Unit No Name Subject Expert I Physical World and Measurement Mr. S. L. Verma, KV Aizawl II Kinematics Mr. S. L. Verma, KV Aizawl III Law of Motion Mr. Anurag Jayaswal KV Silchar IV Work, Energy, Power Mr. Anurag Jayaswal KV Silchar V Motion of System of Particles and Rigid Body Mr. Harendra Kumar KV Lumding VI Gravitation Mr. Harendra Kumar KV Lumding VII Properties of Bulk Matter Mrs. Salam KV Langjing VIII Thermodynamics Mrs. Laxmi Salam KV Langjing IX Behaviour of Perfect Gas and Kinetic Theory of gases Mr. B. K. Jha, KV Karimganj X Oscillations and Mr. B. K. Jha, KV Karimganj ***************** MATHEMATICAL TOOLS Physical constants:- Mass of an electron (Me) = 9.1x10-31kg.
For Θ =450 H = u2 /4g = 1/4 of the Rmax. Q6. State the parallelogram law of vector addition. Derive an expression for magnitude and direction of resultant of the two vectors. Ans. The addition of two vector and is resultant = + And R = ( A2 + B2+ 2AB CosΘ)1/2 And tan β = B SinΘ/ ( A + B CosΘ) , Θ is the angle between vector and vector , And β is the angle which makes with the direction of . Q7. A gunman always keeps his gun slightly tilted above the line of sight while shooting. Why, Ans. Because bullet follow parabolic trajectory constant downward acceleration. Q8. Derive the relation between linear velocity and angular velocity. Ans : Derive the expression V = r ω Q9. What do you mean by rectangular components of a vector? Explain how a vector can be resolved into two rectangular components in a plane . Q10. The greatest height to which a man can a stone is h, what will be the longest distance upto which he can throw the stone ? Ans: we know that Hmax.= Rmax /2 So h = R/2 Or R = 2h Short answer questions (3 marks)
F = N F N For vertical direction F1 - mg – fs = ma fs mg iii) Case - 3 Body pushed against the vertical plane by inclined force and accelerates vertically upward. FCosθ F For horizontal direction θ N = FSinθ (since body is at rest) N For vertical direction FCosθ - mg - fs = ma fs mg MEMORY MAP
f = 1 = ωa T 2 Centripetal Acceleration – When a body performs uniform circular motion its speed remains constant but velocity continuously changes due to change of direction. Hence a body is continuously accelerated and the acceleration experienced by the body is known as centripetal acceleration (that is the acceleration directed towards the center). v2 a P2 r v1 ∆v ∆v O θ ∆s C B R v2 θ v1 v θ v P1 A Consider a particle performing uniform circular motion with speed v. When the particle changes its position from P1 to P2 its velocity changes from v1 to v2 due to change of direction. The change in velocity from P1 to P2 is ∆v which is directed towards the center of the circular path according to triangle law of subtraction of vectors. From figure ∆OP1P2 and ∆ABC are similar, hence applying the condition of similarity BC = P1 P1 AB O P1 or, ∆v = ∆s v r or, ∆v = v∆s r Dividing both sides by ∆t, ∆v = v∆s ∆t r∆t Taking limit ∆t 0 both sides, Lim ∆v = v Lim ∆θ
For lowermost B net force towards the center is centripetal force. Tension TB acts towards the center of the circular path whereas weight mg acts away from it. Hence, TB – mg = mvB2 r Putting, vB = √5rg TB – mg = m5gr r or, TB = 6mg in critical condition of vertical circular motion of a body tied to a string velocities at topmost and lowermost be √(rg) and √(5rg) respectively and tensions in the strings be 0 and 6mg respectively. General Condition for Slagging of String in Vertical Circular Motion For the body performing vertical circular motion tied to a string, slagging of string occurs in the upper half of the vertical circle. If at any instant string makes angle θ with vertical then applying net force towards center is equal to centripetal force, we have mgCos θ θ θ T v mg T + mgCos θ = mv2
.Moment of inertia and torque. The moment of inertia of a rigid body about an axis of rotation is numerically equal to the external torque required to produce a unit angular acceleration in the body BOUT THE GIVEN AXIS. MATHEMATICALLY: .Law of conservation of angular momentum. If no external torque acts on a system, the total angular momentum of the system remains unchanged. Mathematically: For translational equilibrium of a body, ==0 For rotational equilibrium of a rigid body, 1.The following table gives a summary of the analogy between various quantities describing linear motion and rotational motion. s.no. 1. 2. 3. 4. 5. 6. Linear motion Distance/displacement (s) Linear velocity, Linear acceleration, Mass (m) Linear momentum, Force, Also, force s.no. 1. 2. 3. 4. 5. 6. Rotational motion or angular displacement () Angular velocity, Angular acceleration= Moment of inertia () Angular momentum, Torque, Also, torque,
7. 8. 9. 10. 11. 12. Translational KE, Work done, Power, Linear momentum of a system is conserved when no external acts on the system. Equation of translator motion have their usual meaning. 7. 8. 9. 10. 11. 12. Rotational KE, Work done, Power, Angular momentum of a system is conserved when no external torque acts on the system Equations of rotational motion have usual meaning.