BITSAT 2024 Shift 1 Paper

To find the speed of the mass just before it touches the spring, we begin by determining the spring constant, $k$, using Hooke's Law, where the force $F$ is proportional to the compression $x$ :$F=kx$Given that a force of 100 N compresses the spring by 1 m , the spring constant $k$ is calculated as:$k=\;\frac{F}{x}=\;\frac{100\ \mathrm{N}}{1\ \mathrm{m}}=100\ \mathrm{N}/\mathrm{m}$Next, we apply the principle of energy conservation. The initial potential and kinetic energy will equal the final potential energy stored in the compressed spring. The relevant energy conservation equation is:$\;\frac{1}{2}mv^2+mgh=\;\frac{1}{2}kx_{\max}^2$where:$m$ is the mass of the block ( 10 kg ),$g$ is the acceleration due to gravity $(10\ \text{m}/\text{s}^2)$,$h$ is the vertical height change, which is the same as the compression in this scenario ( $x_{\max}=2\ \text{m}$ ),$v$ is the speed just before touching the spring,$x_{\text{max}}$ is the maximum compression of the spring ( 2 m ).Rearranging the equation for $v$ gives:$v=\sqrt{\;\frac{k x_{\max}^2}{m}-2gh}$Substituting the known values:$v=\sqrt{\;\frac{(100\ \mathrm{N/m})(2\ \mathrm{m})^2}{10\ \mathrm{kg}}-2(10\ \mathrm{m/s}^2)\frac{2}{2}}$This simplifies to:$v=\sqrt{\;\frac{(100)(4)}{10}-40}$ $v=\sqrt{40-20}=\sqrt{20}\ \text{m/s}$Thus, the speed of the mass just before it touches the spring is $\sqrt{20}\ \text{m/s}$