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 Nm-1 and are placed on a frictionless horizontal surface. The block A was given an initial velocity of 0.15 ms-1 in the direction shown in the figure. The maximum compression of the spring during the 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 \[0.15m{{s}^{-1}},\]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, \[0.15m{{s}^{-1}},\] 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 \[{{m}_{A}}u=({{m}_{A}}+{{m}_{B}})v\] 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\[x=\sqrt{\frac{0.027}{k}}k=\sqrt{\frac{0.027}{10.8}}=0.05m\]