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Question #43690

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The Question

A radioactive sample of instantaneous mass $m$ is attached to a light horizontal spring of spring constant $k$ . The system oscillates on a smooth horizontal surface in a viscous medium that exerts a drag force $F_d = -cv$ , where $v$ is the instantaneous velocity of the sample and $c$ is a positive constant. The sample undergoes radioactive decay with a decay constant $\lambda$ , such that its mass decreases according to $\frac{dm}{dt} = -\lambda m$ . The decay products are continuously ejected such that they are left completely stationary with respect to the horizontal surface. If $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$ represents the instantaneous mechanical energy of the oscillating sample, which of the following expressions gives the rate of change of this mechanical energy, $\frac{dE}{dt}$ ?

Options

A.
$\left( \frac{\lambda m}{2} - c \right) v^2$
B.
$-\left( \frac{\lambda m}{2} + c \right) v^2$
C.
$-c v^2$
D.
$\left( \lambda m - c \right) v^2$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Radioactive Decay Law Dynamics|Mechanical Energy Conservation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Newton's Second Law to variable mass systems and its intersection with the Work-Energy Theorem.

Students commonly fall into the trap of writing $F_{net} = ma \implies m\frac{dv}{dt} = -kx - cv$ , ignoring the thrust force arising from the ejected mass.

This leads to the incorrect energy derivative $-\left(\frac{\lambda m}{2} + c\right)v^2$ .

Another common mistake is assuming the rate of change of mechanical energy is simply the power of the non-conservative drag force ( $-cv^2$ ), failing to account for the energy carried away by the lost mass and the work done by the thrust force.

The correct approach requires using $m\frac{dv}{dt} = F_{ext} + u_{rel}\frac{dm}{dt}$ to properly account for the momentum transfer.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Nuclei - Class 12
Concept	Radioactive Decay Law Dynamics
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free