A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision

v₂f = 0.16 m/s (to the right)

Explanation:

Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.The initial momentum is due to the small cart only, because the big one is at rest.Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :

       

The final momentum can be written as follows:

       

        As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:

       

As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current

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