Deriving the final velocities of two colliding bodies ( 4. System of Particles and Rotational Motion
All derivations are explained fully in NCERT Part 1 & Part 2. You can read online or download as PDF from NCERT website.
(P1−P2)Δmρ=12Δm(v22−v12)+Δm⋅g(h2−h1)open paren cap P sub 1 minus cap P sub 2 close paren the fraction with numerator delta m and denominator rho end-fraction equals one-half delta m open paren v sub 2 squared minus v sub 1 squared close paren plus delta m center dot g of open paren h sub 2 minus h sub 1 close paren Divide the entire equation by and multiply by
∫uvv⋅dv=a∫0sdsintegral from u to v of v center dot d v equals a integral from 0 to s of d s all important derivations of physics class 11 pdf download
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Deriving the expression for terminal velocity ( 7. Thermodynamics and Kinetic Theory Mayer’s Relation: Proving
W=GMm[−1x]R∞=GMm[0−(−1R)]=GMmRcap W equals cap G cap M m open bracket negative 1 over x end-fraction close bracket sub cap R raised to the infinity power equals cap G cap M m open bracket 0 minus open paren negative the fraction with numerator 1 and denominator cap R end-fraction close paren close bracket equals the fraction with numerator cap G cap M m and denominator cap R end-fraction Deriving the final velocities of two colliding bodies ( 4
i.e., P+12ρv2+ρgh=Constanti.e., cap P plus one-half rho v squared plus rho g h equals Constant Ascent Formula (Capillary Rise) When a capillary tube of radius is dipped in a liquid of density and surface tension , the liquid rises to a height
tanθ+μ1−μtanθ=v2Rgthe fraction with numerator tangent theta plus mu and denominator 1 minus mu tangent theta end-fraction equals the fraction with numerator v squared and denominator cap R g end-fraction
) about any arbitrary axis equals the moment of inertia about a parallel axis passing through the centre of mass ( Icmcap I sub cm end-sub ) plus the product of total mass ( ) and the square of the perpendicular distance ( ) between the axes: Solving for final velocities
) against this restoring force to change the spring's length by an infinitesimal distance
The horizontal range is the total horizontal distance covered during the time of flight R=ux⋅Tcap R equals u sub x center dot cap T
Fi=miai=mi(riα)(since a=rα)cap F sub i equals m sub i a sub i equals m sub i open paren r sub i alpha close paren space open paren since a equals r alpha close paren Torque acting on this single particle:
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u1+v1=v2+u2⟹u1−u2=v2−v1u sub 1 plus v sub 1 equals v sub 2 plus u sub 2 ⟹ u sub 1 minus u sub 2 equals v sub 2 minus v sub 1 (Velocity of approach equals velocity of separation). Solving for final velocities