Question

A light spring of length $l$ 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

• A. $\frac{l}{k}$
• B. $l-\frac{m g}{k}$
• C. $l+\frac{k}{m}$
• D. $l-k m g$

Answer 1

Answer: (B)

As the ball of mass m falls on the spring, the spring gets compressed by say $\Delta x$. At this moment, the net downward force $=m g-k \Delta x$ As long this force is downward, the acceleration is positive, the velocity of the ball goes on increasing. It will become maximum, when $a=0$. Thereafter $a$ reverses direction and $v$ starts decreasing. $a=0 \Rightarrow F_{\text {net }}=0$
$m g-k \Delta x=0$ or $\Delta x=\frac{m g}{k}$
At this moment, height of the ball from the table
$=l-\Delta x=l-\frac{m g}{k}$