Conservation of energy and momentum in a gauss gun

In summary: The summary is In summary, the speaker is doing a project on the Gauss gun and Newton's cradle using MatLab. They are facing problems with the gauss gun, specifically with the use of different types of balls (hollow and solid) and the conservation of momentum. They have calculated the velocities of the balls but found that momentum is not conserved, so they are now considering the torque applied to each ball during collision. They have assumed perfectly elastic collisions and no losses due to friction or air resistance. They are seeking help for calculating torque and angular acceleration during the collision and whether the same torque is applied to the ball leaving the pack.
  • #1
billy_button
1
0
Hi, I'm doing a project on the Gauss gun and Newton's cradle. I'm trying to produce little models in MatLab and have come across some problems. Namely with the gauss gun, using different types of balls, hollow and solid (I've not even got round to adding acceleration from a magnet yet).

Currently I have a hollow ball hitting the pack and releasing a solid ball at the end. I have calculated the velocities of both, assuming energy is conserved, from the linear and rotational KE's, resulting in the solid ball leaving the pack with a higher velocity than the hollow ball hit it with. This means however, that momentum has not been conserved (off the top of my head I think the solid ball gained momentum??).

I'm now assuming that it must really be conserved and I need to look at the torque applied to stop and start each ball? Does this sound right and if so 1)how do I calculate the torque or angular acceleration at the instant of the collision? and 2)would the same torque be applied to the ball leaving the pack?

For 'ease' of calculations I have assumed perfectly elastic collisions and no losses due to friction/air resistance.

I'd be most appreciative of any help.
 
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  • #2
Why would you use torque in a linear arrangement of balls? Torque around what?

Momentum must be conserved, this should be one of the constraints in the equations, for each collision.
 

Related to Conservation of energy and momentum in a gauss gun

1. How does a gauss gun conserve energy and momentum?

A gauss gun operates on the principle of electromagnetic induction, where a series of magnets and conductive rails are used to accelerate a projectile. By carefully arranging the magnets and rails, the energy and momentum of the projectile is conserved as it moves down the barrel.

2. Can a gauss gun violate the law of conservation of energy and momentum?

No, a gauss gun does not violate the laws of conservation. The energy and momentum of the system are always conserved, but it may appear that the projectile is gaining energy as it accelerates down the barrel. This is due to the conversion of potential energy into kinetic energy.

3. How efficient is a gauss gun in conserving energy and momentum?

A gauss gun can be very efficient in conserving energy and momentum, with some designs achieving over 90% efficiency. However, factors such as friction and air resistance can reduce this efficiency.

4. What happens to the energy and momentum of the projectile after it leaves the gauss gun?

The energy and momentum of the projectile remain constant after it leaves the gauss gun. However, it may experience a decrease in speed due to air resistance and other external factors.

5. Are there any real-life applications of the conservation of energy and momentum in gauss guns?

Yes, gauss guns have been used in various applications such as launching satellites into orbit, accelerating particles in particle accelerators, and studying the behavior of materials under high velocities.

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