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Question #43712

Difficulty: 8/10 | 🟢 Free

The Question

A spacecraft of mass MM (in kg\text{kg}) is traveling in deep space along the positive xx-direction with a velocity v0v_0 (in m/s\text{m/s}). It is propelled by a rocket engine that ejects exhaust gas at a constant mass rate μ=−dMdt\mu = -\frac{dM}{dt} (in kg/s\text{kg/s}) with a constant relative speed uu (in m/s\text{m/s}) in the negative xx-direction. At t=0t=0, the spacecraft encounters a localized 'fusion barrier' that exerts an instantaneous repulsive impulse of magnitude JJ (in N⋅s\text{N}\cdot\text{s}) in the negative xx-direction. Assuming the mass MM remains constant during the infinitesimally short duration of the impulse, what is the instantaneous change in the rate of change of the spacecraft's kinetic energy, Δ(dKdt)=dKdt∣t=0+−dKdt∣t=0−\Delta \left( \frac{dK}{dt} \right) = \left. \frac{dK}{dt} \right|_{t=0^+} - \left. \frac{dK}{dt} \right|_{t=0^-}?

Options

  • A.

    μJM(v0−u−J2M) W\frac{\mu J}{M} \left( v_0 - u - \frac{J}{2M} \right) \, \text{W}

  • B.

    −μuJM W-\frac{\mu u J}{M} \, \text{W}

  • C.

    μJM(u−v0−J2M) W\frac{\mu J}{M} \left( u - v_0 - \frac{J}{2M} \right) \, \text{W}

  • D.

    μJM(v0+u+J2M) W\frac{\mu J}{M} \left( v_0 + u + \frac{J}{2M} \right) \, \text{W}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Fusion Barriers|Recoil and Thrust Forces|Power-Velocity Relationship

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Work-Energy Theorem to a Variable Mass System (VMS).

A common trap is conflating the thrust power (P=Fthrustv=μuvP = F_{thrust} v = \mu u v) with the actual rate of change of the spacecraft's kinetic energy.

Students often incorrectly assume dKdt=Mvdvdt=μuv\frac{dK}{dt} = M v \frac{dv}{dt} = \mu u v, forgetting the crucial mass-loss term 12dMdtv2=−12μv2\frac{1}{2} \frac{dM}{dt} v^2 = -\frac{1}{2} \mu v^2.

This term accounts for the kinetic energy carried away by the exhaust mass in the ground frame.

When the impulsive barrier abruptly changes the spacecraft's velocity, the kinetic energy partition between the spacecraft and its exhaust shifts discontinuously.

Ignoring the mass-loss term leads directly to the incorrect option −μuJM-\frac{\mu u J}{M}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterNuclei - Class 12
ConceptFusion Barriers
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree