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Ballistic Pendulum A bullet of mass \(10 \mathrm{~g}\) strikes a ballistic pendulum of mass \(2.0 \mathrm{~kg}\). The center of mass of the pendulum rises a vertical distance of \(12 \mathrm{~cm}\). Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Short Answer

Expert verified
The bullet's initial speed is 309.44 m/s.

Step by step solution

01

- Convert Units

Convert the given masses and distance to SI units. The mass of the bullet is given as 10 g, which is 0.01 kg, and the distance of 12 cm, which is 0.12 m.
02

- Apply Conservation of Energy

Use the conservation of energy to determine the speed of the pendulum and bullet immediately after the collision. Calculate the potential energy (PE) at the highest point using the formula: \[ PE = mgh \] where \( m \) is the combined mass of the pendulum and bullet, \( g \) is acceleration due to gravity (9.8 m/s²), and \( h \) is the vertical height (0.12 m). \[ PE = (2.01 \, kg) \times (9.8 \, m/s^2) \times (0.12 \, m) \] \[ PE = 2.36028 \, J \]
03

- Convert Potential Energy to Kinetic Energy

At the highest point, the potential energy (PE) is converted from the kinetic energy (KE). Set PE equal to KE and solve for the speed (v) just after the collision using: \[ KE = \frac{1}{2} mv^2 \] \[ 2.36028 \, J = \frac{1}{2} \times 2.01 \, kg \times v^2 \] Solving for \( v \), \[ v = \frac{\text{sqrt}(2 \times 2.36028 \, J / 2.01 \, kg)} \] \[ v = 1.539 \, m/s \]
04

- Apply Conservation of Momentum

Use the conservation of momentum to determine the initial speed of the bullet. The momentum before collision equals the momentum after collision. Initial momentum = momentum just after collision: \[ m_b v_b = (m_b + M_p) V \] where \( m_b = 0.01 \, kg \), \( v_b \) is the initial speed of the bullet, \( M_p = 2.0 \, kg \), and \( V = 1.539 \, m/s \). \[ 0.01 \, kg \times v_b = 2.01 \, kg \times 1.539 \, m/s \] Solving for \( v_b \), \[ v_b = \frac{(2.01 \times 1.539) \, kg \times m/s}{0.01 \, kg} \] \[ v_b = 309.44 \, m/s \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 the context of a ballistic pendulum, the kinetic energy of the moving bullet and pendulum system is converted into potential energy as the system rises to its highest point.
Using the formula for potential energy, \( PE = mgh \) (where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height), we compute the potential energy when the pendulum reaches its highest point. By setting this value equal to the kinetic energy (\( KE = \frac{1}{2} mv^2 \)) just after the collision, we can solve for the system's speed immediately after impact.
This approach is essential for understanding how energy transformations govern the motion in the ballistic pendulum problem.
conservation of momentum
Conservation of momentum is a fundamental principle in physics stating that the total momentum of a closed system remains constant if no external forces act upon it. In a ballistic pendulum, this principle helps us determine the speed of the bullet before collision.
When the bullet embeds into the pendulum, the combined mass moves together. Using the equation for conservation of momentum:
\( m_b v_b = (m_b + M_p) V \)
where \( m_b \) is the mass of the bullet, \( v_b \) is the bullet's initial velocity, \( M_p \) is the mass of the pendulum, and \( V \) is the combined velocity just after impact, we can solve for the bullet's initial speed.
Understanding how momentum is conserved during collisions helps explain a wide range of real-world phenomena, from car crashes to space travel.
potential energy
Potential energy is the energy possessed by an object due to its position relative to other objects. In the ballistic pendulum experiment, as the pendulum with the embedded bullet rises, it reaches a maximum height. At this peak, its kinetic energy is entirely converted into gravitational potential energy.
This energy can be calculated using: \( PE = mgh \)
Incorporating the combined mass of the system and the height attained, potential energy allows us to understand the relationship between height and energy in vertically moving systems.
Grasping how potential energy is calculated and used can elucidate various scenarios from roller coasters to hydroelectric power plants.
kinetic energy
Kinetic energy is the energy an object possesses due to its motion. For the ballistic pendulum, the kinetic energy of the system right after the collision can be calculated using: \( KE = \frac{1}{2} mv^2 \)
This energy results from the initial motion of the bullet and pendulum. As the system ascends, its kinetic energy converts to potential energy.
This transformation demonstrates the direct relationship between an object's speed and its kinetic energy. Understanding kinetic energy and its role in motion is crucial for a variety of applications, from vehicle safety to athletic performance analysis.
collision physics
Collision physics encompasses the study of how objects interact during impacts. In the context of a ballistic pendulum, we consider an inelastic collision, where the bullet embeds into the pendulum, and they move together post-collision.
During such collisions, kinetic energy is not conserved due to deformation and heat, but momentum remains conserved.
Applying collision physics effectively involves understanding both conservation principles and the real-world effects of collisions, whether in engineering safety features or analyzing sports impacts.
It is crucial not only to predict outcomes but also to design systems that manage the energy during impacts safely.

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Most popular questions from this chapter

Cable Breaks The cable of the 1800 \(\mathrm{kg}\) elevator cab in Fig. \(10-54\) snaps when the cab is at rest at the first floor, where the cab bottom is a distance \(d=3.7 \mathrm{~m}\) above a cushioning spring whose spring constant is \(k=0.15 \mathrm{MN} / \mathrm{m}\). A safety device clamps the cab against guide rails so that a constant frictional force of \(4.4 \mathrm{kN}\) opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance \(x\) that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Alpha Particle An alpha particle (mass \(4 \mathrm{u}\) ) experiences an elastic head-on collision with a gold nucleus (mass \(197 \mathrm{u}\) ) that is originally at rest. (The symbol u represents the atomic mass unit.) What percentage of its original kinetic energy does the alpha particle lose?

Mount Everest The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

Voyager 2 Spacecraft Voyager 2 (of mass \(m\) and speed \(v\) relative to the Sun) approaches the planet Jupiter (of mass \(M\) and speed \(V_{J}\) relative to the Sun) as shown in Fig. \(10-60 .\) The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this slingshot encounter, which can be analyzed as a collision? Assume \(v=12 \mathrm{~km} / \mathrm{s}\) and \(V_{J}=13 \mathrm{~km} / \mathrm{s}\) (the orbital speed of Jupiter). The mass of Jupiter is very much greater than the mass of the spacecraft \((M \gg m)\).

Block Dropped on a Spring A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) 2.5 N/cm (Fig. 10-32). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

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