A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40 J within 1 s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is

We need to find the minimum length of the swimming pool that a bullet must travel so that it completely comes to rest, given its initial kinetic energy, its reduced kinetic energy after a set time, and its mass.

1. Find the Initial and Final Velocities

Kinetic energy ($$\text{KE}$$) is defined by the formula:

$$\text{KE} = \frac{1}{2}mv^2$$

Given the mass of the bullet, $$m = 200\text{ g} = 0.2\text{ kg}$$.

• Initial velocity ($$u$$) where $$\text{KE}_{\text{initial}} = 90\text{ J}$$: $$90 = \frac{1}{2} \times 0.2 \times u^2$$ $$90 = 0.1 \times u^2 \implies u^2 = 900 \implies u = 30\text{ m/s}$$
• Velocity ($$v$$) after $$t = 1\text{ s}$$ where $$\text{KE}_{\text{final}} = 40\text{ J}$$: $$40 = \frac{1}{2} \times 0.2 \times v^2$$ $$40 = 0.1 \times v^2 \implies v^2 = 400 \implies v = 20\text{ m/s}$$

2. Find the Retardation (Acceleration)

Using the first equation of motion ($$v = u + at$$) for the time interval of $$1\text{ s}$$:

$$20 = 30 + a(1)$$

$$a = 20 - 30 = -10\text{ m/s}^2$$

The retardation offered by the water in the swimming pool is $$10\text{ m/s}^2$$.

3. Calculate the Minimum Length of the Pool to Stop the Bullet

To find the total distance ($$s$$) required for the bullet to completely come to rest, its final velocity at rest ($$v_{\text{rest}}$$) must be $$0\text{ m/s}$$.

Using the third equation of motion:

$$v_{\text{rest}}^2 = u^2 + 2as$$

$$0^2 = 30^2 + 2(-10)s$$

$$0 = 900 - 20s$$

$$20s = 900 \implies s = \frac{900}{20} = 45\text{ m}$$

Therefore, the minimum length of the pool the bullet has to travel is 45 m, which corresponds to Option A.