Q.

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 that path.

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a

P → 1, 2, 3, 4, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 5

b

P → 1, 2, 3, 4, 5; Q → 3, 5; R → 2, 3, 4, 5; S → 2, 5

c

P → 2, 3, 4; Q → 5; R → 1, 2, 4; S → 2, 5

d

P → 1, 2, 3, 5; Q → 2, 5; R →2, 3, 4, 5; S → 2, 5

answer is A.

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Detailed Solution

(P) r→(t)=αti^+βtj^                           ⇒V→=dr→(t)dt=αi^+βj^{ Constant }⇒a→=dv→dt=0                                   P→=mv→ (Remain constant) K=12mv2{ Remain constant }       F→=ma→=-∂U∂xi^+∂U∂yj^=0⇒U→ Constant ∴E=K+U= Constant dL→dt=τ→=r→×F→=0⇒L→=constant  (Q) r→=αcos⁡(ωt)i^+βsin⁡(ωt)j^⇒v→=dr→dt=-αωsinωti^+βωcosωtj^⇒a→=dv→dt=−αω2cos⁡(ωt)i^−βω2sin⁡(ωt)j^⇒a→=−ω2r→ Path of the particle is elliptical                 ∴F∝r⇒U∝r2∵ΔU=-∫F→.dr→=∫mω2rdr=mω2r22U depends on r hence it will change with timeTotal energy remain constant because force is central  (R) r→(t)=α(cos⁡ωi^+sin⁡(ωt)j^)⇒v→(t)=dr→(t)dt=α(−ωsin⁡(ωt)i^+ωcos⁡(ωt)j^)⇒a→(t)=−αω2[cos⁡(ωt)i^+sin⁡(ωt)j^] (S) r→=αti^+β2t2j^⇒v→=αi^+βtj^⇒a→=dv→dt=βj^ Constant F→=ma→ Constant τ→=r→×F→=mαβtk^L→=∫τ→dt=12mαβt2k^    Depends on timeΔU=−∫F→⋅dr→=−m∫0t βj^⋅(αi^+βtj^)dt=-12mβ2t2 ΔK=12mv2=12mα2+β2t2∴E=ΔK+ΔU=12mα2[ Remain constant]
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