I'm not really sure if I'm on the right track or not...Calvin Pitts said:
So would that be as simple as adding the momentums together and calling it the momentum of the system?scottdave said:
In that last line you are equating the net momentum of two different systems... why should they be equal??Calvin Pitts said:
The mechanical energy supplied by the explosion of the gun-powder in the gun is the same in both scenarios. The momentum is conserved during the shot, but the system is the earth+gun+projectile in the first case, and the gun+projectile in the second case. You can use conservation of momentum in the second case (total momentum is the same before and after the shot) , and you get the energy of the system from the muzzle speed in the first case.Calvin Pitts said:
ehild said:
So wait, if the total momentum is the same before and after, wouldn't it just be zero? Because I thought the momentum before the shot would be zero because nothing is moving.ehild said:
Yes. Now you write the relation between v1 and v2 using conservation of momentum. And yes, the total momentum is zero before the shot.Calvin Pitts said:
Correct, the sum before and after is zero. But remember that momentum is a vector quantity, having both magnitude and direction...Calvin Pitts said:
ehild said:
Yes. m1v1= -m2v2.Calvin Pitts said:
ehild said:Just post a picture of your work, ugly or not, it is useful.
The result is correct, although I can not follow your notations. Sooner or later you learn to type in the Maths.Calvin Pitts said:
ehild said:The result is correct, although I can not follow your notations. Sooner or later you learn to type in the Maths.
Not bad for a first thread!
I like the way you presented it, showing that the first scenario is a different system (because of the external force). I think the way I read the question, and the way that I said it may have been misleading into suggesting that the momentum in the first situation is the same as the second situation, when in fact, the momentum before the explosion is the same as after the explosion.Hiero said:
A gun-projectile elastic collision is a type of collision in which a gun fires a projectile (such as a bullet) and the projectile collides with a target or surface, bouncing off without any loss of kinetic energy.
The elasticity of the projectile determines how much energy is conserved during the collision. In an elastic collision, the projectile will bounce off the target with the same velocity and kinetic energy as before the collision. In an inelastic collision, some of the energy is lost and the projectile may not bounce off with the same velocity.
The outcome of a gun-projectile elastic collision can be influenced by factors such as the mass and velocity of the projectile, the angle of impact, and the elasticity of the materials involved. These factors can affect the amount of kinetic energy transferred during the collision.
The conservation of momentum states that the total momentum of a system remains constant in the absence of external forces. In a gun-projectile elastic collision, the total momentum of the system (gun + projectile) before and after the collision is equal, as long as there are no external forces acting on the system.
Yes, a gun-projectile elastic collision can be perfectly elastic if there is no loss of kinetic energy during the collision. However, in real-world scenarios, some energy will always be lost due to factors such as air resistance and friction, making the collision technically inelastic.