial frames of reference. If an observer is into a linearly accelerating non-inertial frame, then in order to use Newton's laws of motion to explain the behaviour of a particle or a system of particles, a pseudo force is applied opposite to the acceleration of the observer. Mathematically, Fpseudo ρ = −ma m = mass of the particle a = acceleration of the observer General expression for force : As p = mv ρ ⇒ F = d ρ (mv ) = dt dm ρ v + dt mdv dt Case - I : When mass = constant, ⇒ dm = 0 dt ρ dv ρ F = m ma dt Case - II : In variable mass system like Rocket propulsion, firing of bullets etc. Relative speed is constant in these cases ρ ρ dm F = v dt CONSERVATION OF LINEAR MOMENTUM According to Newton's third law ρ fext → ∝ dp dt If net external force acting on the particle or system of particles is zero then the total momentum of the particle or system of particles remain conserved. → dp = → 0 or p = constant dt Applications of Conservation of Momentum A block of mass m is at rest in a gravity free space suddenly explodes into two parts in the ratio 1 : 3, if lighter particle has a speed of 9 m/s just after explosion, then the speed of heavier particle can be calculated as shown. → p initial= 0 → ⎛ m ⎞ ˆ ⎛ 3m ⎞ → p final = ⎜ 4 ⎟9i + ⎜ ⎝ ⎠ ⎝ → → p initial = p final 0 = m 9iˆ + 3m → 4 4 → ˆ v = −3i m/s A man of mass M standing on a frictionless floor, throws a ball of mass m with a speed u as shown in figure. The velocity of man, after he throws the ball is v → p initial = 0 → ˆ → p final = mui + M v → → As p initial = p final → − miˆ v = M A ball of mass m moving with constant speed of u m/s is caught by man of mass M standing on a frictionless floor. The velocity acquired by the man is v → ˆ p initial= mui → → p final = (m + M ) v → → p initial = p final → muiˆ v = (m + M ) If a gun of mass M fires a bullet of mass m with speed v, then recoil speed of the gun is given by V = mv/M. ROCKET PROPULSION It is an example of variable mass-system. In a rocket, combustion chamber carries fuel and oxidising agent. When the fuel burns, a hot jet of gases emerges out forcefully from the small hole in the tail of the rocket. Let u be the velocity of emerging gases relative to the rocket and (ΔM/Δt) be the rate of fuel consumption, then the upthrust on the rocket is F = u (ΔM/Δt). Let M0 be the initial mass (rocket + fuel), M and v be the mass and velocity of the rocket at any instant t, then ρ ρ ⎛ M0 ⎞ ν = −u loge ⎜ M ⎟ APPLICATIONS OF NEWTON’S LAWS OF MOTION A machine gun fires n bullet per second with speed u and mass of each bullet is m. The Force required to keep the gun stationary is ρ F = nmv m,u M Bullets moving with a speed v hit a wall When the bullets come to rest in wall m v Force on wall Fwall = nmv When the bullets rebound elastically Fwall = 2nmv Liquid jet of area A moving with speed v hits a wall A v Force required be a pump to move the liquid with this speed is F = v dm dt = v × ρAv = ρAv2 As jet hits a vertical wall and does not rebound, the force exerted by it on the wall is, Fwall = ρAv2 When water rebounds elastically, Fwall = 2ρAv2 For oblique impact as shown, Fwall = 2ρAv2 cosθ Liquid jet v v θ θ wall The blocks shown are being pushed by a force F. F1, F2 are contact forces between M1 & M2 and M2 & M3 respectively a a = F FBD of M 1 2 3 3 a F1 – F2 = M2a ⇒ F1 = (M3 + M2)a FBD of M2 a F2 = M3a M3 ⎛ M2 + M3 ⎞ or F2 = 1 M2 M3 F, F1 = ⎜ ⎝ 1 + M2 ⎟ F + M3 ⎠ The strings are massless. Let T1 and T2 be the tensions in the two strings and ‘a’ be the acceleration. a = F , T = M2 + M3 , T = M3F F M1 + M2 + M3 M1 + M2 + M3 M1 + M2 + M3 Tension in the block at x from left end is given as (a) Mx F Tx L M (b) T = F(L − x ) x L (c) T = F1x + F2 (L − x) L F2 M F1 x L L x T T ceiling massless string FBD of M w.r.t. ground Mg When system is stationary T – Mg = 0 System moves up with acceleration ‘a’ T = M(g + a) System moves down with acceleration a′ T = M(g – a′) Uniform rope of mass Ms. FBD of lower portion Tx x Ms gx L Stationary system T = Ms gx x L If the rope is accelerating upwards, then Tx = Ms x (g + a) L If the rope is accelerating downwards, then Tx IMPORTANT PROBLEMS Pulley mass systems Stationary pulley a = M2 − M1 g M2 + M1 = Ms x (g − a) L massless T T a a ⎛ M1M2 ⎞ M g T = 2⎜ ⎝ M1 ⎟g + M2 ⎠ 1 M2 g Pulley is moving upward with acceleration a0 ⎛ M M ⎞ a0 T = 2⎜ 1 2 ⎟ (g + a0 ) ⎝ M1 + M2 ⎠ ⎛ M2 − M1 ⎞ ar T ar = ⎜ ⎝ 2 + M ⎟ (g + a0 ) 1 ⎠ T ar a1 = – (a0 a1 + ar) a2 = ar – a0 M1 a2 M1 a0 ar = relative acceleration of M1 and M2 w.r.t. pulley