Question

A small block of mass $100\text{g}$ is pressed against a horizontal spring fixed at one end to compress it by $5\text{cm}$. The spring constant is $100\text{N/m}$. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground ?
$(\text{Take,}h=2\text{m},g=10{\text{m/s}}^2)$

• A. $0.5\text{m}$
• B. $1\text{m}$
• C. $2\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is B $1\text{m}$

Given:
$m=100\text{g};x=5\text{cm};K=100\text{N/m}$
$h=2\text{m};g=10{\text{m/s}}^2$

Let the speed of the block is $v$, when it leaves the spring.

The total energy of the spring block system should remain constant.

$\therefore \frac{1}{2}K{x}^2=\frac{1}{2}m{v}^2$

$100(0.05{)}^2=0.1v^2$

$v^2=\frac{100}{0.1}\times 25\times {10}^{-4}$

$\therefore v=\sqrt{2.5}\text{m/s}$

When the block loses contact with the spring, it performs horizontal projectile motion as shown below.

So, range of projectile motion is given by,
$\text{Range}\left(R\right)=v\times \text{time of flight}\left(T\right)$

$R=v\sqrt{\frac{2h}{g}}=\sqrt{2.5}\times \sqrt{\frac{2\times 2}{10}}=\sqrt{\frac{25\times 4}{10\times 10}}$

$\therefore R=1\text{m}$

Therefore, option $\left(B\right)$ is correct.