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Motion question

2008 · Shift 2 · Q45
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Motion question

2008 · Shift 2 · Q45

JEE AdvancedPhysicsMotionMCQ+3 / −1
STATEMENT-1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary. STATEMENT-2: If the observer and the object are moving at velocities v→1{\overrightarrow v _1}v1​ and v→2{\overrightarrow v _2}v2​ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is v→2{\overrightarrow v _2}v2​-v→1{\overrightarrow v _1}v1​.
  1. A
    STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT ,- 1
  2. B
    STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1
  3. C
    STATEMENT - 1 is True, STATEMENT - 2 is False
  4. D
    STATEMENT - 1 is False, STATEMENT - 2 is True
View written solutionFree

Correct answer: B

Analysis of Statement-1

STATEMENT-1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.

  1. Nearby Objects: Let the velocity of the train (and the observer) with respect to the ground be v⃗T\vec{v}_{T}vT​. Let a nearby object (e.g., a tree) be stationary with respect to the ground, so its velocity is v⃗obj=0\vec{v}_{obj} = 0vobj​=0. The velocity of the object with respect to the observer is given by the relative velocity formula: v⃗obj,obs=v⃗obj−v⃗T=0−v⃗T=−v⃗T\vec{v}_{obj, obs} = \vec{v}_{obj} - \vec{v}_{T} = 0 - \vec{v}_{T} = -\vec{v}_{T}vobj,obs​=vobj​−vT​=0−vT​=−vT​ The negative sign indicates that the object appears to move in the direction opposite to the train's motion. This part of the statement is correct.

  2. Distant Objects: The perception of motion depends not just on the linear relative velocity but on the angular velocity of the object with respect to the observer. The angular velocity, ω\omegaω, is the rate at which the line of sight to the object rotates. The relationship between linear velocity (v⊥v_{\perp}v⊥​, the component perpendicular to the line of sight) and angular velocity is ω=v⊥r\omega = \frac{v_{\perp}}{r}ω=rv⊥​​, where rrr is the distance to the object.

  3. For both a nearby object and a distant object (like a far-off mountain), if they are stationary on the ground, their linear velocity relative to the observer is the same, i.e., −v⃗T-\vec{v}_{T}−vT​.

  4. However, the distance rrr to these objects is very different. Let rnearr_{near}rnear​ be the distance to the nearby object and rdistantr_{distant}rdistant​ be the distance to the distant object. We have rdistant≫rnearr_{distant} \gg r_{near}rdistant​≫rnear​.

  5. The angular velocity of the nearby object is ωnear≈∣v⃗T∣rnear\omega_{near} \approx \frac{|\vec{v}_{T}|}{r_{near}}ωnear​≈rnear​∣vT​∣​. The angular velocity of the distant object is ωdistant≈∣v⃗T∣rdistant\omega_{distant} \approx \frac{|\vec{v}_{T}|}{r_{distant}}ωdistant​≈rdistant​∣vT​∣​.

  6. Since rdistantr_{distant}rdistant​ is very large, ωdistant\omega_{distant}ωdistant​ is very small, approaching zero. Our eyes perceive this very small angular velocity as the object being stationary.

  7. Therefore, Statement-1 is a correct description of the phenomenon of parallax and is factually true.

Analysis of Statement-2

STATEMENT-2: If the observer and the object are moving at velocities v→1{\overrightarrow v _1}v1​ and v→2{\overrightarrow v _2}v2​ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is v→2{\overrightarrow v _2}v2​-v→1{\overrightarrow v _1}v1​.

  1. This is the fundamental definition of relative velocity in classical mechanics.
  2. Let r⃗1\vec{r}_1r1​ be the position vector of the observer and r⃗2\vec{r}_2r2​ be the position vector of the object in a laboratory frame.
  3. The position of the object with respect to the observer is r⃗21=r⃗2−r⃗1\vec{r}_{21} = \vec{r}_2 - \vec{r}_1r21​=r2​−r1​.
  4. Differentiating with respect to time to find the relative velocity: v⃗21=dr⃗21dt=dr⃗2dt−dr⃗1dt=v⃗2−v⃗1\vec{v}_{21} = \frac{d\vec{r}_{21}}{dt} = \frac{d\vec{r}_2}{dt} - \frac{d\vec{r}_1}{dt} = \vec{v}_2 - \vec{v}_1v21​=dtdr21​​=dtdr2​​−dtdr1​​=v2​−v1​
  5. Thus, Statement-2 is a correct physical principle.

Relationship between Statement-1 and Statement-2

  1. We have established that both Statement-1 and Statement-2 are true.
  2. Now we must determine if Statement-2 is the correct explanation for Statement-1.
  3. Statement-2 explains why nearby objects appear to move backward. Using the formula from Statement-2, the relative velocity of any stationary object (near or far) is −v⃗T-\vec{v}_{T}−vT​.
  4. However, Statement-2 alone does not explain why distant objects appear stationary. According to Statement-2, a distant stationary mountain has the same linear relative velocity (−v⃗T)(-\vec{v}_{T})(−vT​) as a nearby tree. This contradicts the observation that the mountain appears stationary.
  5. The full explanation for Statement-1 requires not only the concept of relative linear velocity (from Statement-2) but also the concept of angular velocity and how it depends on distance (rrr). The apparent stationarity of distant objects is due to their very small angular velocity, which Statement-2 does not address.
  6. Therefore, while Statement-2 is a true statement and is a part of the explanation, it is not the complete or correct explanation for the entirety of Statement-1.

Conclusion

  • Statement-1 is True.
  • Statement-2 is True.
  • Statement-2 is NOT a correct explanation for Statement-1.

This corresponds to option B.

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