Question

A 5.00-g bullet is shot \(through\) a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.

Short answer

The bullet exits the block at 395.4 m/s.

Step-by-step solution

Understand the Conservation of Energy

When the bullet goes through the block, the block gains some kinetic energy which is then converted into potential energy as it rises. The potential energy gained by the block corresponds to the change in height, which is given as 0.38 cm or 0.0038 m. We'll first calculate the potential energy gained by the block.

Calculate Potential Energy of the Block

The potential energy gained by the block as it rises can be calculated using the formula: \[ PE = mgh \]where \(m = 1 \) kg (mass of the block), \(g = 9.81 \) m/se (acceleration due to gravity), and \(h = 0.0038 \) m (height). This gives:\[PE = 1 \times 9.81 \times 0.0038 = 0.037278\, \text{J}\]

Use Conservation of Momentum for Bullet and Block Collision

Before the collision, only the bullet has kinetic energy as it moves towards the block with an initial speed. As the bullet exits the block, the speed of the bullet changes, and the block gains some velocity. We use the principle of conservation of momentum:\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]where \(m_1 = 0.005\) kg, \(v_1 = 450\) m/s, \(m_2 = 1\) kg, and \(v_2 = 0\) m/s (since the block is initially at rest). \(v_1'\) is the final velocity of the bullet and \(v_2'\) is the velocity of the block after the bullet passes through.

Calculate Kinetic Energy of the Block After Collision

The kinetic energy of the block right after the bullet exits it is equal to the potential energy gained because energy is conserved. Therefore, the kinetic energy \(KE = 0.037278\) J is:\[ KE = \frac{1}{2} m_2 v_2'^2 \]Solving for \(v_2'\), we have:\[ v_2'^2 = \frac{2 \times 0.037278}{1} = 0.074556 \]\[ v_2' = \sqrt{0.074556} = 0.273\, \text{m/s} \]

Apply Conservation of Momentum to Find Bullet's Final Speed

Using the momentum conservation equation:\[ 0.005 \times 450 = 0.005 \times v_1' + 1 \times 0.273 \]\[ 2.25 = 0.005v_1' + 0.273 \]Solving for \(v_1'\),\[ 0.005v_1' = 2.25 - 0.273 \]\[ 0.005v_1' = 1.977 \]\[ v_1' = \frac{1.977}{0.005} = 395.4\, \text{m/s} \]

Conclusion

Therefore, the speed of the bullet as it emerges from the block is 395.4 m/s.

Key concepts

Kinetic Energy

Kinetic energy is a form of energy associated with objects in motion. It quantifies the energy an object possesses due to its movement, calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the object and \( v \) is its velocity.
For example, when our bullet with a mass of 5 grams moves at an initial speed of 450 m/s, it has kinetic energy. When it collides with the block, some of this kinetic energy is transferred, illustrating the concept of energy conservation.
Kinetic energy plays a crucial role during collisions, as seen when the bullet transfers energy to the block, causing the block to rise. This rise is a testament to kinetic energy converting into potential energy, which we will discuss further below.

• Energy due to motion
• Calculated as \( \frac{1}{2} m v^2 \)
• Transfers between objects during collisions

Potential Energy

Potential energy is the energy stored in an object due to its position or configuration. In the context of our exercise, the block gains potential energy as it rises when the bullet transfers some of its kinetic energy. This potential energy can be calculated using the formula:\[ PE = mgh \]where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( h \) is the height the block has risen.
By rising 0.38 cm or 0.0038 m, the block stores energy in the form of gravitational potential energy, which fundamentally comes from the initial kinetic energy of the bullet.
Potential energy often contrasts with kinetic energy because it’s related to an object's position rather than its motion. In our scenario, even though the block is momentarily at rest at its peak height, the energy is still present in the form of potential.

• Stored energy due to position
• Calculated as \( mgh \)
• Converted from kinetic energy

Conservation of Energy

The principle of conservation of energy is pivotal in physics. It states that energy in a closed system is constant, transforming from one form to another but never lost. In our exercise, this concept is beautifully illustrated as the system shifts back and forth between kinetic and potential energy.
Initially, the bullet possesses kinetic energy by virtue of its high speed. As it moves through the block, some kinetic energy is transferred to the block, causing it to move and subsequently rise.
As the block rises, kinetic energy is transformed into potential energy, visible in its increased height. This conversion underscores the conservation of energy, where the total energy before and after the interaction remains the same, albeit transformed from kinetic to potential form.

• Energy cannot be created or destroyed
• Transforms between kinetic and potential
• Total energy remains constant in a closed system