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According to the Momentum Conservation Principle:

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

Lets say we have a 2 carts of a certain mass. We launch 2 balls of equal mass with equal speed, 1 into each cart, 1st made from plasticine ( inelastic collision), the 2nd made from rubber (elastic collision). Does it mean that a cart hit by rubber ball would have after an impact a greater speed?

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What exactly you mean by launching a ball "into" a cart is important. If this means that the rubber ball lands on a horizontal surface of the cart and bounces back out, the cart doesn't have to move at all (the momentum exchange can be purely in the vertical direction for a perfectly elastic collision) while it does with the sticky ball.

If what you mean is the balls impact the cart from behind, pushing it forward, then yes, since the rubber ball bounces back, conservation of momentum implies that the car impacted by it moves faster.

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AFAIK collision is defined by two objects coming into contact, hence the name.

This may be of assistance in terms of determining an inelastic versus elastic collision: https://en.wikipedia.org/wiki/Coefficient_of_restitution

Other than that, here are your two situations that I can think of and can work with -

Case 1 - Elastic

$$m_{\text{cart}}v_{\text{cart}_0} + m_{\text{ball}}v_{\text{ball}_0} = m_{\text{ball}}v_{\text{ball}_0}$$

(Cart is at rest; ball is assumed to not roll.) Due to conservation of momentum -

$$m_{\text{ball}}v_{\text{ball}_0} = m_{\text{cart}}v_{\text{cart}} + m_{\text{ball}}v_{\text{ball}}$$

Our second equation of use is conservation of kinetic energy -

$$\frac{1}{2}m_{\text{cart}}v^2_{\text{cart}_0} + \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0} = \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0}$$ $$\frac{1}{2}m_{\text{ball}}v^2_{\text{ball}_0} = \frac{1}{2}m_{\text{cart}}v^2_{\text{cart}} + \frac{1}{2}m_{\text{ball}}v^2_{\text{ball}}$$

You can work your way to a result from there.

Case 2 - Perfectly Inelastic

$$m_{\text{ball}}v_{\text{ball}_0} = \left(m_{\text{cart}} + m_{\text{ball}} \right)v$$

Kinetic energy is not conserved here, but the final velocity can be found quite easily.

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