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Question:

In the List-I below, four different paths of a particle are given as functions of time. In these functions, α and β are positive constants of appropriate dimensions and α≠β. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned; →p is the linear momentum, →L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for the path. List - I List - II P. →r(t)=αt^i+βt^j 1. →p Q. →r(t)=αcos(ωt)^i+βsin(ωt)^j 2. →L R. →r(t)=α(cos(ωt)^i+sin(ωt)^j) 3. K S. →r(t)=αt^i+β2t^2^j 4. U 5. E

P→1,2,3,4,5; Q→2,5; R→2,3,4,5; S→5

P→2,3,4; Q→5; R→1,2,4; S→2,5

P→1,2,3,4,5; Q→3,5; R→2,3,4,5; S→2,5

P→1,2,3,5; Q→2,5; R→2,3,4,5; S→2,5

Solution:

(P) →r(t)=αt^i+β^tj →v=d→r(t)dt=α^i+β^jconstant →a=→dvdt=0 →P=m→v(remain constant) k=1/2mv^2remain constant →F=−[∂U∂x^i+∂U∂Y^i]=0 ⇒U→constant E = K + U d→Ldt=→T=→r×→F=0 →L=constant (Q) →r=−αcos(ωt)^i+βsin(ωt)^j →v=d→rdt=−αωsin(ωt)^i+βωcos(ωt)^j →a=d→vdt=−αω^2cos(ωt)^i−βω^2sin(ωt)^j=−ω^2[αcos(ωt)^i+βsin(ωt)^j] →a=−ω^2→r →T=→r×→F=0 →r and →Fare parallel △U=−∫→F.dr=+∫0rmω^2.r.dr △U=mω^2[r^2/2] U∝r^2 r=√α^2cos^2(ωt)+β^2sin^2(ωt) r is a function of time (t) U depends on r hence it will change with time Total energy remain constant because force is central (R) →r(t)=α(cosωt^i+sin(ωt)^j) →v(t)=d→r(t)dt=α[−ωsin(ωt)^i+ωcos(ωt)^j] |→v|=αω(Speed remains constant) →a(t)=d→v(t)dt=α[−ω^2cos(ωt)^i−ω^2sin(ωt)^j]=−αω^2[cos(ωt)^i+sin(ωt)^j] →a(t)=−ω^2(→r) →T=→F×→r=0 |→r|=α(remain constant) Force is central in nature and distance from fixed point is central. Potential energy remains constant Kinetic energy is also constant (speed is constant) (S) →r=αt^i+β2t^2^j →v=d→rdt=αt^i+βt^j(speed of particle depends on ‘t’) →a=d→vdt=β^jconstant →F=m→aconstant △U=−∫→F.d→r=−m∫0tβ^j(α^i+β^tj)dt U=−mβ^2t^2/2 k=1/2mv^2=1/2m(α^2+β^2t^2) E=k+U=1/2mα^2[remain constant].