This problem demonstrates conservation of linear momentum in an explosion-type interaction. When external forces on a system are negligible, the total momentum remains constant throughout any internal interaction. Initially, both carts are at rest, so p_initial = 0. After the spring releases, the system momentum must still be zero: m₁v₁ + m₂v₂ = 0. Since the carts are identical and cart 1 moves left at 3.0 m/s, we have m(-3.0 m/s) + m(v₂) = 0. Choice A (3.0 m/s left) would give the same direction for both carts, violating momentum conservation. Solving: v₂ = +3.0 m/s (rightward). For explosion problems starting from rest, remember that objects must move in opposite directions with momenta that cancel out.

This problem demonstrates conservation of linear momentum in a perfectly inelastic collision with one object initially at rest. When external forces are negligible, the total momentum of the system remains constant before and after the collision. Initially, only the 3.0 kg cart moves: p_initial = (3.0 kg)(2.0 m/s) + (1.0 kg)(0 m/s) = 6.0 kg·m/s rightward. After sticking together, the combined mass is 4.0 kg, so p_final = (4.0 kg)v_final. Choice D (6.0 m/s) incorrectly uses the momentum value as velocity. Setting p_initial = p_final: 6.0 = 4.0v_final, giving v_final = 1.5 m/s rightward. To find the final velocity in sticky collisions, divide the initial total momentum by the combined final mass.

This question assesses the conservation of linear momentum in an explosion-like release of a compressed spring. In a system where external forces are negligible, the total linear momentum remains constant and is zero if the system starts at rest. When Cart 1 moves left at 3.0 m/s, its momentum is -1.2 kg·m/s (right as positive). To conserve zero total momentum, Cart 2 must have +1.2 kg·m/s, resulting in 2.0 m/s right. Option B, 3.0 m/s right, could stem from assuming equal speeds without considering mass ratios. A transferable strategy is to assume initial momentum is zero for stationary systems and solve for opposite momenta in explosions or separations.

This question assesses the conservation of linear momentum in an inelastic head-on collision. In a system where external forces are negligible, the total linear momentum remains constant, with initial equaling final. The initial momentum is 1.0 kg × 5.0 m/s east plus 4.0 kg × (-1.0 m/s) west, totaling +1.0 kg·m/s east. After sticking, the 5.0 kg combined mass has velocity 1.0 / 5.0 = 0.20 m/s east. Option B, 0.20 m/s west, might result from switching the direction signs in the calculation. A transferable strategy is to assign consistent signs for directions and verify the net momentum direction matches the initial calculation.

This question assesses the conservation of linear momentum in a collision between two carts. In a system where external forces are negligible, such as on a frictionless track, the total linear momentum remains constant before and after the collision. The initial momentum is 2.0 kg × 1.5 m/s east plus 1.0 kg × (-0.50 m/s) west, totaling +2.5 kg·m/s. This value is conserved, so the total momentum after is also +2.5 kg·m/s. Option D, 0 kg·m/s, might be selected if one ignores the directions and assumes cancellation. A transferable strategy is to calculate the net initial momentum considering directions and recognize it persists unchanged in isolated systems.

This problem tests conservation of linear momentum in a perfectly inelastic collision. When external forces are negligible, the total momentum of a system remains constant before and after any interaction. Initially, the system has momentum from only the moving puck: p_initial = (0.20 kg)(4.0 m/s) + (0.30 kg)(0 m/s) = 0.80 kg·m/s eastward. After the collision, both pucks stick together with combined mass 0.50 kg, so p_final = (0.50 kg)v_final must equal 0.80 kg·m/s. Choice A (0.80 m/s) incorrectly uses the momentum value as the velocity. Solving for v_final: v_final = 0.80 kg·m/s ÷ 0.50 kg = 1.6 m/s. To solve momentum conservation problems, always write p_initial = p_final and carefully track the masses before and after the collision.

This question assesses the conservation of linear momentum during an interaction where two objects push off each other. In a system where external forces are negligible, such as on level ice, the total linear momentum of the system remains constant and is zero since both skaters start at rest. When Skater A moves west at 2.8 m/s, their momentum is -140 kg·m/s (taking east as positive). To conserve momentum, Skater B must have +140 kg·m/s, resulting in a velocity of 2.0 m/s east. Option A, 2.8 m/s east, might be selected if one assumes equal speeds regardless of mass differences. A transferable strategy is to define a consistent direction for positive momentum and ensure the vector sum remains constant for the isolated system.

This question assesses the conservation of linear momentum in an inelastic collision. In a system where external forces are negligible, such as on frictionless ice, the total linear momentum of the system remains constant before and after the collision. The initial momentum is calculated as the 2.0 kg cart's momentum of 6.0 kg·m/s east, since the 1.0 kg cart is at rest. After the carts stick together, their combined mass is 3.0 kg, so the final velocity is found by dividing the conserved momentum by the total mass, yielding 2.0 m/s east. Option A, 1.0 m/s east, might result from incorrectly averaging the velocities instead of conserving momentum. A transferable strategy is to always calculate the total initial momentum of the system and set it equal to the total final momentum to solve for unknown velocities.

This question assesses the conservation of linear momentum in an elastic collision. In a system where external forces are negligible, the total linear momentum of the system remains constant, equal before and after the collision. The initial momentum is 0.20 kg × 4.0 m/s east plus 0.30 kg × (-2.0 m/s) west, totaling +0.20 kg·m/s. Since momentum is conserved regardless of the collision type, the total momentum after is also +0.20 kg·m/s. Option B, 0 kg·m/s, might arise from mistakenly assuming the momenta cancel out completely without calculation. A transferable strategy is to compute the initial total momentum vectorially and recognize it must equal the final total for isolated systems.

This question assesses the skill of applying conservation of linear momentum to an inelastic collision between two carts. The total system momentum remains constant because external forces are negligible during the collision on the frictionless track. The initial momentum is (0.60 kg)(2.0 m/s) + (0.40 kg)(0 m/s) = 1.2 kg·m/s to the right. After colliding and sticking, the final momentum is (1.0 kg) v, so v = 1.2 m/s to the right. A common distractor like 2.0 m/s might come from ignoring the mass of the second cart and using only the initial velocity. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.