question_answer

Two rectangular \[0.15\text{ }m{{s}^{-1}}\] blocks A and B of masses 2 kg and 3 kg respectively are connected by a spring of spring constant \[10.8\text{ }N{{m}^{-1}}\] and are placed on a fricrionless horizontal surface. The block A was given an initial velocity of \[0.15\text{ }m{{s}^{-1}}\] in the direction shown in the figure. The maximum compression of the spring during motion is

A)  0.01 m                                 

B)  0.02 m

C)  0.05 m                                 

D)  0.03 m

Correct Answer: C

Solution :

As the block A moves with velocity \[0.15\text{ }m{{s}^{-1}}\],  spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, ie, \[0.15\text{ }m{{s}^{-1}}\]. Let this velocity be v. Now, spring is in a state of maximum compression. Let \[x\] be the maximum compression at this stage. According to the law of conservation of linear momentum, we get                          \[\xrightarrow{0.15\,\,m{{s}^{-1}}}\]                                 \[{{m}_{A}}u=({{m}_{A}}+{{m}_{B}})v\]                 or            \[v=\frac{{{m}_{A}}u}{{{m}_{A}}+{{m}_{B}}}\]                                 \[=\frac{2\times 0.15}{2+3}=0.06\,m{{s}^{-1}}\] According to the law of conservation of energy.                 \[\frac{1}{2}{{m}_{A}}{{u}^{2}}=\frac{1}{2}({{m}_{A}}+{{m}_{B}}){{v}^{2}}+\frac{1}{2}k{{x}^{2}}\]                 \[\frac{1}{2}\,{{m}_{A}}{{u}^{2}}-\frac{1}{2}\,({{m}_{A}}+{{m}_{B}}){{v}^{2}}=\frac{1}{2}\,k{{x}^{2}}\] \[\frac{1}{2}\times 2\times {{(0.15)}^{2}}-\frac{1}{2}\,(2+3)\,{{(0.6)}^{2}}=\frac{1}{2}k{{x}^{2}}\]      \[0.0225-0.09=\frac{1}{2}\,k{{x}^{2}}\]or \[0.0135=\frac{1}{2}\,k{{x}^{2}}\] or            \[x=\sqrt{\frac{0.027}{k}}=\sqrt{\frac{0.027}{10.8}}=0.05\,m\]