Find the initial speed, in m/s, of the bullet

Question

A ballistic pendulum is a simple, but practical, device that can be used to measure the speed of a bullet. The pendulum consists of a block of wood with mass 4.30 kg, at the end of a thin wire so that the center of the block is 83.4 cm below the top of the wire. With the block at rest and the wire vertical, a bullet of mass 29.0 grams is fired at the block. It embeds in the block and causes the block to swing so that it momentarily comes to rest when the wire makes an angle of 35.6^O with the vertical. Find the initial speed, in m/s, of the bullet.

Answer 1

Hi Maleeha,

To determine the height of the pendulum in meters, we can use the equation:

h = L (1 - cos(θ)) = 0.834 m (1 - cos(35.6°)) = 0.834 m (0.186) ≈ 0.155 m

To determine the velocity of the bullet, we can use the general closed system energy balance equation:Δ

ΔE_sys = ΔU + ΔKE + ΔPE = Q - W;

Where ΔE_sys is the change in total energy for the entire system

Where ΔU is the change in internal energy for the entire system

Where ΔKE is the change in kinetic energy for the entire system

Where ΔPE is the change in potential energy for the entire system

Since we are not told of any heat or work energy transfer in or out of the system, we can assume ΔU ≈ 0 & Q = W = 0.

This Leaves: ΔKE + ΔPE = 0

Expanded: KE₁ - KE₂ + PE₁ - PE₂ = 0

Rearranged: KE₁ + PE₁ = KE₂ + PE₂

At the beginning, nothing has changed in elevation, ∴ PE₁ = 0

After the pendulum stops, there is no movement, ∴ KE₂ = 0

This leaves: KE₁ = PE₂

For kinetic energy, it has a formula as KE = 1/2 * m * v²

For potential energy, it has a formula as PE = mgh

For "m" (the mass), this is the combined mass of the pendulum block and the bullet. We will label the combined mass, M = m_b+m_p

for g, the gravitational constant, we will use g = 9.81 m/s²

This gives us: 1/2 * M * v² = M * g * h

Recall, this equation is just after the collision happens, so v will not be the answer here.

M is divided out of both sides, algebraically, leaving: 1/2 * v² = g * h

Isolating v: v = √(2 * g * h)

v = √(2 * 9.81 m/s² *0.155 m) ≈ 1.744 m/s

Using conservation of momentum:

m_bv_b = M * v

v_b = M * v / m_b = (m_b+m_p) * v / m_b

v_b = (0.029 kg + 4.30 kg) * 1.744 m/s / 0.029 kg

v_b ≈ 260 m/s

The bullet exited the chamber at 260 m/s

Hope this Helps!!!

-Andrew Evans