/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 162 A block \(A\) of mass \(m\) conn... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 162

A block \(A\) of mass \(m\) connected with a spring of force constant \(k\) is executing SHM. The position \((x)\) and time \((t)\) equation of the block is \(x=x_{0}+a \sin \omega t\). An identical block \(B\) moving towards negative \(x\)-axis with velocity \(v_{0}\) collides elastically with block \(A\) at time \(t=0\). Then (A) Displacement time equation of \(A\) after collision will be \(x=x_{0}-v_{0} \sqrt{\frac{m}{k}} \sin \omega t\). (B) Displacement time equation of \(A\) after collision will be \(x=x_{0}+v_{0} \sqrt{\frac{m}{k}} \sin \omega t\). (C) Velocity of \(B\) just after collision will be \(a \omega\) towards positive \(x\)-direction. (D) Velocity of \(B\) just after collision will be \(v_{0}\) towards positive \(x\)-direction.

Short Answer

Expert verified
The correct options are (A) and (D). After the collision, the displacement-time equation of block A becomes \(x=x_{0}-a \omega \sqrt{\frac{m}{k}} \sin \omega t\), and the velocity of block B just after collision is \(v_{0}\) towards positive x-direction.

Step by step solution

01

Analyzing the Displacement-time Equation for Block A

Since block A is undergoing simple harmonic motion, the equation for its motion is given as \(x = x_{0} + a \sin \omega t\). After an elastic collision, the sign of the velocity of the block will change, which means in the displacement equation, the sign of \(a \sin \omega t\) term also changes. The velocity of block A at \(t=0\) can be given as \(v_{A}=a \omega\) from the equation for the instantaneous velocity in SHM. The term \(\sqrt{\frac{m}{k}}\) is the time period of SHM of block A, which appears in the equation for displacement in SHM. Thus the equation for the displacement of block A after collision will be \(x=x_{0}-v_{A} \sqrt{\frac{m}{k}} \sin \omega t = x_{0}-a \omega \sqrt{\frac{m}{k}} \sin \omega t\).
02

Compute the Velocity of Block B After the Collision.

Since the collision is elastic, we have conservation of momentum and kinetic energy. The initial momentum is \(P_{i}=mv_{0} - ma\omega\) and the initial kinetic energy is \(K_{i}=\frac{1}{2}m v_{0}^{2}\). After the collision, considering that the velocity of A has been reversed, the final momentum is \(P_{f}=mv_{A} + mv_{B}\) and the final kinetic energy is \(K_{f}= \frac{1}{2}m v_{A}^{2}+\frac{1}{2} m v_{B}^{2}\). Equating the initial and final momenta and solving the resulting equations, one obtains \(v_{B}=v_{0}\).

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