A bullet of mass 0.062 kg traveling horizontally at a speed of 150 m/s embeds itself in a block of mass 90 kg that is sitting at rest on a nearly frictionless surface
What is the speed of the block after the bullet embeds itself in the block?
conserve momentum:
.062*150 + 90*0 = (.062+90)v
v = 0.103 m/s
Teri 9211
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the bullet embeds itself in the block is equal to the total momentum after.
The momentum of an object is given by the product of its mass and velocity. Let's denote the initial velocity of the bullet as v₁, the initial velocity of the block as v₂, and the final velocity of the combined bullet-block system as v.
Before the collision:
Momentum of the bullet = mass of the bullet × velocity of the bullet
P(bullet) = m(bullet) × v₁
Momentum of the block = mass of the block × velocity of the block
P(block) = m(block) × v₂
After the collision, the bullet embeds itself in the block, so they move together as one system. Therefore, the total mass of the system is the sum of the masses of the bullet and the block.
Total momentum after the collision:
P(total) = (m(bullet) + m(block)) × v
According to the conservation of momentum principle, the total momentum before the collision is equal to the total momentum after the collision.
P(bullet) + P(block) = P(total)
m(bullet) × v₁ + m(block) × v₂ = (m(bullet) + m(block)) × v
Now we can substitute the given values into the equation. m(bullet) = 0.062 kg, m(block) = 90 kg, v₁ = 150 m/s, and v₂ = 0 m/s since the block is at rest.
(0.062 kg × 150 m/s) + (90 kg × 0 m/s) = (0.062 kg + 90 kg) × v
9.3 kg m/s = 90.062 kg × v
Divide both sides of the equation by 90.062 kg to solve for v:
v = 9.3 kg m/s / 90.062 kg
v ≈ 0.103 m/s
Therefore, the speed of the block after the bullet embeds itself in the block is approximately 0.103 m/s.
To find the speed of the block after the bullet embeds itself in it, we can use 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 on it. Mathematically, this can be expressed as:
Initial momentum of the system = Final momentum of the system
The initial momentum of the system is the momentum of the bullet, and the final momentum of the system is the combined momentum of the bullet and the block.
The momentum (p) of an object is calculated by multiplying its mass (m) with its velocity (v):
p = m * v
Given:
Mass of the bullet (m1) = 0.062 kg
Velocity of the bullet (v1) = 150 m/s
Mass of the block (m2) = 90 kg (before the collision)
Velocity of the block (v2) = ? (after the collision)
Using the principle of conservation of momentum, we can write the equation:
(m1 * v1) + (m2 * v2) = (m1 + m2) * v3
Where v3 is the final velocity of the combined system after the collision.
Plugging in the values, we have:
(0.062 kg * 150 m/s) + (90 kg * v2) = (0.062 kg + 90 kg) * v3
Simplifying further:
9.3 kg * v2 = 90.062 kg * v3
Now, since the block was initially at rest, the initial velocity of the block (v2) is 0:
9.3 kg * 0 = 90.062 kg * v3
Solving for v3 gives us the final velocity of the combined system:
v3 = (9.3 kg * 0) / 90.062 kg
v3 = 0 m/s
Therefore, the speed of the block after the bullet embeds itself in the block is 0 m/s, indicating that the block does not move after the collision.