a bullet of mass 0.015kg strikes ballistic pendulum of mass 2.0kg .the center of mass of the pendulum rises a vertical distance of 0.12m.assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

To calculate the bullet's initial speed, we can use the principle of conservation of momentum.

The initial momentum of the bullet before it strikes the pendulum can be calculated using the equation:

m₁ * v₁ = p₁

where,
m₁ = mass of the bullet = 0.015 kg (given)
v₁ = initial velocity of the bullet (to be determined)
p₁ = momentum of the bullet

The final momentum of the system, after the bullet embeds in the pendulum, can be calculated using the equation:

(m₁ + m₂) * v₂ = p₂

where,
m₂ = mass of the pendulum = 2.0 kg (given)
v₂ = final velocity of the bullet and pendulum after the collision (to be determined)
p₂ = momentum of the bullet and pendulum after the collision

Since the bullet remains embedded in the pendulum after collision, they move together as a combination of masses. Therefore, the final velocity (v₂) can be related to the rise in center of mass using the equation:

m₂ * v₂ = (m₁ + m₂) * vᵢ

where,
vᵢ = initial velocity of the bullet and pendulum

Now, we can equate the initial momentum to the final momentum:

m₁ * v₁ = (m₁ + m₂) * vᵢ

Substituting the given values, we have:

0.015 kg * v₁ = (0.015 kg + 2.0 kg) * vᵢ

Simplifying further:

v₁ = (2.015 kg / 0.015 kg) * vᵢ

Now, we need to calculate the rise in the center of mass (h), which is given as 0.12 m (given). This rise in center of mass can be related to the initial velocity using the equation:

h = (vᵢ^2) / (2g)

where,
g = acceleration due to gravity = 9.8 m/s²

Rearranging the equation, we have:

vᵢ^2 = 2gh

Substituting the given values, we have:

vᵢ^2 = 2 * 9.8 m/s² * 0.12 m
vᵢ^2 = 2.352 m²/s²

Taking the square root of both sides, we get:

vᵢ = √(2.352 m²/s²)
vᵢ ≈ 1.534 m/s

Now, we can substitute this value back into the equation for v₁:

v₁ = (2.015 kg / 0.015 kg) * vᵢ
v₁ ≈ 2.015 * 1.534 m/s
v₁ ≈ 3.095 m/s

Therefore, the bullet's initial speed is approximately 3.095 m/s.

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