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 $
BETA
