This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Cart A has momentum p₁ = (1.0 kg)(+8.0 m/s) = +8.0 kg⋅m/s, and Cart B has p₂ = (3.0 kg)(-2.0 m/s) = -6.0 kg⋅m/s, giving total initial momentum p_before = +8.0 - 6.0 = +2.0 kg⋅m/s. After the perfectly inelastic collision, the carts stick together with combined mass (4.0 kg), so v_f = p_before / (m_A + m_B) = +2.0 / 4 = +0.50 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and correctly calculates the common final velocity for the combined mass. Choice D makes a sign error by treating Cart B moving left as having positive momentum, when leftward motion should be negative in the chosen coordinate system, leading to incorrect total momentum (e.g., +8.0 + 6.0 = +14.0, v_f = +14.0 / 4 = +3.5 m/s, but adjusted signs yield -1.50 if further errors). When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m₁v₁ᵢ + m₂v₂ᵢ = (1.5 kg)(8.0 m/s) + (3.0 kg)(0 m/s) = 12 + 0 = 12 kg⋅m/s; after the inelastic collision objects separate: using momentum conservation p_after = p_before: (1.5 kg)(-2.0 m/s) + (3.0 kg) $v_{2f}$ = 12 kg⋅m/s, so -3 + 3 $v_{2f}$ = 12, 3 $v_{2f}$ = 15, $v_{2f}$ = 5.0 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and properly solves the momentum equation for the unknown velocity. Choice A incorrectly assumes the perfectly inelastic collision formula (common final velocity) when the objects actually bounce apart, leading to an incorrect result. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Satellite A has momentum p₁ = (300 kg)(2.0 m/s) = 600 kg⋅m/s, and Satellite B has momentum p₂ = (100 kg)(-4.0 m/s) = -400 kg⋅m/s, giving total initial momentum p_before = 600 - 400 = 200 kg⋅m/s; after the perfectly inelastic collision, they stick together with combined mass (400 kg), so v_f = p_before / (m₁ + m₂) = 200 / 400 = 0.50 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and correctly calculates the common final velocity for the combined mass. Choice C makes a sign error by treating the leftward motion as positive in the chosen coordinate system, leading to incorrect total momentum. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. A key distinction: in perfectly inelastic collisions, objects stick together and move with a common final velocity v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂), while in other collisions objects bounce apart with different final velocities that must be determined by both momentum conservation and additional information about the collision type.

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m₁v₁ᵢ + m₂v₂ᵢ = (0.20 kg)(10 m/s) + (0.40 kg)(-2.0 m/s) = 2 - 0.8 = 1.2 kg⋅m/s; after the inelastic collision objects separate: using momentum conservation p_after = p_before: (0.20 kg)(-4.0 m/s) + (0.40 kg) $v_{B f}$ = 1.2 kg⋅m/s, so -0.8 + 0.4 $v_{B f}$ = 1.2, 0.4 $v_{B f}$ = 2, $v_{B f}$ = 5.0 m/s. Choice C is correct because it properly applies momentum conservation with correct signs for directions and properly solves the momentum equation for the unknown velocity. Choice D incorrectly assumes kinetic energy is conserved and uses ½mv² before = ½mv² after, but kinetic energy is only conserved in elastic collisions, not in the inelastic collision described. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Astronaut A has momentum p₁ = (90 kg)(1.0 m/s) = 90 kg⋅m/s, and Astronaut B has momentum p₂ = (60 kg)(-2.0 m/s) = -120 kg⋅m/s, giving total initial momentum p_before = 90 - 120 = -30 kg⋅m/s; after the perfectly inelastic collision, they stick together with combined mass (150 kg), so v_f = p_before / (m₁ + m₂) = -30 / 150 = -0.20 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and correctly calculates the common final velocity for the combined mass. Choice A makes a sign error by treating the leftward motion as positive, leading to incorrect total momentum. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. A key distinction: in perfectly inelastic collisions, objects stick together and move with a common final velocity v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂), while in other collisions objects bounce apart with different final velocities that must be determined by both momentum conservation and additional information about the collision type.

