A body of mass $0.15 \mathrm{~kg}$ moving with a velocity of $15 \mathrm{~ms}^{-1}$ comes to rest, when it…

A body of mass $0.15 \mathrm{~kg}$ moving with a velocity of $15 \mathrm{~ms}^{-1}$ comes to rest, when it hits a spring that is fixed at another end. If the force constant of the spring is $1500 \mathrm{Nm}^{-1}$, then the compression in the spring is
  1. $0.15 \mathrm{~m}$
  2. $0.1 \mathrm{~m}$
  3. $0.2 \mathrm{~m}$
  4. $0.5 \mathrm{~m}$

Solution

Given, mass of the body, $m=0.15 \mathrm{~kg}$ Velocity of the body, $v=15 \mathrm{~ms}^{-1}$ Spring constant, $k=1500 \mathrm{Nm}^{-1}$ According to law of conservation of energy, the change in kinetic energy is equal to the change in potential energy, i.e. $ \begin{aligned} \frac{1}{2} m v^2-0 & =0+\frac{1}{2} k x^2 \\ \Rightarrow \quad x^2 & =\frac{m}{k} v^2 \Rightarrow x=v \sqrt{\frac{m}{k}}=15 \times \sqrt{\frac{0.15}{1500}} \\ & =0.15 \mathrm{~m} \end{aligned} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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