Can Kinetic Energy Be Completely Lost in a Collision?

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A system cannot completely lose its kinetic energy in a collision, as demonstrated by elastic collisions where objects bounce off each other, retaining kinetic energy. Inelastic collisions, such as two equal balls of damp clay that deform and stick together, do result in some kinetic energy being converted into other forms, but the total energy is conserved. The discussion highlights the distinction between elastic and inelastic collisions regarding energy retention. Ultimately, while kinetic energy can be transformed, it cannot be entirely lost in a collision. Understanding these principles is crucial in physics.
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It is not possible for a system to lose all of its kinetic energy in a collision.

A) True
B) False

This question seems really tricky, but I believe the answer to be A: true. Even if two identical masses are approaching each other head-on at the same velocity (two billiard balls), when they collide they bounce off each other and continue with KE. Is this right?
 
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TwinGemini14 said:
It is not possible for a system to lose all of its kinetic energy in a collision.

A) True
B) False

This question seems really tricky, but I believe the answer to be A: true. Even if two identical masses are approaching each other head-on at the same velocity (two billiard balls), when they collide they bounce off each other and continue with KE. Is this right?

That is true for elastic collisions.

What about inelastic collisions - say 2 equal balls of damp clay that deform and stick together?
 
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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