Understand the momentum and velocit

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In an elastic collision between two equally massive carts moving at equal velocities in opposite directions, they will bounce off each other, conserving both momentum and kinetic energy. The total momentum before and after the collision remains the same, and the kinetic energy is also conserved, meaning the carts will have the same speed but in opposite directions after the collision. In contrast, during an inelastic collision, the carts will collide and stick together, resulting in a loss of kinetic energy, although momentum is still conserved. The combined mass will move with a reduced velocity after the collision. Understanding these principles clarifies the differences between elastic and inelastic collisions.
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Hi guys, i don't understand the momentum and velocity and when they say elasticity in the following question...

What will happen when equally massive carts collide elastically with one another at equal velocity?

What will happen when the collision is inelastic?

I am very confused. please help me.
 
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It means "What will happen if two carts with equal mass and same magnitude of velocity ( obviously in opposite direction!) collide elastically? What will happen when the collision is inelastic?"
 
An elastic collision is one in which there is no loss in the kinetic energy of the system, i.e. sum of kinetic energies of carts before collision = sum of kinetic energies of carts after collision.
 
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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