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A shell of mass \(m\) is fired from a cannon of mass \(n m\) (without the shell) with a relative speed of \(v_{0}\). The shell is projected horizontally. At the time of firing. (1) The total energy released is \(\frac{1}{2} \frac{m n}{(n+1)} v_{0}^{2}\) (2) The speed of shell is \(\frac{n v_{0}}{n+1}\) (3) The speed of cannon is \(\frac{v_{0}}{n+1}\) (4) The total energy released is \(\frac{1}{2} \frac{m v_{0}^{2}}{n+1}\)

Short Answer

Expert verified
Correct options are (1), (2), and (3).

Step by step solution

01

Understanding the Problem

Given a shell of mass \(m\) and a cannon with mass \(nm\), we need to find velocities and total energy. The given relative speed of the shell with respect to the cannon is \(v_0\). The goal is to derive expressions for the shell's speed, cannon's speed, and verify the total energy released at firing.
02

Conservation of Momentum

Due to conservation of momentum, the initial momentum (which is 0) equals the final momentum. Therefore:\[ m v_s = n m v_c \]where \(v_s\) is the speed of the shell, and \(v_c\) is the speed of the cannon.
03

Expressing Shell Velocity

We need to express \(v_s\) (shell velocity) using the relation:\[ v_s - v_c = v_0 \]Substitute \(v_c = \frac{v_s}{n}\) from the momentum equation:\[ v_s - \frac{v_s}{n} = v_0 \]Solve for \(v_s\):\[ v_s = \frac{n v_0}{n + 1} \]
04

Expressing Cannon Velocity

Substitute \(v_s = \frac{n v_0}{n+1}\) back into \(v_s = v_c + v_0\) to find \(v_c\). Using previously derived momentum relation:\[ v_c = \frac{v_s}{n} = \frac{v_0}{n+1} \]
05

Calculating Total Energy Released

Using the expression for kinetic energy, the total energy released when the shell is fired is:\[ KE_{total} = \frac{1}{2} m v_s^2 + \frac{1}{2} (nm) v_c^2 \]Plug \(v_s = \frac{n v_0}{n+1}\) and \(v_c = \frac{v_0}{n+1}\):\[ KE_{total} = \frac{1}{2} m \left(\frac{n v_0}{n+1}\right)^2 + \frac{1}{2} (nm) \left(\frac{v_0}{n+1}\right)^2 \]
06

Simplifying Energy Calculations

The energy becomes:\[ KE_{total} = \frac{1}{2} m \frac{n^2 v_0^2}{(n+1)^2} + \frac{1}{2} nm \frac{v_0^2}{(n+1)^2} \]Combine terms:\[ KE_{total} = \frac{1}{2} \frac{m n(n+1) v_0^2}{(n+1)^2} = \frac{1}{2} \frac{m n v_0^2}{(n+1)} \]
07

Verification of given options

Verify given options with the results:1) The total energy released matches the option (1): \(\frac{1}{2} \frac{m n}{(n+1)} v_{0}^{2} \).2) The speed of the shell is \(\frac{n v_0}{n+1}\), matching option (2).3) The speed of the cannon is \(\frac{v_0}{n+1}\), matching option (3).4) Option (4) does not match our calculations. It is incorrect.

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Key Concepts

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

Kinetic Energy
When discussing the concept of kinetic energy in the context of a shell fired from a cannon, it is essential to understand how energy is conserved and transferred in such a system.
Kinetic energy is defined as the energy that an object possesses due to its motion, expressed mathematically as \( KE = \frac{1}{2} mv^2 \).
Here, \(m\) represents the mass of the object and \(v\) its velocity.
In this scenario, when a shell is fired from a cannon, both the shell and the cannon gain kinetic energy.
- **Shell's Contribution**: The shell, having a velocity \(v_s\), will have a kinetic energy given by \( \frac{1}{2} m v_s^2 \).
- **Cannon's Contribution**: The cannon moves in the opposite direction with velocity \(v_c\), contributing a kinetic energy of \( \frac{1}{2} (nm) v_c^2 \), where \(n\) is the mass factor of the cannon compared to the shell.
The total kinetic energy released is the sum of these contributions when the system is isolated from external forces, such as friction or air resistance.
Therefore, the total energy equation simplifies to \( KE_{total} = \frac{1}{2} \frac{mnv_0^2}{n+1} \).
This demonstrates how energy is conserved and shared between the shell and the cannon at the time of firing.
Relative Speed
Relative speed is a crucial concept when analyzing the motion of two objects in the same system.
It indicates how fast one object is moving in comparison to another.
In this exercise, the relative speed \(v_0\) is defined as the speed of the shell with respect to the cannon.
This means that if the cannon was initially at rest, the shell would appear to speed off at this relative velocity in relation to the cannon.
To calculate the relative speed in this context, we consider the velocities of both the shell and cannon:
  • The equation \( v_s - v_c = v_0 \) represents the relative speed, where \( v_s \) is the velocity of the shell, and \( v_c \) is the velocity of the cannon.
It is important to note that the velocities here are considered immediately after firing, so inertia or other forces aren't altering their state.
The conservation of momentum helps solve for these velocities, ensuring the actual speed observed accounts for both the forward motion of the shell and the recoil of the cannon.
Through this analysis, we observe how conservation laws apply in a dynamic setting.
Shell Velocity
One pivotal goal in this exercise is determining the velocity of the shell, noted as \( v_s \). Understanding this helps in analyzing the dynamics of the firing event.
Using the conservation of momentum and the definition of relative speed, the expression for shell velocity is derived:- First, we insert the relation from momentum conservation \( mv_s = n mv_c \).- By rearranging to isolate \( v_s \), and substituting \( v_c \) with a suitable term, we derive the expression for \( v_s \): \( v_s = \frac{nv_0}{n+1} \).This equation demonstrates how the mass ratio between the cannon and shell impacts the velocity of the shell.
The velocity is derived from the shared energy and momentum in this isolated system.
Here, each component's final speed depends directly on the mass involved and the shared energy source (the explosive force within the cannon).
This velocity interpretation provides insights into the shell's resultant motion relative to the ground after it is fired.
Cannon Velocity
Determining the cannon's velocity \( v_c \) is necessary to understand the dynamics of the system as a whole.
After the shell is fired, the cannon moves in the opposite direction due to the principle of conservation of momentum.
From our derivations, we found \( v_c \) using the relationships inherent in the system:- Knowing \( v_s \) from previous calculations, and using the relative speed equation, the cannon velocity is given by \( v_c = \frac{v_0}{n+1} \).By integrating such calculations, the cannon's velocity becomes crucial in comprehending the entire motion system.
It represents how much recoil is imparted to the cannon, a direct consequence of the momentum transfer involved in the explosive event.
Since the cannon is considerably heavier than the shell, its movement is significantly slower, illustrating a key concept of how mass distribution affects velocity upon impact.
By exploring these velocities, the lesson in the conservation of energy and momentum is fully realized, showcasing their practical application in physics.

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Most popular questions from this chapter

A frog sits on the end of a long board of length \(L=5 \mathrm{~m}\). The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed (in \(\mathrm{m} / \mathrm{s}\) ), i.e., relative to ground that allows the frog to do the trick? The board and the frog have equal masses.

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