question_answer

A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant \[k=50\,n/m.\]The maximum compression of the spring would be:

A)  0.15 m

B)  0.12 m

C)  1.5 m                          

D)  0.5 m

Correct Answer: A

Solution :

 By the law of conservation of energy, i.e., \[\Delta \Kappa +\Delta U=0\] \[\left( 0-\frac{1}{2}m{{v}^{2}} \right)+\left( \frac{1}{2}k{{x}^{2}}-0 \right)=0\] \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}=\frac{m{{v}^{2}}}{k}\] \[\Rightarrow \]\[x=\sqrt{\left( \frac{m{{v}^{2}}}{k} \right)}\] \[\Rightarrow \]\[x=\sqrt{\left( \frac{0.5\times 1.5\times 1.5}{50} \right)}\] \[\therefore \] = 0.15 m