Question

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the \(2.00-\mathrm{g}\) bullet into a \(2.00-\mathrm{kg}\) wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of \(0.500 \mathrm{~cm}\) above its initial position. What is the velocity of the bullet on leaving the gun's barrel?

Short answer

Answer: The velocity of the bullet on leaving the gun's barrel is approximately 315 m/s.

Step-by-step solution

Identify the relevant variables

Let: - \(m_b\) be the mass of the bullet - \(m_w\) be the mass of the wooden block - \(v_b\) be the initial velocity of the bullet - \(h\) be the maximum height reached by the block after the bullet is embedded in it Given values: - \(m_b = 2.00 \,\mathrm{g} = 0.002 \,\mathrm{kg}\) - \(m_w = 2.00 \,\mathrm{kg}\) - \(h = 0.500 \,\mathrm{cm} = 0.0050 \,\mathrm{m}\)

Apply the conservation of momentum

First, consider the conservation of momentum for the bullet-block system. The initial momentum of the system equals the final momentum of the system: \((m_b + m_w)v_f = m_bv_b\) Here, \(v_f\) is the final velocity of the bullet and the block when they are stuck together.

Determine the final velocity of the system

To find the final velocity, \(v_f\), we'll use the conservation of energy. The initial mechanical energy of the system is equal to the final mechanical energy of the system: \(KE_0 + PE_0 = KE_f + PE_f\) Initially, the kinetic energy of the block is zero, and the potential energy of the block is also zero: \(\frac{1}{2}(m_b + m_w)v_f^2 = (m_b + m_w)gh\) Rearrange the equation to solve for \(v_f\): \(v_f = \sqrt{2gh}\) Substitute the given values and calculate the result: \(v_f = \sqrt{2(9.81 \,\mathrm{m/s^2})(0.0050 \,\mathrm{m})} \approx 0.313 \,\mathrm{m/s}\)

Calculate the initial velocity of the bullet

We can now use the conservation of momentum equation to find the initial velocity of the bullet: \(v_b = \frac{(m_b + m_w)v_f}{m_b}\) Substitute the given values and calculated result: \(v_b = \frac{(0.002 \,\mathrm{kg} + 2.00 \,\mathrm{kg})(0.313 \,\mathrm{m/s})}{0.002 \,\mathrm{kg}} \approx 315 \,\mathrm{m/s}\) Thus, the velocity of the bullet on leaving the gun's barrel is approximately 315 \(\mathrm{m/s}\).

Key concepts

Conservation of Momentum

In physics, the principle of conservation of momentum tells us that the total momentum of a closed system remains constant over time if external forces are not acting on it.

Let's apply this concept to our exercise where we have a bullet and a wooden block. Before the bullet is fired, both the bullet and the block are at rest, meaning their combined momentum is zero. Upon firing, the bullet moves forward, and when it embeds itself into the block, the block will start moving. At this moment, the momentum of the system must remain the same as it was initially – which was zero. So, the bullet's momentum before the collision (its mass times its velocity) is equal to the combined momentum of the bullet and block after the collision.

This principle is crucial for solving our exercise: by understanding that momentum is conserved, we can relate the unknown muzzle velocity of the bullet to the measurable velocity of the block and bullet after impact. Proper application of this principle will lead to an accurate calculation of the bullet's initial velocity.

Kinetic Energy

Kinetic energy is the energy of motion. Any object that's moving has kinetic energy, and it's given by the equation:
\[ KE = \frac{1}{2}mv^2 \]
where m is the mass and v is the velocity of the moving object.

Considering our exercise, the bullet, upon leaving the gun's barrel, possesses kinetic energy due to its velocity. As soon as the bullet embeds into the wooden block, part of its kinetic energy is transferred to the block, causing it to move and swing upward.

This transferred kinetic energy is then converted into potential energy as the block rises to its maximum height. Understanding how kinetic energy plays a role in the system's changes allows us to track the energy transformations and thus determine the initial kinetic energy of the bullet, which can then be used to calculate its muzzle velocity.

Potential Energy

Potential energy is the energy stored within a system due to the position or configuration of its parts. In the context of our problem, we consider gravitational potential energy, which is given by the equation:
\[ PE = mgh \]
where m is the mass, g is the acceleration due to gravity, and h is the height.

As the bullet-block system rises after the bullet is embedded, it gains height and thus potential energy. At the system's highest point, all the kinetic energy from the moment after the collision is now converted into potential energy. This energy 'potential' tells us how high the block can rise, which depends on its mass and the height attained.

By calculating the potential energy at the block's highest point, we can deduce the amount of kinetic energy it had immediately after the collision and before it started rising. With this information linked to the conservation of momentum and conservation of energy principles, we can solve for the bullet's muzzle velocity with precision.