Question

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 12.0 \(g\) is fired at the block with a horizontal velocity \(v_0\). The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 4.80 N. What was the initial speed \(v_0\) of the bullet?

Short answer

The initial speed of the bullet \( v_0 \) is approximately 232 m/s.

Step-by-step solution

Understand the Problem

We're dealing with a conservation of momentum problem followed by energy conservation. First, the bullet embeds itself in the block, and then the combined mass rises to a certain height due to the momentum transferred by the bullet. Our task is to find the initial speed \( v_0 \) of the bullet.

Apply Conservation of Energy

First, we need to determine the velocity of the block and bullet immediately after the collision. When the block rises to 0.800 m, the kinetic energy has transformed into potential energy and some tension energy. Use: \[ mgh = \frac{1}{2} mv^2 \] with \( h = 0.800 \text{ m} \). Calculate the velocity \( v \) of the block when it reaches this height.

Calculate Tension Contribution to Energy

The tension in the cord at the highest point is 4.80 N. Since the tension does not affect the vertical position directly, we relate it through forces in a circular motion. The tension also contributes as \( T - mg = \frac{mv^2}{r} \), where \( v \) is the velocity just before reaching the 0.8 m height and \( r \) is the radius here equal to \( 1.6 - 0.8 = 0.8 \). Solve for \( v \).

Solve for Initial Velocity After Collision

Apply the conservation of energy at the top point: \( mgh = \frac{1}{2} mv^2 \). Calculate \( v \) using \( g = 9.8 \text{ m/s}^2 \).

Apply Conservation of Momentum

Using the found velocity \( v \) after collision, apply the conservation of momentum. The initial momentum \( mv_0 \) equals the final momentum \( (m_1 + m_2)v \): \( 0.012 v_0 = (0.800 + 0.012)v \). Solve for the initial velocity \( v_0 \) of the bullet.

Key concepts

Conservation of Momentum

In any closed system with no external forces, the momentum before an event is equal to the momentum after. This principle is crucial in understanding collisions like the one between the bullet and the block in our exercise. Momentum is the product of mass and velocity, and in this scenario, we calculate the momentum of both bullet and block as they interact.

• Momentum before collision is given by the bullet: \( \text{momentum} = m_{bullet} \times v_0 \).
• Momentum after the collision includes both the bullet and block: \( (m_{bullet} + m_{block}) \times v \).

We utilize these equations to link the pre and post-collision movements, enabling us to find the bullet's initial velocity. This step is essential as it helps in transforming the system's kinetic profile without any energy being lost, just changing forms.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, initial kinetic energy converts to potential and rotational energy eventually.

• Initially, the bullet's kinetic energy: \( \frac{1}{2} m_{bullet} v_0^2 \).
• As the bullet embeds into the block, the system's kinetic energy gradually transforms to potential energy: \( m_{total}gh \).

Understanding this transition is crucial to determining how high the block swings after absorbing the bullet's impact. The energy transformation helps explain why the system behaves as observed, providing a robust technique to calculate unknown quantities such as velocities.

Circular Motion

Circular motion comes into play once the bullet and block move as a single entity. Post-collision, the block and bullet swing on a cord, presenting a classic example of circular dynamics.

• In circular motion, forces ensure the rotating paths are consistently followed.
• Tension in the cord is pivotal, contributing alongside gravitational force to maintain circular motion.
• The centripetal force required is provided by tension, calculated using: \( F = \frac{mv^2}{r} \), where \(r\) is the effective length they swing through.

Centripetal and tension forces are crucial in keeping everything in motion. By analyzing these forces, you understand the balance needed for sustained circular trajectories.

Kinetic Energy

Kinetic energy is a form of energy associated with motion. Initially, the bullet possesses kinetic energy, which is partially transferred to the wood block upon impact, causing both to move jointly.

• Kinetic energy at collision onset is \( \frac{1}{2} mv^2 \).
• As the block rises, kinetic energy decreases and converts to potential energy.

After the merger of bullet and block, their composite kinetic energy is less than that of the lone bullet but is pivotal in translating the motion into potential energy, explaining how high they rise in the absence of external work.

Potential Energy

Potential energy relates to the position of an object within a force field, here primarily gravity. As the block and bullet rise, their system gains potential energy at a loss of kinetic energy.

• Potential energy is given by: \( PE = mgh \).
• Higher elevation translates to more potential energy stored.
• This energy increase implies work done against gravity.

Understanding potential energy allows us to see how energy conservation remains intact throughout the motion. As kinetic energy diminishes, the corresponding rise in potential energy reflects the height reached, making it a key aspect of analyzing the post-collision system behavior.