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

Hint:

On a frictionless surface, all stored spring energy converts into kinetic energy.

Answer 1

The Correct Option is A

Concept:
Energy conservation: \[ \text{Spring PE} = \text{Kinetic Energy} \] \[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]

Step 1: Convert all units properly
\[ x = 10 \text{ cm} = 0.1 \text{ m}, \quad m = 20g = 0.02 \text{ kg} \]

Step 2: Substitute values
\[ \frac{1}{2}(50)(0.1)^2 = \frac{1}{2}(0.02)v^2 \]

Step 3: Solve step by step
\[ 25 \times 0.01 = 0.01 v^2 \] \[ 0.25 = 0.01 v^2 \] \[ v^2 = 25 \]

Step 4: Final answer
\[ v = 5 \text{ m/s} \] \[ \boxed{5 \text{ m/s}} \]