Rotational Motion Question 74

Question: A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity $ 4m{s^{-1}} $ . It collides with a horizontal spring of force constant $ 200N{m^{-1}} $ . The maximum compression produced in the spring will be

[AIPMT (S) 2012]

Options:

A) 0.5 m

B) 0.6 m

C) 0.7 m

D) 0.2 m

Show Answer

Answer:

Correct Answer: B

Solution:

  • Loss in $ KE= $ Gain in spring energy

    $ \frac{1}{2}mv^{2}[ 1+\frac{K^{2}}{R^{2}} ]=\frac{1}{2}kx _{\max }^{2} $

    where k is the force constant.

    Given, $ v=4m/s,m=3kg,k=200N/m $

    For solid cylinder, $ \frac{K^{2}}{R^{2}}=\frac{1}{2} $

    $ \therefore $ $ \frac{1}{2}\times 3\times {{(4)}^{2}}[ 1+\frac{1}{2} ]=\frac{1}{2}\times 200\times x _{\max }^{2} $

    The maximum compression in the spring

    $ x _{\max }^{{}}=0.6m $



NCERT Chapter Video Solution

Dual Pane