A 10 g-bullet moving horizontally at 375 m/s penetrates a 3 kg wood block resting on a frictionless horizontal surface. If the bullet slows down to 300 m/s after emerging from the block, what is the speed of the block immediately after the bullet emerges
To solve this problem, we can apply the principle of conservation of momentum. The momentum before the bullet enters the block is equal to the momentum after the bullet emerges from the block.
Let's denote the initial velocity of the block as v and the final velocity of the block as V. Considering that the bullet is moving horizontally, and the block is initially at rest, the momentum before the bullet enters the block is given by:
Initial momentum = Bullet momentum + Block momentum
= (mass of bullet) * (velocity of bullet) + (mass of block) * (velocity of block)
= (0.01 kg) * (375 m/s) + (3 kg) * (0 m/s)
= 3.75 kg m/s
The momentum after the bullet emerges from the block is given by:
Final momentum = Bullet momentum + Block momentum
= (mass of bullet) * (velocity of bullet) + (mass of block) * (velocity of block)
= (0.01 kg) * (300 m/s) + (3 kg) * V
= 3 kg * V + 3 kg * (300 m/s)
= 3 kg * (V + 300 m/s)
Since the principle of conservation of momentum tells us that the initial momentum is equal to the final momentum, we can set up an equation:
3.75 kg m/s = 3 kg * (V + 300 m/s)
Simplifying the equation, we get:
3.75 kg m/s = 3 kg * V + 900 kg m/s
Subtracting 900 kg m/s from both sides, we have:
3.75 kg m/s - 900 kg m/s = 3 kg * V
-896.25 kg m/s = 3 kg * V
Dividing both sides by 3 kg, we find:
V = -896.25 kg m/s / 3 kg
V = -298.75 m/s
Therefore, the speed of the block immediately after the bullet emerges is approximately 298.75 m/s in the opposite direction.
To determine the speed of the block immediately after the bullet emerges, we can apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces act upon it. In this case, the system consists of the bullet and the block.
Let's denote the initial velocity of the bullet as v1, the final velocity of the bullet as v2, and the final velocity of the block as V. We are given that the mass of the bullet (m1) is 10 grams (0.01 kg) and the mass of the block (m2) is 3 kg.
The initial momentum of the system (before the bullet enters the block) is given by:
Initial momentum = (mass of bullet) × (initial velocity of bullet)
= m1 × v1
The final momentum of the system (after the bullet emerges from the block) is given by:
Final momentum = (mass of bullet) × (final velocity of bullet) + (mass of block) × (final velocity of block)
= m1 × v2 + m2 × V
Since momentum is conserved, the initial momentum and the final momentum of the system should be the same:
m1 × v1 = m1 × v2 + m2 × V
Now, let's substitute the given values into the equation and solve for V:
0.01 kg × 375 m/s = 0.01 kg × 300 m/s + 3 kg × V
3.75 kg·m/s = 3 kg·m/s + 3 kg × V
Subtracting 3 kg·m/s from both sides:
0.75 kg·m/s = 3 kg × V
Dividing both sides by 3 kg:
0.25 m/s = V
Therefore, the speed of the block immediately after the bullet emerges from it is 0.25 m/s.