A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5 cm (figure following). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?

#### Solution

\[\text{Given}: \]

\[\text{ Mass of the block, m = 100 g = 0 . 1 kg } \]

\[\text{ Compression in the spring, x = 5 cm = 0 . 05 m }\]

\[\text{ Spring constant, k = 100 N/m }\]

Let *v* be the velocity of the block when it leaves the spring.

Applying the law of conservation of energy,

Elastic potential energy of the spring = Kinetic energy of the block

\[\frac{1}{2}\text{m} \nu^2 = \frac{1}{2}\text{kx}^2 \]

\[ \Rightarrow \nu = \text{x}\sqrt{\frac{\text{k}}{\text{m}}}\]

\[ = \left( 0 . 05 \right) \times \sqrt{\frac{\left( 100 \right)}{0 . 1}}\]

\[ = 1 . 58 \text{ m/s}\]

For the projectile motion,

\[\theta = 0^\circ, \text{ y } = - 2\]

\[\text{ Now, y }= \left( \text{ u } \cdot \sin \theta \right) \text{ t } - \frac{1}{2}\text{gt}^2 \]

\[ - 2 = \left( - \frac{1}{2} \right) \times \left( 9 . 8 \right) \times \text{t}^2\]

\[\Rightarrow \text{t = 0 . 63 \sec,} \]

\[\text{ So, x }= \left( \text{u \cos } \theta \right) \text{ t }\]

\[ = \left( 1 . 58 \right) \times \left( 0 . 36 \right) = 1 \text{ m } \]

Therefore, the block hits the ground at 1 m from the free end of the spring in the horizontal direction.