Problem 78 - University Physics with Modern Physics

Question

A small wooden block with mass 0.800 \(\mathrm{kg}\) is suspended from the lower end of a light cord that is 1.60 \(\mathrm{m}\) long. The block is initially at rest. A bullet with mass 12.0 \(\mathrm{g}\) 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 \(0.800 \mathrm{m},\) the tension in the cord is 4.80 \(\mathrm{N} .\) What was the initial speed \(v_{0}\) of the bullet?

Step-by-Step Solution

Answer

The bullet's initial speed was approximately 272.8 m/s.

Step 1: Convert Units

Convert the mass of the bullet into kilograms for consistency with the block mass. The bullet mass is 12.0 g, so \( \text{bullet mass} = 0.012 \text{ kg} \).

Step 2: Calculate Initial Potential Energy

At the highest point of the swing (0.800 m), calculate the potential energy. The mass of the bullet and block together is \(0.800\,\text{kg} + 0.012\,\text{kg} = 0.812\,\text{kg}\). Thus, the potential energy at 0.800 m is \(PE = mgh = 0.812 \times 9.8 \times 0.800\).

Step 3: Calculate Tension's Contribution

At the maximum height, the total energy consists of potential energy and work done by the tension. The tension (4.80 N) contributes only to centripetal force and does not perform work. Therefore, at the highest point, potential energy equals the kinetic energy at the bottom plus work done.

Step 4: Use Conservation of Momentum for Initial Collision

Use the conservation of momentum at the point when the bullet hits the block. Let \(v\) be the velocity of the block-bullet system just after the collision:\[ m_{bullet} \times v_0 = (m_{bullet} + m_{block}) \times v \] Substitute known values to express \(v_0\) in terms of \(v\).

Step 5: Find Speed from Energy Conservation

From Step 2, let the kinetic energy at the bottom of the swing equal the potential energy at 0.800 m:\[ \frac{1}{2} \cdot 0.812 \cdot v^2 = 0.812 \cdot 9.8 \cdot 0.800 \]Solve for \(v\).

Step 6: Calculate Initial Speed of Bullet

With \(v\) known, use Step 4's momentum equation to solve for \(v_0\):\[ v_0 = \frac{(m_{block} + m_{bullet}) \cdot v}{m_{bullet}} \]

Step 7: Solve and Verify

After solving, \(v_0 \approx 272.8 \text{ m/s} \). Recheck calculations by verifying energy conservation and momentum principles are upheld throughout the steps.

Key Concepts

Potential Energy

Potential energy is the energy possessed by an object due to its position relative to other objects. In this exercise, the bullet and block together possess potential energy when they elevate during the swing. The formula for potential energy is:

• \[ PE = mgh \]
• Where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately 9.8 \( \mathrm{m/s^2} \)), and \( h \) is the vertical height above the reference point.

When the system reaches its highest point, the potential energy is at maximum because the block and embedded bullet have moved upward by 0.800 m. Understanding potential energy helps to identify how high the object has swung by conservation of mechanical energy, thus giving clues about kinetic energy and initial speeds.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. The basic formula for kinetic energy is:

• \[ KE = \frac{1}{2} mv^2 \]
• Where \( m \) represents the mass of the object and \( v \) its velocity.

In the context of this exercise, once the bullet embeds into the block, the entire system must overcome its initial inertia to swing. Immediately after the collision, the potential energy is transformed into kinetic energy, which is then partly converted into potential energy as the system swings upwards. Therefore, analyzing kinetic energy gives insights into the movement of the system right after the collision, aiding in determining the initial velocity of the bullet.

Collision Dynamics

Collision dynamics involve analyzing the effects and movements when two objects collide. In this exercise, we see an inelastic collision where the bullet becomes embedded in the block, highlighting several key dynamics:

• **Conservation of Momentum**: The total momentum before and after the collision remains constant. Hence, \[ m_{bullet} \cdot v_0 = (m_{bullet} + m_{block}) \cdot v .\]
• **Understanding Inelastic Collision**: In these collisions, objects stick together post-impact, and kinetic energy is not conserved although momentum is.

Applying these dynamics helps us derive the speed of the block-bullet system post-impact from which the initial speed of the bullet, \( v_0 \), is calculated. These dynamics are essential in solving real-world collision problems effectively.