A 0.190 kg projectile is fired with a velocity of +705 m/s at a 2.00 kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is +40.0 m/s. Find the velocity with which the projectile exits from the block.
To find the velocity with which the projectile exits from the block, we can use the law of conservation of momentum.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity: momentum = mass × velocity.
Let's assume that the velocity of the projectile after it passes through the block is v, and the mass of the wooden block is M.
The momentum of the projectile before the collision is given by: momentum_before = mass_projectile × velocity_projectile.
The momentum of the block before the collision is given by: momentum_block_before = mass_block × velocity_block.
The momentum of the projectile after the collision is given by: momentum_after = mass_projectile × velocity.
The momentum of the block after the collision is given by: momentum_block_after = mass_block × velocity_block_after.
Since the total momentum before the collision is equal to the total momentum after the collision, we can write the equation:
momentum_before + momentum_block_before = momentum_after + momentum_block_after.
mass_projectile × velocity_projectile + mass_block × velocity_block_before = mass_projectile × velocity + mass_block × velocity_block_after.
Now we can plug in the given values:
mass_projectile = 0.190 kg,
velocity_projectile = 705 m/s,
mass_block = 2.00 kg,
velocity_block_before = 0 m/s (since it's at rest),
velocity_block_after = 40.0 m/s.
0.190 kg × 705 m/s + 2.00 kg × 0 m/s = 0.190 kg × v + 2.00 kg × 40.0 m/s.
135.11 kg·m/s = 0.190 kg × v + 80.0 kg·m/s.
135.11 kg·m/s - 80.0 kg·m/s = 0.190 kg × v.
55.11 kg·m/s = 0.190 kg × v.
To find the velocity with which the projectile exits the block, we divide both sides of the equation by 0.190 kg:
v = 55.11 kg·m/s / 0.190 kg
v ≈ 290.06 m/s.
Therefore, the velocity with which the projectile exits the block is approximately 290.06 m/s.
To find the velocity with which the projectile exits from the block, we can apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the projectile and the block make up an isolated system since there are no external forces acting on them.
Mathematically, the conservation of momentum can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 * V1) + (m2 * V2)
Where:
m1 and v1 are the mass and velocity of the projectile before collision,
m2 and v2 are the mass and velocity of the block before collision,
m1 and V1 are the mass and velocity of the projectile after collision, and
m2 and V2 are the mass and velocity of the block after collision.
In this case, the mass and velocity of the projectile before collision (m1 and v1) are given as 0.190 kg and +705 m/s, respectively. The mass and velocity of the block before collision (m2 and v2) are given as 2.00 kg and +0 m/s, respectively. The velocity of the block after collision (V2) is given as +40.0 m/s.
Substituting these values into the equation above, we can solve for the velocity of the projectile after collision (V1):
(0.190 kg * 705 m/s) + (2.00 kg * 0 m/s) = (0.190 kg * V1) + (2.00 kg * 40.0 m/s)
134 kg·m/s = 0.190 kg·V1 + 80.0 kg·m/s
Subtracting 80.0 kg·m/s from both sides:
134 kg·m/s - 80.0 kg·m/s = 0.190 kg·V1
54 kg·m/s = 0.190 kg·V1
Dividing both sides by 0.190 kg:
V1 = 54 kg·m/s / 0.190 kg
V1 ≈ 284.21 m/s
Therefore, the velocity with which the projectile exits from the block is approximately +284.21 m/s.