Laws Of Motion Question 422
Question: A light spring of length I and spring constant k is placed vertically on a table. A small ball of mass m falls on it. The height h from the surface of the table at which the ball will have the maximum velocity is
Options:
A) $ \frac{l}{k} $
B) $ l-\frac{mg}{k} $
C) $ l+\frac{k}{mg} $
D) $ l-kg $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Equation of motion for the ball at the moment when the spring is compressed by $ \Delta x, $
$ ma=mg-k\Delta x $
As long as the acceleration of the ball is positive, its velocity increases. At the moment, when the acceleration vanishes, the velocity of the ball attains the maximum value.
The spring is then compressed by an amount $ \Delta l $ given by $ mg-k\Delta l=0 $ or $ \Delta l=\frac{mg}{k} $ So, when ball attains the maximum velocity, its height from the table $ h=l-\frac{mg}{k} $
BETA
