question_answer

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

A)  0.01 m 

B)                                         0.02m

C)  0.05m                  

D)         0.03m

Correct Answer: C

Solution :

As the block\[A\]moves with velocity\[0.15\,\,m{{s}^{-1}}\], it compresses the spring which pushes\[B\]towards right.\[A\]goes on compressing the spring till the velocity acquired by\[B\]becomes equal to the velocity of\[A\],\[ie\],\[0.15\,m{{s}^{-1}}\]. Let the 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                 \[{{m}_{A}}u=({{m}_{A}}+{{m}_{B}})v\] 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.                 \[\frac{1}{2}{{m}_{A}}{{u}^{2}}=\frac{1}{2}({{m}_{A}}+{{m}_{B}}){{v}^{2}}=\frac{1}{2}k{{x}^{2}}\]                 \[\frac{1}{2}{{m}_{A}}{{u}^{2}}-\frac{1}{2}({{m}_{A}}+{{m}_{B}}){{v}^{2}}=\frac{1}{2}k{{x}^{2}}\]                 \[\frac{1}{2}\times 2\times {{(0.15)}^{2}}-\frac{1}{2}(2+3){{(0.06)}^{2}}=\frac{1}{2}k{{x}^{2}}\]                 \[0.0225-0.009=\frac{1}{2}k{{x}^{2}}\,\,or\,\,0.0135=\frac{1}{2}k{{x}^{2}}\] or            \[x=\sqrt{\frac{0.027}{k}}=\sqrt{\frac{0.027}{10.8}}=0.05m\]