JEE Challenger
More from Motion in a Plane
PhyHardAdv 2018 • Paper 2

Matching Particle Paths with Conserved Physical Quantities

In the provided List-I, four distinct motion paths of a particle are expressed as functions of time. In these expressions, α\alpha and β\beta are positive constants with appropriate physical dimensions such that α≠β\alpha \neq \beta. The force acting on the particle in each scenario is either conservative or zero. In List-II, five physical attributes are given: p⃗\vec{p} represents linear momentum, L⃗\vec{L} represents angular momentum with respect to the origin, KK is kinetic energy, UU is potential energy, and EE is total mechanical energy. Match each trajectory in List-I with the corresponding quantities in List-II that remain conserved for that motion.

List-IList-IIP. r⃗(t)=αti^+βtj^1. p⃗Q. r⃗(t)=αcos⁡ωti^+βsin⁡ωtj^2. L⃗R. r⃗(t)=α(cos⁡ωti^+sin⁡ωtj^)3. KS. r⃗(t)=αti^+β2t2j^4. U5. E\begin{array}{ll} \text{List-I} & \text{List-II} \\ \text{P. } \vec{r}(t) = \alpha t \hat{i} + \beta t \hat{j} & \text{1. } \vec{p} \\ \text{Q. } \vec{r}(t) = \alpha \cos \omega t \hat{i} + \beta \sin \omega t \hat{j} & \text{2. } \vec{L} \\ \text{R. } \vec{r}(t) = \alpha (\cos \omega t \hat{i} + \sin \omega t \hat{j}) & \text{3. } K \\ \text{S. } \vec{r}(t) = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j} & \text{4. } U \\ & \text{5. } E \end{array}

Options

A

P→1,2,3,4,5;Q→2,5;R→2,3,4,5;S→5\text{P} \rightarrow 1, 2, 3, 4, 5; \quad \text{Q} \rightarrow 2, 5; \quad \text{R} \rightarrow 2, 3, 4, 5; \quad \text{S} \rightarrow 5

Correct
B

P→1,2,3,4,5;Q→3,5;R→2,3,4,5;S→2,5\text{P} \rightarrow 1, 2, 3, 4, 5; \quad \text{Q} \rightarrow 3, 5; \quad \text{R} \rightarrow 2, 3, 4, 5; \quad \text{S} \rightarrow 2, 5

C

P→2,3,4;Q→5;R→1,2,4;S→2,5\text{P} \rightarrow 2, 3, 4; \quad \text{Q} \rightarrow 5; \quad \text{R} \rightarrow 1, 2, 4; \quad \text{S} \rightarrow 2, 5

D

P→1,2,3,5;Q→2,5;R→2,3,4,5;S→2,5\text{P} \rightarrow 1, 2, 3, 5; \quad \text{Q} \rightarrow 2, 5; \quad \text{R} \rightarrow 2, 3, 4, 5; \quad \text{S} \rightarrow 2, 5

Topics & Concepts

Motion in a PlaneKinematicsAngular MomentumWork-Energy Theorem

Step-by-Step Solution

To determine which physical quantities are conserved for each motion path in List-I, we evaluate the velocity v⃗(t)=dr⃗dt\vec{v}(t) = \frac{d\vec{r}}{dt}, acceleration a⃗(t)=dv⃗dt\vec{a}(t) = \frac{d\vec{v}}{dt}, and the net force F⃗(t)=ma⃗(t)\vec{F}(t) = m\vec{a}(t) for each path.

