Question

A 0.0397 kg bullet leave a rifle and embeds itself in a 5.24 kg stationary block of wood the wood bullet combination flies away with a velocity of 2.4 m/s what was the initial velocity of the bullet in m/s

Answer 1

The initial velocity of the bullet is approximately 12.9606 m/s.

Step-by-step explanation:

To solve this problem, we can apply the principle of conservation of momentum. The initial momentum of the bullet and the final momentum of the bullet-wood combination are equal.

Let's denote the initial velocity of the bullet as 'v'.

Using the equation:

(mass of the bullet) x (initial velocity of the bullet) = (mass of the bullet-wood combination) x (final velocity of the bullet-wood combination)

Substituting the given values:

(0.0397 kg)(v) = (0.0397 kg + 5.24 kg)(2.4 m/s)

Simplifying the equation and solving for 'v', we get:

v ≈ 12.9606 m/s

So, the initial velocity of the bullet is approximately 12.9606 m/s.