A car of mass $1000 \mathrm{~kg}$ having a velocity of $10 \mathrm{~ms}^{-1}$ collides a horizontally…

A car of mass $1000 \mathrm{~kg}$ having a velocity of $10 \mathrm{~ms}^{-1}$ collides a horizontally mounted spring. If the spring constant is $4000 \mathrm{Nm}^{-1}$, then the maximum compression of the spring is
  1. $25 \mathrm{~m}$
  2. $15 \mathrm{~m}$
  3. $5 \mathrm{~m}$
  4. $10 \mathrm{~m}$

Solution

Mass of car, $m=1000 \mathrm{~kg}$ Velocity, $v=10 \mathrm{~m} / \mathrm{s}$ Spring constant, $\mathrm{k}=4000 \mathrm{Nm}^{-1}$ Apply the conservation of energy $\frac{1}{2} m v^2=\frac{1}{2} k(\Delta x)^2$ $\Delta \mathrm{x}=\sqrt{\frac{\mathrm{mv}^2}{\mathrm{k}}}=\sqrt{\frac{1000 \times 100}{4000}}$ Compression of the spring $\Delta x=5 \mathrm{~m}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

Practice more Work Power Energy questions on Aicharya