Final answer:
To find the speed of the block after a bullet becomes embedded in it, we use the conservation of momentum formula. With the mass of the bullet at 6.00 grams and its speed at 113 m/s, and the mass of the block at 2.00 kg, the final velocity of the block is calculated to be 0.339 m/s.
Step-by-step explanation:
The scenario describes a classic physics problem involving conservation of momentum when two objects, a bullet and a block, collide and become one united mass. We can find the final velocity of this combined mass by using the formula for conservation of linear momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. The formula is given by m1×v1 + m2×v2 = (m1+m2)×v_final, where m1 and v1 are the mass and velocity of the bullet, and m2 and v2 are the mass and velocity of the block, which is initially at rest.
Using the given values: the mass of the bullet m1 is 6.00 g (which is 0.006 kg for consistency with SI units), the velocity of the bullet v1 is 113 m/s, and the mass of the block m2 is 2.00 kg with an initial velocity v2 of 0 m/s (since it is at rest), we can calculate the final velocity v_final after the bullet embeds itself in the block.
The calculation would be:
(0.006 kg × 113 m/s) + (2.00 kg × 0 m/s) = (0.006 kg + 2.00 kg) × v_final
thus, v_final = (0.006 kg × 113 m/s) / (0.006 kg + 2.00 kg).
Cleaning up the values gives us the final velocity of the block as: v_final = 0.339 m/s, after rounding to three significant figures.