Two blocks A(3kg) and B(2kg) resting on a smooth horizontal surface is connected by a spring of stiffness 480N/m. Initially the spring is undefined and a velocity of 2m/s is imparted to A along the line of the spring away from B .The max extension in meters of the spring subsequent motion is
Asked by m.nilu
18th September 2018,
10:24 AM
Answered by Expert
Answer:
Initially, only block-A is moving with velocity 2 m/s.
Hence initial momentum = 3×2 = 6 kg-m/s ..................(1)
At time of maximum compression of spring, let the common velocity be v m/s
Hence final momentum = (3+2)×v = 5×v kg-m/s ................(2)
By conservation of momentum, we equate (1) and (2) to get common velocity v = 1.2 m/s
Initial kinetic energy = (1/2)×3×2×2 = 6 Joules ................(3)
Final kinetic energy = (1/2)×5×1.2×1.2 = 3.6 Joules ..............(4)
difference in kinetic energy is stored as potential energy in spring,
i.e., (1/2)kx2 = (1/2)×480×x2 = 6 - 3.6 = 2.4 Joules ..........(5)
from (5), we get the maximum displacement, x = 0.1 m
×v
Answered by Expert
27th September 2018,
11:34 PM
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