Projectile Kinetic Energy and Recoil in Spring Gun Experiment

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The discussion focuses on the physics of a spring gun experiment involving projectile kinetic energy and recoil. It emphasizes the conservation of momentum, stating that the initial momentum is zero, which leads to the equation M1V1 = (M2 - M1)V2. The kinetic energy of the projectile is represented as K = 0.5M1V1^2, while the recoil energy of the gun is derived from KE = 0.5(M2 - M1)V2^2. The final answer for the gun's kinetic energy is expressed as (M1/(M2-M1))K. Understanding these principles is crucial for solving the problem accurately.
CoreanJesus

Homework Statement


A projectile of mass M1 is fired horizontally from a spring gun that is initially at rest on a frictionless surface. The combined mass of the gun and projectile is M2. If the kinetic energy of the projectile after firing is K, the gun will recoil with a kinetic energy equal to...

Homework Equations


this is part of the problem.
I know you need the kinetic energy formula but I don't get the other formula.

The Attempt at a Solution


N/A Answer is (M1/(M2-M1))K
 
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Hi,

You will need to take into consideration that the total momentum is conserved.
 
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But what should the initial momentum be? 0?
 
Yes, the initial momentum is zero. That does not mean that the final velocity of the gun and of the projectile will be also zero. Do not forget that velocity (and momentum too) is a vector. So if you have two velocities (or momentum) in the opposite directions, you need to subtract one velocity from the other.
 
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so M1V1=(M2-M1)V2
K=.5M1V1^2
KE=.5(M2-m1)V2^2 and plugin?
 
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Exactly.
 
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Thank you so much!
 

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