JEE Advanced 2024 Paper 1 Q04 Physics Mechanics Spring-Block Medium

JEE Advanced 2024 Paper 1 · Q04 · Spring-Block

A block of mass $5\,\text{kg}$ moves along the $x$-direction subject to the force $F = (-20x + 10)\,\text{N}$, with the value of $x$ in metre. At time $t = 0\,\text{s}$, it is at rest at position $x = 1\,\text{m}$. The position and momentum of the block at $t = (\pi/4)\,\text{s}$ are

  1. A. $-0.5\,\text{m},\ 5\,\text{kg m/s}$
  2. B. $0.5\,\text{m},\ 0\,\text{kg m/s}$
  3. C. $0.5\,\text{m},\ -5\,\text{kg m/s}$
  4. D. $-1\,\text{m},\ 5\,\text{kg m/s}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Equation of motion: $m\ddot x = -20x + 10 \Rightarrow \ddot x = -4(x - \tfrac{1}{2})$. Letting $u = x - \tfrac{1}{2}$, $\ddot u = -4u$ — SHM with $\omega = 2\,\text{rad/s}$ about the equilibrium $x_{eq} = 0.5\,\text{m}$. Initial conditions: $x(0) = 1$, $\dot x(0) = 0$, so $u(0) = 0.5$, $\dot u(0) = 0$. Therefore $u(t) = 0.5\cos(2t)$, i.e., $x(t) = \tfrac{1}{2} + \tfrac{1}{2}\cos(2t)$, $\dot x(t) = -\sin(2t)$. At $t = \pi/4$: $x = \tfrac{1}{2} + \tfrac{1}{2}\cos(\pi/2) = 0.5\,\text{m}$, $\dot x = -\sin(\pi/2) = -1\,\text{m/s}$. Momentum $p = m\dot x = 5 \cdot (-1) = -5\,\text{kg·m/s}$.

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