Question

One end Of a spring initially in natural length $l_0$ lying on a horizontal frictionless table is fixed and the Other end is fastened to a small puck of mass $m.$ The puck is given velocity in a direction perpendicular to the spring of magnitude $V_0$. In the course of motion the maximum elongation of the spring is $\frac{l_0}{10}$. The force constant of the spring is: $\text{(Take}\frac{m{v}_0^2}{l_0^2}=121\text{units})$

Answer 1

At the fixed end the net torque will be zero as the line of action of the forces passes through it. Since , the net torque is zero angular momentum will remain conserved Also at maximum elongation, vet of izluck will be tangent to the spring . So

$m{v}_{0}l=mv(l_0+\frac{l_0}{10})$

$\Rightarrow v=\frac{10}{11}{V}_0\left(\frac{l_0}{l_0}\right)=\frac{10V_0}{11}$

$\Delta KE=\frac{1}{2}m[V^2-V_0^2]=\frac{m}{2}[\frac{100V_0^2}{14}-V_0^2]$

$\Delta KE=\left(\frac{-21}{242}\right)m{v}_0^2$

$\Delta PE=\frac{K}{2}[{\left(\frac{1}{10}{l}_o\right)}^2-(0{)}^2]=\frac{K{l}_0^2}{200}$

As the spring force is conservation in nature

$\Delta KE+\Delta PE=0$

$\Rightarrow \frac{-21}{242}m{v}_0^2+\frac{K{l}_0^2}{200}=0$

$\Rightarrow K=\frac{\left(21\right)\left(200\right)}{\left(242\right)}\left[\frac{m{v}_0^2}{l_0^2}\right]$

$K=\frac{\left(21\right)}{\left(242\right)}\left(200\right)\left(121\right)=2.1KN/m$