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Question

Physics Question on System of Particles & Rotational Motion

A solid cylinder of mass 3kg3\, kg is rolling on a horizontal surface with velocity 4ms14\, ms^{-1}. It collides with a horizontal spring of force constant 200Nm1200\, N\, m^{-1}. The maximum compression produced in the spring will be

A

0.5 m

B

0.6 m

C

0.7 m

D

0.2 m

Answer

0.6 m

Explanation

Solution

12mv2[1+K2R2]=12kxmax2\frac{1}{2} mv ^{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right]=\frac{1}{2} kx _{\max }^{2}
Putting m=3kg,v=4m/s,K2R2=1/2m =3 kg , v =4 m / s , \frac{ K ^{2}}{ R ^{2}}=1 / 2
(For solid cylinder), k=200Nm1k =200\, Nm ^{-1} we get
xmax=0.6mx _{\max }=0.6\, m