Analysis for Path P:

r⃗(t)=αti^+βtj^\vec{r}(t) = \alpha t \hat{i} + \beta t \hat{j}

  • Velocity and Acceleration: v⃗(t)=dr⃗dt=αi^+βj^(constant)\vec{v}(t) = \frac{d\vec{r}}{dt} = \alpha \hat{i} + \beta \hat{j} \quad (\text{constant}) a⃗(t)=dv⃗dt=0⃗  ⟹  F⃗=0⃗\vec{a}(t) = \frac{d\vec{v}}{dt} = \vec{0} \implies \vec{F} = \vec{0}

  • Conserved Quantities:

    1. Linear Momentum (p⃗\vec{p}): p⃗=mv⃗=m(αi^+βj^)=\vec{p} = m \vec{v} = m(\alpha \hat{i} + \beta \hat{j}) = constant. (Conserved)
    2. Angular Momentum (L⃗\vec{L}): L⃗=r⃗(t)×p⃗=(αti^+βtj^)×m(αi^+βj^)=0⃗=\vec{L} = \vec{r}(t) \times \vec{p} = (\alpha t \hat{i} + \beta t \hat{j}) \times m(\alpha \hat{i} + \beta \hat{j}) = \vec{0} = constant. (Conserved)
    3. Kinetic Energy (KK): K=12mv2=12m(α2+β2)=K = \frac{1}{2} m v^2 = \frac{1}{2} m (\alpha^2 + \beta^2) = constant. (Conserved)
    4. Potential Energy (UU): Since F⃗=0⃗=−∇U\vec{F} = \vec{0} = -\nabla U, U=U = constant. (Conserved)
    5. Total Mechanical Energy (EE): E=K+U=E = K + U = constant. (Conserved)

P→1,2,3,4,5\mathbf{P \rightarrow 1, 2, 3, 4, 5}

Analysis for Path Q:

r⃗(t)=αcos⁡(ωt)i^+βsin⁡(ωt)j^(α≠β)\vec{r}(t) = \alpha \cos(\omega t) \hat{i} + \beta \sin(\omega t) \hat{j} \quad (\alpha \neq \beta)

  • Velocity and Acceleration: v⃗(t)=−αωsin⁡(ωt)i^+βωcos⁡(ωt)j^\vec{v}(t) = -\alpha \omega \sin(\omega t) \hat{i} + \beta \omega \cos(\omega t) \hat{j} a⃗(t)=−ω2(αcos⁡(ωt)i^+βsin⁡(ωt)j^)=−ω2r⃗(t)\vec{a}(t) = -\omega^2 (\alpha \cos(\omega t) \hat{i} + \beta \sin(\omega t) \hat{j}) = -\omega^2 \vec{r}(t)

  • Conserved Quantities:

    1. Linear Momentum (p⃗\vec{p}): p⃗\vec{p} changes direction and magnitude over time. (Not Conserved)
    2. Angular Momentum (L⃗\vec{L}): Since force F⃗=ma⃗=−mω2r⃗\vec{F} = m\vec{a} = -m\omega^2 \vec{r} is a central force, the torque τ⃗=r⃗×F⃗=0⃗\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}. Thus: L⃗=r⃗(t)×mv⃗(t)=mαβωk^=constant. (Conserved)\vec{L} = \vec{r}(t) \times m\vec{v}(t) = m \alpha \beta \omega \hat{k} = \text{constant. \quad \textbf{(Conserved)}}
    3. Kinetic Energy (KK): K=12mω2(α2sin⁡2ωt+β2cos⁡2ωt)K = \frac{1}{2} m \omega^2 (\alpha^2 \sin^2\omega t + \beta^2 \cos^2\omega t) varies with time since α≠β\alpha \neq \beta. (Not Conserved)
    4. Potential Energy (UU): U=12mω2(α2cos⁡2ωt+β2sin⁡2ωt)U = \frac{1}{2} m \omega^2 (\alpha^2 \cos^2\omega t + \beta^2 \sin^2\omega t) varies with time since α≠β\alpha \neq \beta. (Not Conserved)
    5. Total Mechanical Energy (EE): Since F⃗\vec{F} is conservative, E=K+U=12mω2(α2+β2)=E = K + U = \frac{1}{2} m \omega^2 (\alpha^2 + \beta^2) = constant. (Conserved)

