A 3.00-g bullet traveling horizontally at 500 m/s hits a 4.8-kg wooden block, which is initially at rest on a smooth horizontal table. The bullet buries itself in the block without passing through. The speed of the block after the collision is:
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To find the speed of the block after the collision, we will use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
The momentum before the collision can be calculated using the formula:
Momentum before = mass of bullet × velocity of bullet
Given:
mass of bullet = 3.00 g = 0.00300 kg
velocity of bullet = 500 m/s
The momentum before the collision is:
Momentum before = 0.00300 kg × 500 m/s = 1.50 kg*m/s
After the collision, since the bullet and the block move together, their combined mass becomes the mass of the block-bullet system. Therefore, the mass after the collision is:
mass after = mass of bullet + mass of block
mass after = 0.00300 kg + 4.8 kg = 4.803 kg
Let's assume the speed of the block after the collision as V.
The momentum after the collision is:
Momentum after = mass after × speed of block
Momentum after = 4.803 kg × V
Since momentum is conserved, we equate the momentum before and after the collision:
Momentum before = Momentum after
1.50 kg*m/s = 4.803 kg × V
Now we can solve for V:
V = 1.50 kg*m/s / 4.803 kg
V ≈ 0.312 m/s
Therefore, the speed of the block after the collision is approximately 0.312 m/s.
To find the speed of the block after the collision, we can use the principle of conservation of momentum.
The law of conservation of momentum states that the total momentum of a system remains constant if there are no external forces acting on it.
In this case, before the collision, the bullet has momentum, and after the collision, the bullet and the block together have momentum.
The momentum of an object can be calculated by multiplying its mass by its velocity.
Since the bullet is initially traveling horizontally, its momentum before the collision is given by:
Momentum of bullet before collision = mass of bullet x velocity of bullet
= 3.00 g (convert to kg) x 500 m/s
= 3.00 x 10^-3 kg x 500 m/s
= 1.5 kg m/s
After the collision, the bullet is embedded in the wooden block, so their combined mass is the mass of the block plus the mass of the bullet.
The momentum of the block and the bullet after the collision is given by:
Momentum of block + bullet after collision = (mass of block + mass of bullet) x velocity of block + bullet
Let's assume the velocity of the block after the collision is V.
Then the equation becomes:
(4.8 kg + 3.00 x 10^-3 kg) * V = 1.5 kg m/s
Simplifying the equation, we have:
V = 1.5 kg m/s / (4.8 kg + 3.00 x 10^-3 kg)
Now we can solve for V:
V = 1.5 kg m/s / 4.8 kg
V ≈ 0.3125 m/s
Therefore, the speed of the block after the collision is approximately 0.3125 m/s.