To disable ballistic missiles during the early boost phase of their flight, an "electromagnetic rail gun," to be carried in low-orbit Earth satellites, has been proposed. The gun might fire a \(2.38-\mathrm{kg}\) maneuverable projectile at \(10.0 \mathrm{~km} / \mathrm{s}\). The kinetic energy carried by the projectile is sufficient on impact to disable a missile even if it carries no explosive. (A weapon of this kind is a "kinetic energy" weapon.) The projectile is accelerated to muzzle speed by electromagnetic forces. Suppose instead that we wish to fire the projectile using a spring (a "spring" weapon). What must the force constant be in order to achieve the desired speed after compressing the spring \(1.47 \mathrm{~m} ?\)

Short Answer

Expert verified
By substituting the calculated kinetic energy and given displacement into the formula \(k = \frac{2*U}{x^2}\), the spring constant required to achieve the same speed as the 'electromagnetic rail gun' can be calculated.

Step by step solution

01

Calculate the kinetic energy of the projectile

To calculate the kinetic energy of the projectile, apply the formula \(KE = 0.5 * m * v^2\), where m is the mass of the projectile and v is its velocity. Here, the mass m is given as 2.38 kg and the velocity v as 10.0 km/s, which should be converted to m/s by multiplying it by 1000.
02

Find the Force constant \(k\)

Knowing that the spring potential energy (stored energy) when it reaches the maximum displacement (maximum compression or elongation) is equal to the kinetic energy of the projectile, apply the formula \(U = 0.5 * k * x^2\), where U is the potential energy (which is equivalent to the kinetic energy calculated in Step 1), k is the spring constant to be found, and x is the displacement. In our case, the displacement x as given is 1.47 m. Rearranging the formula to find \(k\), we get \(k = \frac{2*U}{x^2}\)

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Key Concepts

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

Ballistic Missile Defense
The concept of Ballistic Missile Defense (BMD) is vital for ensuring national and global security by intercepting and neutralizing incoming missiles, particularly during their initial boost phase. This is the period right after launch when the missile is accelerating and is, theoretically, most vulnerable.

In the context of the proposed exercise, we discuss a kinetic energy weapon, specifically an 'electromagnetic rail gun', which is planned to be stationed on satellites in low earth orbit. The principle behind using a kinetic energy weapon is that it inflicts damage through the immense kinetic energy of a projectile, thus requiring no explosives. Its high-speed impact has the potential to destroy or disrupt the target.

The key to effective BMD with kinetic energy weapons lies in precision and the immense speed at which projectiles must be launched to intercept missiles potentially traveling at comparable velocities. The technical challenges to achieving these speeds and accuracy are significant but imperative for the defense strategy.
Electromagnetic Rail Gun
An electromagnetic rail gun is an advanced weapon utilizing magnetic fields to launch projectiles at extremely high velocities. This device consists of two parallel rails and a sliding armature that holds the projectile. When electric current flows through the rails, it creates a magnetic field which exerts a force on the armature, thus propelling the projectile.

The velocity achievable by such weapons makes them a prime candidate for kinetically-based interception strategies in ballistic missile defense. The main attraction of rail guns lies in their ability to fire projectiles at velocities much greater than those possible with conventional chemical propellants, up to several kilometers per second. This immense speed provides the kinetic energy required to damage or destroy a target upon impact, as described in our exercise scenario.

The efficiency and effectiveness of rail guns promise a leap in defense capabilities, with their long-range, high velocity, and lack of explosive ordnance reducing risks of collateral damage.
Spring Constant Calculation
The spring constant, denoted by letter k, is a measure of a spring's stiffness and is a vital parameter in the calculation for the proposed spring-based weapon system. The spring constant essentially determines how much force is needed to compress or extend a spring by a certain distance.

To find the required spring constant for a projectile to achieve a specific speed, as in our exercise, we must equate the kinetic energy of the projectile to the potential energy stored in the spring. Using the formula for spring potential energy, \(U = 0.5 * k * x^2\), where U is the energy and x is the compression or elongation distance, we can solve for the spring constant k.

In practical applications, a higher spring constant means a stiffer spring, and achieving the desired projectile speed would require significant force to compress. This principle, while simple in concept, involves complex engineering challenges when designing mechanisms capable of compressing springs with incredibly high force constants.

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Most popular questions from this chapter

A body is rolling horizontally without slipping with speed \(v\). It then rolls up a hill to a maximum height \(h\). If \(h=3 v^{2} / 4 g\) what might the body be?

The potential energy corresponding to a certain two-dimensional force is given by \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right) .(a)\) Derive \(F_{x}\) and \(F_{y}\) and describe the vector force at each point in terms of its coordinates \(x\) and \(y .(b)\) Derive \(F_{r}\) and \(F_{\theta}\) and describe the vector force at each point in terms of the polar coordinates \(r\) and \(\theta\) of the point. ( \(c\) ) Can you think of a physical model of such a force?

Show that for the same initial speed \(v_{0}\) the speed \(v\) of a projectile will be the same at all points at the same elevation, regardless of the angle of projection. Ignore air drag.

A pendulum is made by tying a \(1.33-\mathrm{kg}\) stone to a string \(3.82 \mathrm{~m}\) long. The stone is projected perpendicular to the string, away from the ground, with the string at an angle of \(58.0^{\circ}\) with the vertical. It is observed to have a speed of \(8.12 \mathrm{~m} / \mathrm{s}\) when it passes its lowest point. (a) What was the speed of the stone when projected? (b) What is the largest angle with the vertical that the string will reach during the stone's motion? (c) Using the lowest point of the swing as the zero of gravitational potential energy, calculate the total mechanical energy of the system.

In one dimension, the magnitude of the gravitational force of attraction between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$F_{x}(x)=G \frac{m_{1} m_{2}}{x^{2}}$$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty,(b)\) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=\) \(x_{1}+d ?\)

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