For a spring block system as shown in figure, a time varying force F=5tN is applied on mass 2kg. After 10seconds, velocity of 3kg mass is 30m/s. Find the velocity of 2kg mass at this instant.
A
40m/s
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B
80m/s
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C
803m/s
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D
20m/s
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Solution
The correct option is B80m/s From the fundamental definition: Fext=macom....(i)
Considering (blocks+spring) as system Fext=5tN.
Spring force is an internal force for the system, hence velocity of centre of mass at that instant can be given as, 5t=(msys)dvcomdt
Integrating w.r.t time with limits t=0 to t=10s (msys)∫vcom0dvcom=∫1005tdt ⇒(3+2)vcom=[5t22]100 ∴vcom=510×(102−0) =50m/s...(ii)
Considering both blocks to be moving along direction of force F, at t=10s, velocity of 3kg mass, v1=30m/s and let velocity of 2kg mass be v2m/s. ⇒vcom=m1v1+m2v2m1+m2...(iii)
From Eq. (ii) and (iii), 50=(3×30)+(2×v2)3+2 ∴v2=250−902=80m/s