Q→2,5\mathbf{Q \rightarrow 2, 5}

Analysis for Path R:

r⃗(t)=α(cos⁡(ωt)i^+sin⁡(ωt)j^)\vec{r}(t) = \alpha (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j})

  • Velocity and Acceleration: v⃗(t)=αω(−sin⁡(ωt)i^+cos⁡(ωt)j^)\vec{v}(t) = \alpha \omega (-\sin(\omega t) \hat{i} + \cos(\omega t) \hat{j}) a⃗(t)=−ω2r⃗(t)\vec{a}(t) = -\omega^2 \vec{r}(t)

  • Conserved Quantities:

    1. Linear Momentum (p⃗\vec{p}): Direction of v⃗(t)\vec{v}(t) changes continuously. (Not Conserved)
    2. Angular Momentum (L⃗\vec{L}): L⃗=r⃗(t)×mv⃗(t)=mα2ωk^=\vec{L} = \vec{r}(t) \times m\vec{v}(t) = m \alpha^2 \omega \hat{k} = constant. (Conserved)
    3. Kinetic Energy (KK): K=12mα2ω2=K = \frac{1}{2} m \alpha^2 \omega^2 = constant. (Conserved)
    4. Potential Energy (UU): U=12mω2r2=12mω2α2=U = \frac{1}{2} m \omega^2 r^2 = \frac{1}{2} m \omega^2 \alpha^2 = constant. (Conserved)
    5. Total Mechanical Energy (EE): E=K+U=E = K + U = constant. (Conserved)

R→2,3,4,5\mathbf{R \rightarrow 2, 3, 4, 5}

Analysis for Path S:

r⃗(t)=αti^+β2t2j^\vec{r}(t) = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}

  • Velocity and Acceleration: v⃗(t)=αi^+βtj^\vec{v}(t) = \alpha \hat{i} + \beta t \hat{j} a⃗(t)=βj^  ⟹  F⃗=mβj^(constant force)\vec{a}(t) = \beta \hat{j} \implies \vec{F} = m \beta \hat{j} \quad (\text{constant force})

  • Conserved Quantities:

    1. Linear Momentum (p⃗\vec{p}): p⃗(t)=m(αi^+βtj^)\vec{p}(t) = m(\alpha \hat{i} + \beta t \hat{j}) changes with time. (Not Conserved)
    2. Angular Momentum (L⃗\vec{L}): L⃗=r⃗(t)×mv⃗(t)=12mαβt2k^\vec{L} = \vec{r}(t) \times m\vec{v}(t) = \frac{1}{2} m \alpha \beta t^2 \hat{k} changes with time. (Not Conserved)
    3. Kinetic Energy (KK): K=12m(α2+β2t2)K = \frac{1}{2} m (\alpha^2 + \beta^2 t^2) changes with time. (Not Conserved)
    4. Potential Energy (UU): U=−mβy=−12mβ2t2U = -m \beta y = -\frac{1}{2} m \beta^2 t^2 changes with time. (Not Conserved)
    5. Total Mechanical Energy (EE): Since the force F⃗=mβj^\vec{F} = m \beta \hat{j} is conservative: E=K+U=12mα2+12mβ2t2−12mβ2t2=12mα2=constant. (Conserved)E = K + U = \frac{1}{2} m \alpha^2 + \frac{1}{2} m \beta^2 t^2 - \frac{1}{2} m \beta^2 t^2 = \frac{1}{2} m \alpha^2 = \text{constant. \quad \textbf{(Conserved)}}

S→5\mathbf{S \rightarrow 5}

Conclusion:

  • P→1,2,3,4,5\text{P} \rightarrow 1, 2, 3, 4, 5
  • Q→2,5\text{Q} \rightarrow 2, 5
  • R→2,3,4,5\text{R} \rightarrow 2, 3, 4, 5
  • S→5\text{S} \rightarrow 5

This corresponds to Option A.

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