Maximum Compression of a Spring in a Collision?

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In a perfectly inelastic collision with a wall, an object on a spring compresses the spring to its maximum extent. The discussion highlights the challenge of calculating the force exerted during the collision, noting that traditional methods like Newton's Second Law may not apply directly. Instead, conservation of energy can be utilized, equating the kinetic energy of the moving system to the potential energy stored in the spring at maximum compression. This leads to the equation mv² = kx², allowing for the determination of maximum compression by isolating x. The conversation emphasizes the importance of using physical intuition and energy conservation in solving the problem.
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Homework Statement


An object with mass m is being held on a spring in equilibrium position with spring constant k. This system(both the object and the spring) is moving in one direction with a velocity v. The system then collides perfectly inelastically with a wall. What is the maximum compression of the spring?

Homework Equations


F=-kx
Conservation of Momentum(?)
Newton's Third Law of Motion

The Attempt at a Solution


By Newton's Third Law, the force the wall exerts on the system is equivalent in magnitude to the force the system pushes on the wall. However, the force appears incalculable. Newton's Second Law cannot really be used here because it concerns net force on an object. Therefore, it appears that the only way out is to calculate the impulse the object is experiencing and then to take the time derivative of that impulse. This does not seem to accomplish anything either, since the impulse is mv, and when one takes the time derivative of that, it comes out to be ma, which doesn't help my cause one bit. I feel at an impasse with this problem. Help would be much appreciated.
 
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Perhaps we can use some physical intuition to find a relevant equation.

Perfectly inelastic generally means the two objects "stick" together and for most situations means that you can't rely on conservation of energy. However, in this case you could argue that the energy is completely converted into potential energy of the spring and hence we could try using conservation of energy.

With this tool at your hands, the solution is simple.
 
Coto said:
Perhaps we can use some physical intuition to find a relevant equation.

Perfectly inelastic generally means the two objects "stick" together and for most situations means that you can't rely on conservation of energy. However, in this case you could argue that the energy is completely converted into potential energy of the spring and hence we could try using conservation of energy.

With this tool at your hands, the solution is simple.

Ah! I never thought of it that way. So what you're saying is that mv²=kx² since the 1/2's cancel.
 
Correct. All you need to do is isolate for x then.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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