/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A wooden block of mass \(M\) res... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 35

A wooden block of mass \(M\) resting on a frictionless horizontal surface is attached to a rigid rod of length \(\ell\) and of negligible mass (Fig. P11.35). The rod is pivoted at the other end. A bullet of mass \(m\) traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) hits the block and becomes embedded in it. (a) What is the angular momentum of the bullet-block system? (b) What fraction of the original kinetic energy is lost in the collision?

Short Answer

Expert verified
The angular momentum of the bullet-block system is \(m v \ell\). The fraction of the original kinetic energy lost in the collision is \(\frac{M}{M+m}\).

Step by step solution

01

Calculate the initial angular momentum

The initial angular momentum \(L_i\) of the system is the momentum of the bullet at the moment of impact. Angular momentum is calculated using the formula \(L = mvr\), where \(m\) is the mass of the bullet, \(v\) is its velocity, and \(r\) is the distance to the pivot point. In this case, \(r = \ell\), the length of the rod. So, \(L_i = m v \ell\).
02

Calculate the final angular momentum

After the impact, the bullet and the block move together as a single body that rotates around the pivot. The final angular momentum \(L_f\) is calculated using the formula \(L_f = I \omega\), where \(I\) is the moment of inertia of the system and \(\omega\) is the angular velocity. The moment of inertia \(I\) of the system (composed of the bullet and the block) is given by \(I = (M + m) \ell^2\), assuming both masses are concentrated at the end of the rod. Angular velocity \(\omega\) is related to the linear speed \(v\) by the relation \(\omega = v / \ell\). Hence, \(L_f = (M+m)\ell^2 * [v/\ell] = (M + m) v \ell\).
03

Apply Conservation of Angular Momentum

According to the principle of conservation of angular momentum, the initial and final angular momentum should be equal \(L_i = L_f\). This gives \(m v \ell = (M + m) v \ell\), which simplifies to \(v = v\). This confirms that angular momentum is conserved in the collision.
04

Calculate the initial kinetic energy

The initial kinetic energy \(K_i\) of the system is the kinetic energy of the bullet at the moment of impact, which is given by the formula \(K = \frac{1}{2} m v^2\). So, \(K_i = \frac{1}{2} m v^2\).
05

Calculate the final kinetic energy

After the impact, the bullet and the block move together and rotate around the pivot. Their kinetic energy \(K_f\) is given by \(K_f = \frac{1}{2} I \omega^2\). Substituting for \(I\) and \(\omega\) as before, \(K_f = \frac{1}{2}(M + m)\ell^2 * [v/\ell]^2 = \frac{1}{2}(M+m)v^2\).
06

Calculate the fraction of energy lost

The fraction of the original kinetic energy lost in the collision comes from the difference between the initial and final kinetic energies. This fraction is given by \([K_i - K_f]/ K_i = [(\frac{1}{2} m v^2 - \frac{1}{2}(M+m)v^2)/ \frac{1}{2} m v^2] = \frac{M}{M+m}\).

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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. It's crucial in analyzing any dynamic system, particularly in scenarios like a bullet striking a block. The kinetic energy of an object traveling with velocity is given by the formula:
\[ K = \frac{1}{2} m v^2 \]
where:
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object.
Initially, only the bullet has kinetic energy since the block is at rest.
When the bullet embeds into the block, the system (bullet plus block) has a new shared velocity, transforming the way we look at kinetic energy in the aftermath of this impact. The key takeaway is that during the collision energy can be redistributed or lost due to factors like friction or deformation. However, in our scenario, the relationship considers theoretical losses because the surface and rod act perfectly frictionless.
Collision
In physics, a collision is any event where two or more bodies exert forces on each other for a short duration. Collisions can be elastic or inelastic:
  • **Elastic collision:** kinetic energy is conserved.
  • **Inelastic collision:** kinetic energy is not conserved (some is transformed into other energy forms).
In this exercise, when a bullet embeds itself into the block, it becomes an example of a perfectly inelastic collision. This means although they move together post-collision, kinetic energy is not conserved. Instead, some energy is transformed into heat or sound.
Interestingly, angular momentum is conserved even if kinetic energy is not. Understanding this conservation helps in determining resultant properties post-collision, such as angular velocity and system inertia. Conservation laws are foundations of analyzing interactions in closed systems, giving us powerful tools to predict how systems behave after collisions.
Moment of Inertia
Moment of inertia is a property of a body that defines how hard it is to change its rotational motion around an axis. It's analogous to mass in linear motion. In the context of a bullet-block system, it's vital in understanding rotational dynamics around the pivot point.
For our problem setup, the formula for the moment of inertia (I) of the system at the pivot is:
\[ I = (M + m) \ell^2 \]
where:
  • \(M\) is the mass of the block,
  • \(m\) is the mass of the bullet,
  • \(\ell\) is the length of the rod.
This equation suggests that as both the bullet and block are at the rod's end, their combined effect on rotational inertia is significant. It's crucial when calculating the final angular velocity since it aids in connecting it with linear motion.
Being adept at determining moment of inertia allows us to comprehend how a body will rotate when force is applied, such as the impact of a bullet. It solidifies an understanding of rotational dynamics in systems subjected to external interactions.

