This question tests AP Physics C: Mechanics skills, specifically conservation of momentum in elastic collisions. Total momentum is always conserved in collisions when no external forces act. Given: m₁ = 200 kg at v₁ = +1.5 m/s, m₂ = 300 kg at v₂ = -0.5 m/s. Initial momentum: pᵢ = m₁v₁ + m₂v₂ = (200)(1.5) + (300)(-0.5) = 300 + (-150) = 150 kg·m/s. Since momentum is conserved in all collisions (elastic or inelastic), pf = pᵢ = 150 kg·m/s. Choice A correctly shows pᵢ = pf = 150 kg·m/s. Remember that momentum conservation applies to all collision types, not just elastic ones.

This question tests AP Physics C: Mechanics skills, specifically perfectly inelastic collisions. In perfectly inelastic collisions, objects stick together and momentum is conserved but kinetic energy is not. Given: m₁ = 1200 kg at v₁ = +20 m/s, m₂ = 3000 kg at v₂ = 0 m/s. Using conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)vf. Substituting: (1200)(20) + (3000)(0) = (1200 + 3000)vf. This gives: 24000 = 4200vf, so vf = 24000/4200 = 5.71 m/s ≈ 5.7 m/s. Choice A correctly shows vf = +5.7 m/s. Students often forget that in perfectly inelastic collisions, the objects move together with the same final velocity.

This question tests AP Physics C: Mechanics skills, specifically momentum conservation in elastic collisions. Total momentum must be conserved regardless of collision type. Given: m₁ = 500 kg at v₁ = +0.40 m/s, m₂ = 200 kg at v₂ = -1.0 m/s. Initial momentum: pᵢ = m₁v₁ + m₂v₂ = (500)(0.40) + (200)(-1.0) = 200 + (-200) = 0 kg·m/s. Since momentum is conserved in all collisions: pf = pᵢ = 0 kg·m/s. Choice A correctly shows pᵢ = pf = 0 kg·m/s. This is a special case where the initial momenta exactly cancel, resulting in zero total momentum before and after collision.

Perfectly inelastic collision with momentum conservation. Initial momentum: $p_i = m_{car}v_{car} + m_{truck}v_{truck} = (1500)(+18) + (2500)(0) = 27000$ kg⋅m/s. Since external impulse is negligible, momentum is conserved in the collision: $p_f = p_i = 27000$ kg⋅m/s. Choice A correctly states that both initial and final momentum equal 27000 kg⋅m/s. Key concept: in inelastic collisions, kinetic energy is lost but momentum remains constant.

This 2D inelastic collision requires vector addition. Player A's momentum: $\vec{p}_A = (80)(4.0)\hat{i} = 320\hat{i}$ kg⋅m/s (east). Player B's momentum: $\vec{p}_B = (70)(3.0)\hat{j} = 210\hat{j}$ kg⋅m/s (north). Total initial momentum: $\vec{p}_i = 320\hat{i} + 210\hat{j}$ kg⋅m/s. Since they stick together (inelastic) with negligible external impulse, momentum is conserved: $\vec{p}_f = \vec{p}_i = (320\hat{i} + 210\hat{j})$ kg⋅m/s. Choice A is correct. Key insight: momentum conservation applies to each component independently in 2D collisions.

Perfectly inelastic collision where objects stick together. Initial momentum: $p_i = m_{skater}v_{skater} + m_{cart}v_{cart} = (70)(+6.0) + (30)(0) = 420$ kg⋅m/s. After collision, combined mass: $m_{total} = 70 + 30 = 100$ kg. Using momentum conservation: $p_f = p_i$, so $(100)v_f = 420$, giving $v_f = 420/100 = 4.2$ m/s. Choice A is correct. Common mistake: using individual masses instead of combined mass for final velocity calculation.

This question tests AP Physics C: Mechanics skills, specifically elastic head-on collisions between equal masses. When two objects of equal mass collide elastically, they exchange velocities. Given: m₁ = m₂ = 0.16 kg, v₁ᵢ = +2.5 m/s, v₂ᵢ = -1.5 m/s. For equal masses in elastic collision: v₁f = v₂ᵢ and v₂f = v₁ᵢ. Therefore: v₂f = +2.5 m/s. Choice A correctly shows v₂f = +2.5 m/s. This is a special case that simplifies calculations significantly. Students should memorize this result for equal-mass elastic collisions.

Elastic collision between equal-mass satellites. Initial: satellite A (200 kg) at +5.0 m/s, satellite B (200 kg) at -2.0 m/s. For elastic collisions between equal masses, velocities exchange: satellite A takes B's velocity (-2.0 m/s) and satellite B takes A's velocity (+5.0 m/s). Therefore, $v_{2f} = +5.0$ m/s. Choice A is correct. This velocity exchange rule for equal masses simplifies calculations and helps avoid algebraic errors in elastic collision problems.

For elastic collisions between equal masses, velocities exchange. Initial: ball 1 at +2.0 m/s, ball 2 at -1.0 m/s. Since masses are equal (0.16 kg each), in an elastic collision the velocities swap: ball 1 takes ball 2's initial velocity (-1.0 m/s) and ball 2 takes ball 1's initial velocity (+2.0 m/s). Therefore, $v_{2f} = +2.0$ m/s. Choice A is correct. This is a special case of elastic collisions that students should memorize. Watch for sign errors or attempting complex calculations when the simple velocity exchange rule applies.

This question tests understanding of momentum conservation in elastic collisions. First, calculate the initial momentum: $p_i = m_A v_A + m_B v_B = (0.40)(+3.0) + (0.60)(-1.0) = 1.2 - 0.6 = +0.6$ kg⋅m/s. Since the collision is elastic with negligible external impulse, momentum is conserved: $p_f = p_i = +0.6$ kg⋅m/s. Choice B correctly states that both initial and final momentum equal +0.6 kg⋅m/s. Common errors include: forgetting the negative sign for westward velocity, miscalculating the arithmetic, or assuming momentum changes in elastic collisions.