Answer:
approximately 0.3176 m/s
Step-by-step explanation:
To solve this problem, we can apply the principle of conservation of momentum. Assuming the bullet and the block are the only objects involved in the system, we can equate the initial momentum to the final momentum to find the velocity of the block after the bullet passes through it.
The initial momentum of the system is given by the momentum of the bullet before it hits the block. The final momentum of the system is given by the momentum of the bullet after it passes through the block and the momentum of the block itself.
First, let's convert the mass of the bullet from grams to kilograms:
Mass of the bullet = 3.0 g = 3.0 × 10^(-3) kg
The initial momentum of the system is:
Initial momentum = mass of the bullet × initial velocity of the bullet
= (3.0 × 10^(-3) kg) × (550 m/s)
= 1.65 kg·m/s
The final momentum of the system is:
Final momentum = (mass of the bullet × final velocity of the bullet) + (mass of the block × final velocity of the block)
Since the bullet goes completely through the block, the final velocity of the bullet is the same as its initial velocity. Also, as there is no external force acting on the system, the momentum of the system is conserved. Therefore:
Final momentum = initial momentum
(mass of the bullet × final velocity of the bullet) + (mass of the block × final velocity of the block) = 1.65 kg·m/s
Substituting the given values:
(3.0 × 10^(-3) kg) × (190 m/s) + (3.4 kg) × (final velocity of the block) = 1.65 kg·m/s
Solving for the final velocity of the block:
(3.0 × 10^(-3) kg) × (190 m/s) + (3.4 kg) × (final velocity of the block) = 1.65 kg·m/s
0.57 kg·m/s + (3.4 kg) × (final velocity of the block) = 1.65 kg·m/s
(3.4 kg) × (final velocity of the block) = 1.65 kg·m/s - 0.57 kg·m/s
(3.4 kg) × (final velocity of the block) = 1.08 kg·m/s
final velocity of the block = (1.08 kg·m/s) / (3.4 kg)
= 0.3176 m/s
Therefore, the final velocity of the block after the bullet passes through it is approximately 0.3176 m/s.