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Most popular questions from this chapter

A puck of mass \(m\) is attached to a cord passing through a small hole in a frictionless, horizontal surface (Fig. P11.49). The puck is initially orbiting with speed \(v_{i}\) in a circle of \(\mathrm{ra}-\) dius \(r_{i}\). The cord is then slowly pulled from below, decreasing the radius of the circle to \(r\). (a) What is the speed of the puck when the radius is \(r ?\) (b) Find the tension in the cord as a function of \(r .\) (c) How much work \(W\) is done in moving \(m\) from \(r_{i}\) to \(r ?\) (Note: The tension depends on \(r\) ) (d) Obtain numerical values for \(v, T,\) and \(W\) when \(r=0.100 \mathrm{m}\) \(m=50.0 \mathrm{g}, r_{i}=0.300 \mathrm{m},\) and \(v_{i}=1.50 \mathrm{m} / \mathrm{s}.\)

Global warming is a cause for concern because even small changes in the Earth's temperature can have significant consequences. For example, if the Earth's polar ice caps were to melt entirely, the resulting additional water in the oceans would flood many coastal cities. Would it appreciably change the length of a day? Calculate the resulting change in the duration of one day. Model the polar ice as having mass \(2.30 \times 10^{19} \mathrm{kg}\) and forming two flat disks of radius \(6.00 \times 10^{5} \mathrm{m} .\) Assume the water spreads into an unbroken thin spherical shell after it melts.

Comet Halley moves about the Sun in an elliptical orbit, with its closest approach to the Sun being about \(0.590 \mathrm{AU}\) and its greatest distance \(35.0 \mathrm{AU} \quad(1 \mathrm{AU}=\) the Earth-Sun distance). If the comet's speed at closest approach is \(54.0 \mathrm{km} / \mathrm{s}\) what is its speed when it is farthest from the Sun? The angular momentum of the comet about the Sun is conserved, because no torque acts on the comet. The gravitational force exerted by the Sun has zero moment arm.

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(0.529 \times 10^{-10} \mathrm{m}\) around the proton. Assuming the orbital angular momentum of the electron is equal to \(h / 2 \pi,\) calculate (a) the orbital speed of the electron, (b) the kinetic energy of the electron, and (c) the angular frequency of the electron's motion.

A thin uniform rectangular sign hangs vertically above the door of a shop. The sign is hinged to a stationary horizontal rod along its top edge. The mass of the sign is \(2.40 \mathrm{kg}\) and its vertical dimension is \(50.0 \mathrm{cm} .\) The sign is swinging without friction, becoming a tempting target for children armed with snowballs. The maximum angular displacement of the sign is \(25.0^{\circ}\) on both sides of the vertical. At a moment when the sign is vertical and moving to the left, a snowball of mass \(400 \mathrm{g}\), traveling horizontally with a velocity of \(160 \mathrm{cm} / \mathrm{s}\) to the right, strikes perpendicularly the lower edge of the sign and sticks there. (a) Calculate the angular speed of the sign immediately before the impact. (b) Calculate its angular speed immediately after the impact. (c) The spattered sign will swing up through what maximum angle?
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