a , a are accelerations of M and M w.r.t. ground. M2 g M2 a0 (pseudo) 1 2 1 2 (iii) T = M1M2 (sinα + sinβ) M1 + M2 ⎛ M2 sinβ − M1 sinα ⎞ a = ⎜ ⎝ ⎟ g M2 + M1 ⎠ (iv) 2aA = aB …(i) MAg – 2T = MAaA …(ii) T = MBaB …(iii) Two block system : Case - I : Let ‘m’ does not slide down relative to wedge ‘M’ The force required is given by F = (M + m)g tanθ a = g tan θ (in horizontal direction w.r.t. ground) Contact force R between m and M is smooth R Mg cos θ F2 M3 M1 T1 M2 T2 M3 M1 g Case - II : Minimum value of F so that ‘m’ falls freely is given by m F F = Mgcotθ M FRICTION Wedge M moves with acceleration = gcotθ Contact force between M and m is zero. θ Wedge smooth Friction force is of two types Static frictions ‘fs’ Kinetic friction ‘fk’ Static Friction The static friction between two contact surfaces is given by fs < μs N, where N is the normal force between the contact surfaces and μs is a constant which depends on the nature of the surfaces and is called ‘coefficient of limiting friction’. Static friction acts on stationary objects. Its values satisfy the condition fs < μs N μs – Co-efficient of limiting friction Static friction takes its peak value (fs(max) = μsN) when one surface is ‘about to slide’ on the other. Static friction in this case is called limiting friction. Kinetic Friction (μk) It acts on the two contact surfaces only when there is relative slipiry or relative motion between two contact surfaces. fk = μkN The relative motion of a contact surface with respect to each other is opposed by a force given by fk = μk N, where N is the normal force between the contact surfaces and μk is a constant called ‘coefficient of kinetic friction’, which depends, largely, on the nature of the contact surfaces. Friction force Static friction Kinetic friction 45º fs (max) Applied force Graphical representation of variation of friction force with the force applied on a body Angle of Repose Consider a situation in which a block is placed on an inclined plane with co-efficient of friction 'μ' then the maximum value of angle of inclined plane for which the block can remain at rest is defined as angle of repose. Two blocks placed one above the other Case - I : (a) F ≤ μ(M1 + M2)g Both blocks move together with same acceleration a = F M1 + M2 (b) F > μ(M1 + M2)g M2 moves with constant acceleration a2 = μg amax = μg Smooth M moves with acceleration a = F − μM2g 1 1 1 M2 slips backwardon M1. Case - II : (a) F ≤ μ(M1 + M2 )M2 g , both blocks move together with acceleration a = M1 F M1 + M2 with a = μM2 g . 1 (b) F > μ(M1 + M2 )M2g , M1 M1 moves with constant acceleration μ a = μM2 g 1 M M2 moves with acceleration a 1 = F − μM2g 2 M Smooth 2 M2 slips forward on M1. Minimum force required to move a body on a rough horizontal surface F cos θ > μR F ≥ μmg cos θ + μsinθ R F sin θ F θ μ R M F cos θ Fmin = μmg at θ = tan–1(μ) Block on Inclined Plane If µ ≥ tan θ, the block will remain stationary on the inclined plane; and the frictional force acting on the block will be equal to mg sinθ and static in nature f = mg sin θ If a block slides down an inclined plane with constant velocity, the frictional force acting on the block is kinetic in nature and is equal to mg sin θ ⇒ f = mg sin θ If µ < tan θ, the block will slide down the plane with acceleration a equal to a = g sin θ – µ g cos θ Frictional force is kinetic in nature and is given by f = μN = μ mg cos θ (less than mg sin θ) If a block is projected up the plane, retardation a is given by a = g sin θ + µ g cos θ. If mg sin θ exceeds frictional force, the block tends to slide down. The minimum force required to prevent sliding is F min = mg sin θ – µ mg cos θ If you try to push a block up the plane, frictional force and mg sinθ both opposes the upward motion of the block, hence, the minimum force required to move up is given by F = mg sin θ + µ mg cos θ θ DYNAMICS OF CIRCULAR MOTION Neglecting Gravity axis T = Centripetal force = mv 2 r = mω2r O Considering gravity (Conical pendulum) Tsinθ = mω2r …(1) T cosθ l θ Tcosθ = mg θ T T sinθ T mg cos θ …(2) r C For θ to be 90º (i.e., string to be horizontal) mg T = ∞ ∴ It is not possible. Tsinθ = mω2r = mω2lsinθ ⇒ T = mω2l Time period = 2π Vehicle negotiating a curve on a banked road The maximum velocity with which a vehicle can safely negotiate a curve of radius r on a rough inclined road is v = N θ α = v2/R f N cosθ mg f sinθ N θ N cosθ θ N cosθ θ f sinθ f f sinθ mg For a smooth surface μ = 0 ⇒ v = For a horizontal surface, θ = 0 ⇒ v = rg tan θ μrg ❑ ❑ ❑