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Question #42485

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass mm is attached to a horizontal ideal spring of force constant kk and rests at equilibrium (x=0x = 0) on a frictionless surface. A bullet of mass mm moving with a horizontal velocity v0v_0 strikes the block and embeds within it instantaneously. The potential energy U(x)U(x) of the spring-mass system is defined such that U(x0)=0U(x_0) = 0, where x0≠0x_0 \neq 0. If EE represents the total mechanical energy of the system after the collision, which of the following equations correctly describes the phase space trajectory (pp vs xx) of the resulting simple harmonic motion? (Here, pp is the momentum of the combined system).

Options

  • A.

    p24m(E+12kx02)+kx22(E+12kx02)=1\frac{p^2}{4m\left(E + \frac{1}{2}kx_0^2\right)} + \frac{kx^2}{2\left(E + \frac{1}{2}kx_0^2\right)} = 1

  • B.

    p22m(E+12kx02)+kx22(E+12kx02)=1\frac{p^2}{2m\left(E + \frac{1}{2}kx_0^2\right)} + \frac{kx^2}{2\left(E + \frac{1}{2}kx_0^2\right)} = 1

  • C.

    p24mE+kx22E=1\frac{p^2}{4mE} + \frac{kx^2}{2E} = 1

  • D.

    p24m(E−12kx02)+kx22(E−12kx02)=1\frac{p^2}{4m\left(E - \frac{1}{2}kx_0^2\right)} + \frac{kx^2}{2\left(E - \frac{1}{2}kx_0^2\right)} = 1

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Potential Energy Reference|Phase Space Trajectory|Simple Pendulum Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definition of phase space trajectories and potential energy reference frames.

Students often memorize the standard phase space ellipse equation p22mE+x22E/k=1\frac{p^2}{2mE} + \frac{x^2}{2E/k} = 1, which implicitly assumes U=0U=0 at the equilibrium position x=0x=0.

By shifting the zero-potential reference to x=x0x=x_0, the total energy EE is offset by −12kx02-\frac{1}{2}kx_0^2.

Furthermore, the problem incorporates an inelastic collision to test if students correctly update the system's mass to 2m2m in the kinetic energy term p2/2Mp^2/2M.

Memorizers will incorrectly select the standard formula or fail to account for the mass change.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptPotential Energy Reference
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree