Question

A spring of natural length $l$ and spring constant $50\, N/m$ is kept on a horizontal frictionless table with one end attached to a rigid support. First the spring was compressed by $10\, cm$ and then released to hit a ball of mass $20\, g$ kept at a distance l from the rigid support. if after hitting the ball, the natural length of the spring is restored, what is the speed with which the ball moved? (Ignore the air resistance)

• A. 5 m/s
• B. 7 m/s
• C. 25 m/s
• D. 50 m/s
• E. 2500 m/s

Answer 1

Answer: (A)

A spring of spring constant $50\, N / m$ is attached to a rigid wall. A $20\, g$ ball is placed on the horizontal table as shown in figure

If spring is compressed then potential energy
$U=\frac{1}{2} k x^{2}$
When spring hit the ball, it transfer all its potential energy to kinetic energy of ball
$ \frac{1}{2} k x^{2}= KE _{ ball } $
$\Rightarrow \frac{1}{2} k x^{2} =\frac{1}{2} m v^{2} $
$\Rightarrow v=\sqrt{\frac{k x^{2}}{m}}$
Given, $x=20\,cm , m=20\, g$ and $v=50\, N / m$
$ \Rightarrow v =\sqrt{\frac{50 \times\left(20 \times 10^{-2}\right)^{2}}{20 \times 10^{-3}}} $
$=\sqrt{25}=5\, m / s $