Two smooth blocks of mases `m_(1) and m_(2)` attached with an ideal spring of stiffness `k` and kept on hrozontal surface. If `m_(1)` is projected with a horizontal velocity `v_(0)`. Find the maximum compression of the spring.
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Method 1 (using ground frame method) Decide system: Blocks and spring Observation: no external force acting on system in horizontal direction. Conclusion: linear momentum as well as mechanical energy of system will be move with common velocity. Problem solving: Using conservation of linear momentum at initial time and at time of maximum compression of spring. Let the common velocity of system be `V`. `p_(i)=m_(1)v_(0)+0` and `p_(f)=(m_(1)+m_(2))V` As `p_(i)=p_(f)implies m_(1)v_(0)=(m_(1)+m_(2))V`.............i Now using conservation of mechanical energy. `/_\K+/_\U=0` `(K_f-Ki)+/_\U_("spring")=0` `[1/2(m_(1)+m_(2))V^(2)-1/2m_(1)v_(0)^(2)]+1/2kx^(2)=0`...........i From i and ii we get `x=[sqrt((m_(1)m_(2))/((m_(1)_m_(2))k))]v_(0)` Method 2: (work energy theorem from centre of mass frame) The linear momentum of system from CM frame is zero and no external force is acting on the system. We can apply work energy theorem from CM frame `W_(ext)+W_(int)=/_\K_(cm)=(K_(f)-K_(i))_cm` .............iii Initial kinetic energy of system w.r.t CM `(K_(i))_(cm)=1/2((m_(1)m_(2))/(m_(1)+m_(2)))v_0^(2)` Final kinetic energy of system w.r.t CM `(K_(f))_(cm)=1/2mu(v_("net"))^(2)_("final")=0` as both the blocks move with common velocity hence `(v_("net"))_("final")=0` which gives `(K_(f))_(cm)=0` Hence from 1 `0+(-1/2kx^(2))0-1/2((m_(1)m_(2))/(m_(1)+m_(2)))v_(0)^(2)` `impliesx=[sqrt((m_(1)m_(2))/((m_(1)+m_(2)))k)]v_(0)`
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