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m₁v₁ᵢ + m₂v₂ᵢ = (6.0 kg)(4.0 m/s) + (2.0 kg)(0 m/s) = 24 + 0 = 24 kg⋅m/s; after the inelastic collision objects separate: using momentum conservation p_after = p_before: (6.0 kg)(2.0 m/s) + (2.0 kg) $v_{B f}$ = 24 kg⋅m/s, so 12 + 2 $v_{B f}$ = 24, 2 $v_{B f}$ = 12, $v_{B f}$ = 6.0 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and properly solves the momentum equation for the unknown velocity. Choice A incorrectly assumes the perfectly inelastic collision formula (common final velocity) when the objects actually bounce apart, leading to an incorrect result. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Car A has momentum p₁ = m₁v₁ = (1000 kg)(20 m/s) = 20000 kg⋅m/s, and Car B has momentum p₂ = (2000 kg)(10 m/s) = 20000 kg⋅m/s, giving total initial momentum p_before = 20000 + 20000 = 40000 kg⋅m/s; after the perfectly inelastic collision, the cars stick together with combined mass (3000 kg), so v_f = p_before / (m₁ + m₂) = 40000 / 3000 ≈ 13.3 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and correctly calculates the common final velocity for the combined mass. Choice C forgets to multiply velocities by their respective masses, essentially averaging the velocities instead of calculating momentum-weighted velocity—momentum depends on both mass and velocity. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. A key distinction: in perfectly inelastic collisions, objects stick together and move with a common final velocity v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂), while in other collisions objects bounce apart with different final velocities that must be determined by both momentum conservation and additional information about the collision type.

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m₁v₁ᵢ + m₂v₂ᵢ = (1.0 kg)(4.0 m/s) + (1.0 kg)(-1.0 m/s) = 4 - 1 = 3 kg⋅m/s; after the elastic collision objects separate: using momentum conservation p_after = p_before: (1.0 kg)(-1.0 m/s) + (1.0 kg) $v_{Bf}$ = 3 kg⋅m/s, so -1 + $v_{Bf}$ = 3, $v_{Bf}$ = 4.0 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and properly solves the momentum equation for the unknown velocity. Choice C makes a sign error by treating the initial leftward motion as positive, leading to incorrect total momentum. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Car A has momentum p₁ = m_A v_Ai = (1000 kg)(+12 m/s) = +12000 kg⋅m/s, and Car B has momentum p₂ = (1000 kg)(+6 m/s) = +6000 kg⋅m/s, giving total initial momentum p_before = +12000 + 6000 = +18000 kg⋅m/s. After the perfectly inelastic collision, the objects stick together with combined mass (2000 kg), so v_f = p_before / (m_A + m_B) = +18000 / 2000 = +9.0 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and correctly calculates the common final velocity for the combined mass. Choice C forgets to divide by the combined mass properly or adds velocities without masses, leading to an inflated value. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. A key distinction: in perfectly inelastic collisions, objects stick together and move with a common final velocity v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂), while in other collisions objects bounce apart with different final velocities that must be determined by both momentum conservation and additional information about the collision type.

This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m_A v_Ai + m_B v_Bi = (1.0 kg)(+4.0 m/s) + (1.0 kg)(-2.0 m/s) = +4 - 2 = +2 kg⋅m/s. After the collision, using momentum conservation p_after = p_before: (1.0 kg) v_Af + (1.0 kg)(+4.0 m/s) = +2 kg⋅m/s, solving for the unknown velocity gives v_Af + 4 = +2, v_Af = +2 - 4 = -2.0 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and accurately solves the momentum equation for the unknown velocity. Choice D incorrectly assumes kinetic energy conservation alone or doubles a value without basis, leading to an overstated negative velocity. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).