A ballistic pendulum consists of a rectangular wooden block of mass 1.5 kg suspended from a ceiling. A bullet mass 0.025 kg undergoes a completely inelastic collision with the ballistic pendulum. Due to the collision the pendulum rises to a height of 45 cm. Calculate the speed with which bullet hits the pendulum.

While swinging up H = 0.45 m elevation, the block and bullet convert kinetic energy to potential energy, without heat loss. This tells you that

V = sqrt(2gH) = 2.97 m/s
is the velocity of the block and bullet just after impact.

Next, apply conservation of momentum to the inelastic bullet-impact process, assuming that the pendulum does not swing during that brief interval. Let v be the bullet speed, m the bullet mass and M be the block's mass. Momentum conservation tells you that

m v = (M + m) V

Solve for v

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To calculate the speed with which the bullet hits the pendulum, we can use the principle of conservation of momentum.

Conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision. In this case, the bullet and the pendulum block are involved in the collision.

Let's denote the initial speed of the bullet as v_b, the final speed of the bullet and the block together as v_f, and the mass of the pendulum block as M.

The equation for conservation of momentum can be written as:

(m_b * v_b) = (m_b + M) * v_f

Where:
m_b = mass of the bullet (0.025 kg)
v_b = speed of the bullet (unknown)
M = mass of the pendulum block (1.5 kg)
v_f = final speed of the bullet and the block together (unknown)

Now, we need to calculate the initial speed of the bullet (v_b). To do this, we need to consider the conservation of mechanical energy.

When the bullet hits the pendulum, both linear momentum and mechanical energy can be conserved if the collision is completely inelastic. Since the bullet and block stick together after the collision, we can use the principle of conservation of mechanical energy to find v_b.

The mechanical energy before the collision is equal to the mechanical energy after the collision, which includes the kinetic energy of the bullet-block system after the collision and the potential energy gained as the pendulum rises.

The equation for conservation of mechanical energy can be written as:

(1/2) * m_b * v_b^2 = (1/2) * (m_b + M) * v_f^2 + M * g * h

Where:
m_b = mass of the bullet (0.025 kg)
v_b = initial speed of the bullet (unknown)
M = mass of the pendulum block (1.5 kg)
v_f = final speed of the bullet and the block together (unknown)
g = acceleration due to gravity (9.8 m/s^2)
h = height to which the pendulum rises (0.45 m)

Now we have two equations with two unknowns (v_b and v_f). We can solve these equations simultaneously to find the values.

1. Substitute the equation for v_f from the conservation of momentum into the conservation of mechanical energy equation:

(1/2) * m_b * v_b^2 = (1/2) * (m_b + M) * [(m_b * v_b) / (m_b + M)]^2 + M * g * h

2. Simplify the equation:

m_b * v_b^2 = (m_b + M) * [(m_b * v_b) / (m_b + M)]^2 + 2 * M * g * h

3. Solve the equation for v_b using algebraic manipulation:

m_b * v_b^2 = (m_b^2 * v_b^2) / (m_b + M) + 2 * M * g * h

Multiply both sides of the equation by (m_b + M):

m_b * v_b^2 * (m_b + M) = m_b^2 * v_b^2 + 2 * M * g * h * (m_b + M)

Expand the equation:

m_b^2 * v_b^2 + m_b * M * v_b^2 = m_b^2 * v_b^2 + 2 * M * g * h * m_b + 2 * M^2 * g * h

4. Simplify the equation further:

m_b * M * v_b^2 = 2 * M * g * h * m_b + 2 * M^2 * g * h

Divide both sides of the equation by m_b * M:

v_b^2 = 2 * g * h + 2 * M * g * h / (m_b * M)

v_b^2 = 2 * g * (h + M) / m_b

5. Take the square root of both sides of the equation to solve for v_b:

v_b = sqrt(2 * g * (h + M) / m_b)

Now, substitute the known values into the equation and calculate the speed with which the bullet hits the pendulum.