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Question: A 3.00-kg cart with an initial velocity of +4.00 collides inelastically with a 5.00-kg cart with an initial velocity of +2.00 The carts stick together when they collide. What is their speed after colliding?

This is a physics problem of conservation of linear momentum. The linear momentum before the collision is equal to the linear momentum after the collision in an isolated system (one not subject to any external forces). This can be expressed as: m1v1_initial + m2v2_initial = (m1 + m2)v_final where m1 and m2 are the masses of the two carts, v1_initial and v2_initial are their initial velocities, and v_final is their final velocity. Plugging in the given values, we have: (3.00 kg × 4.00 m/s) + (5.00 kg × 2.00 m/s) = (3.00 kg + 5.00 kg) × v_final Solving the equation for v_final gives: v_final = [(3.00 kg × 4.00 m/s) + (5.00 kg × 2.00 m/s)] / (3.00 kg + 5.00 kg) Calculating the values, we get: v_final = [12.00 kg*m/s + 10.00 kg*m/s] / 8.00 kg So: v_final= 22.00 kg*m/s / 8.00 kg = 2.75 m/s Hence, the speed of the combined carts after the collision is 2.75 m/s.

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