a 100-g bullet is fired horizontally into a 14.9kg block of wood resting on a horizontal surface and the bullet becomes embedded in the block. if the muzzle velocity of the bullet is 250 m/s , what is the speed of the bullet immedatly after the impact (neglect surface friction)
To solve this problem, we can apply the law of conservation of linear momentum. According to this law, the total momentum before a collision is equal to the total momentum after the collision.
Let's denote:
m1 = mass of the bullet = 0.100 kg (since it's given as 100 g)
v1 = initial velocity of the bullet = 250 m/s
m2 = mass of the block of wood = 14.9 kg
v2 = velocity of the block of wood after the impact (and the bullet becomes embedded)
Using the conservation of momentum, we can write the equation:
(m1 * v1) + (m2 * 0) = (m1 + m2) * v2
Since the block of wood was initially at rest, its initial velocity (v2) is 0.
Simplifying the equation, we have:
(m1 * v1) = (m1 + m2) * v2
Now, we can substitute the known values:
(0.100 kg * 250 m/s) = (0.100 kg + 14.9 kg) * v2
25 kg·m/s = 15 kg * v2
Dividing both sides by 15 kg:
25 kg·m/s / 15 kg = v2
v2 ≈ 1.67 m/s
Therefore, the speed of the bullet immediately after the impact is approximately 1.67 m/s.
To determine the speed of the bullet immediately after the impact, we need to apply the principle of conservation of momentum. The momentum before the impact is equal to the momentum after the impact.
The momentum of an object can be calculated using the formula: momentum = mass × velocity.
Given:
- The mass of the bullet, m₁ = 100 g = 0.1 kg
- The muzzle velocity of the bullet, v₁ = 250 m/s
- The mass of the block, m₂ = 14.9 kg
- The speed of the bullet immediately after the impact, v₂ (to be determined)
First, let's calculate the momentum before the impact (m₁v₁):
momentum before the impact = m₁ × v₁
momentum before the impact = 0.1 kg × 250 m/s
momentum before the impact = 25 kg·m/s
Since momentum is conserved, the momentum after the impact is also 25 kg·m/s.
Now, let's calculate the momentum after the impact. Since the bullet becomes embedded in the block, we consider the momentum of the bullet-block system together.
momentum after the impact = (m₁ + m₂) × v₂
Substituting the known values:
25 kg·m/s = (0.1 kg + 14.9 kg) × v₂
Simplifying the equation:
25 kg·m/s = 15 kg × v₂
Dividing both sides by 15 kg:
v₂ = 25 kg·m/s / 15 kg
v₂ = 1.67 m/s
Therefore, the speed of the bullet immediately after the impact (neglecting surface friction) is approximately 1.67 m/s.