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A 12.0 -g rifle bullet is fired with a speed of 380 \(\mathrm{m} / \mathrm{s}\) into a ballistic pendulum with maxs 6.00 \(\mathrm{kg}\) , suspended from a cord 70.0 \(\mathrm{cm}\) long (see Example 8.8 in Section 8.3 . Compute (a) the vertical height through which the pendulum rises, (b) the initial kinetic energy of the bullet, and (c) the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendalum.

Short Answer

Expert verified
The pendulum rises 0.0294 m, the bullet's initial KE is 867.6 J, and the system's KE is 1.733 J.

Step by step solution

01

Calculate Initial Kinetic Energy of the Bullet

Use the formula for kinetic energy, \( KE = \frac{1}{2} mv^2 \). Here the mass of the bullet, \( m = 0.012 \ kg \), and the velocity, \( v = 380 \ m/s \). Substitute these values into the formula to find the initial kinetic energy of the bullet: \(KE = \frac{1}{2} \times 0.012 \times (380)^2 = 867.6 \ J.\)
02

Calculate the Speed of Bullet-Pendulum System

Apply the conservation of momentum just after the bullet becomes embedded in the pendulum. The initial momentum of the bullet is equal to the combined momentum of the bullet-pendulum system: \( m_{bullet}v_{bullet} = (m_{system})v_{system} \). Therefore, \(\ (0.012)(380) = (6.012)v_{system} \). Solve for \(v_{system}\): \(v_{system} = \frac{(0.012)(380)}{6.012} = 0.7598 \ m/s.\)
03

Calculate Kinetic Energy of Bullet-Pendulum System

Use the kinetic energy formula again for the combined mass: \( KE_{system} = \frac{1}{2} m_{system} v_{system}^2 \). Here, \( m_{system} = 6.012 \ kg \) and \( v_{system} = 0.7598 \ m/s \). Substitute these values: \( KE_{system} = \frac{1}{2} \times 6.012 \times (0.7598)^2 = 1.733 J.\)
04

Calculate the Height Pendulum Rises

Use the conservation of energy to find the height. After the collision, the kinetic energy is converted into potential energy at the highest point of the swing : \( KE_{system} = PE \). The potential energy formula is \( PE = mgh \), where \( h \) is the height. Set \( KE_{system} = m_{system}gh \) and solve for \( h \): \( h = \frac{KE_{system}}{m_{system}g} = \frac{1.733}{6.012 \times 9.8} \approx 0.0294\ m.\)

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For any moving object, this energy can be calculated using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
Where \( m \) is the mass of the object and \( v \) is its velocity. The faster an object moves, or the heavier it is, the more kinetic energy it has. In the context of the ballistic pendulum, when the bullet is fired, it starts with an initial kinetic energy based purely on its mass and velocity. As worked out in the solution, a bullet with a mass of 12 grams traveling at a speed of 380 meters per second has a kinetic energy of 867.6 Joules. This high kinetic energy is why bullets can penetrate objects and immediately transfer energy upon impact. Kinetic energy's dependence on both mass and speed implies that doubling the speed of an object can quadruple its kinetic energy, highlighting its sensitivity to velocity changes.
Conservation of Momentum
The principle of conservation of momentum is crucial in understanding collisions like the one between the bullet and the pendulum. It states that the total momentum of a closed system remains constant if it is not subjected to external forces. When the bullet hits the pendulum, they stick together, and their total momentum before and after the collision stays the same. This principle can be mathematically expressed as:
  • \( m_{bullet}v_{bullet} = (m_{system})v_{system} \)
In our exercise, the bullet initially carries all the momentum, calculated by multiplying its mass and velocity. After the collision, the bullet and pendulum move as a combined system. By equating their pre- and post-collision momenta, we solve for the velocity of the combined system, which turns out to be 0.7598 m/s. This example exemplifies how momentum conservation aids in determining the changes in motion post-collision, a technique often used in physics to solve intricate problems.
Potential Energy
Potential energy is the stored energy of an object due to its position or arrangement. When dealing with vertical movements under the influence of gravity, the potential energy at a given height can be defined as:
  • \( PE = mgh \)
Where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and \( h \) is the height above the reference point. Following the collision, as the bullet-pendulum system rises to its highest point, kinetic energy is converted entirely into potential energy. Therefore, the initial kinetic energy of the combined system becomes equal to its potential energy at the peak of the swing. From our solution, it was shown that the kinetic energy of 1.733 Joules is converted to potential energy, enabling us to deduce that the pendulum rises to a height of approximately 0.0294 meters. Potential energy is fundamental in predicting how much an object can rise or fall within a gravitational field when kinetic energy transforms.

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Most popular questions from this chapter

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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g .\) $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=a t,\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v,\) find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{s}\) . (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m}\) , find the mass of the raindrop at \(t=3.00 \mathrm{s}\) . For many more intriguing aspects of this problem, see \(\mathbf{K} .\) S. Krane, Amer Jour. Phys, Vol. 49\((1981)\) pp. \(113-117\)

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A 30 -caliber bullet has mass 0.00720 \(\mathrm{kg}\) and a speed of 601 \(\mathrm{m} / \mathrm{s}\) relative to the muzzle when fired from a rifle that has mass 2.80 kg. The loosely held rifle recoils at a speed of 1.85 \(\mathrm{m} / \mathrm{s}\) relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

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