Questions
A block of mass M is suspended from a light spring of force constant k. Another mass m moving upwards with velocity v hits the mass M and gets embedded in it. What will be the amplitude of oscillation of the combined mass ?
detailed solution
Correct option is B
mv=(M+m)v′ or v′=mM+mv Further, maximum P.E. = maximum K.E. ∴ 12kA2=12(M+m)v′2 or A=M+mk1/2×V′ =M+mk1/2×mM+m×V=mv(M+m)kTalk to our academic expert!
Similar Questions
The identical springs of constant ‘K’ are connected in series and parallel as shown in figure A mass M is suspended from them. The ratio of their frequencies of vertical oscillations will be
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