Physics Quiz
Practice Apply Momentum Conservation To Collisions in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
Question 1 / 20
0 of 20 answered
Two carts collide on a low-friction track. Cart 1 has mass m1=1.5kg and initial velocity v1i=+8.0m/s (right). Cart 2 has mass m2=3.0kg and is initially at rest (v2i=0). After an inelastic collision (they bounce apart), Cart 1 moves left at v1f=−2.0m/s. Taking rightward as positive, what is Cart 2's final velocity v2f (in m/s)?
This quiz focuses on Apply Momentum Conservation To Collisions, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
Two carts collide on a low-friction track. Cart 1 has mass m1=1.5kg and initial velocity v1i=+8.0m/s (right). Cart 2 has mass m2=3.0kg and is initially at rest (v2i=0). After an inelastic collision (they bounce apart), Cart 1 moves left at v1f=−2.0m/s. Taking rightward as positive, what is Cart 2's final velocity v2f (in m/s)?
Explanation: 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).
A 1D collision occurs in space (negligible external forces). Satellite A has mass mA=300kg and velocity vAi=+2.0m/s (right). Satellite B has mass mB=100kg and velocity vBi=−4.0m/s (left). They collide and stick together in a perfectly inelastic collision. What is the common final velocity vf (in m/s)?
Explanation: 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.
In space (negligible external forces), Astronaut A (mass mA=90kg) is moving right at vAi=+1.0m/s and grabs Astronaut B (mass mB=60kg) who is moving left at vBi=−2.0m/s. They move together afterward in a perfectly inelastic collision. What is their common final velocity vf (in m/s)?
Explanation: 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.
A 1D inelastic collision occurs on a frictionless track. Object A has mass mA=6.0kg and initial velocity vAi=+4.0m/s. Object B has mass mB=2.0kg and is initially at rest (vBi=0). After the collision, Object A moves at vAf=+2.0m/s. Taking rightward as positive, what is Object B's final velocity vBf (in m/s)?
Explanation: 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).
A rear-end collision occurs on an icy (effectively frictionless) straight road. Car A has mass mA=1000kg and is moving right at vAi=+20m/s. Car B has mass mB=2000kg and is moving right at vBi=+10m/s. The cars lock bumpers in a perfectly inelastic collision and move together. What is the common final velocity vf (in m/s)?
Explanation: 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.
Two carts collide on a frictionless track. Cart A has mass mA=2.0 kg and Cart B has mass mB=2.0 kg. Before the collision, Cart A moves at vAi=+5.0 m/s and Cart B moves at vBi=−3.0 m/s. After an inelastic collision (they bounce apart), Cart A is measured to have vAf=−1.0 m/s. Taking rightward as positive, what is Cart B's final velocity vBf from momentum conservation?
Explanation: 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 = (2.0 kg)(+5.0 m/s) = +10 kg⋅m/s, and Cart B has p₂ = (2.0 kg)(-3.0 m/s) = -6 kg⋅m/s, giving total initial momentum p_before = +10 - 6 = +4 kg⋅m/s. After the inelastic collision where objects separate: (2.0 kg)(-1.0 m/s) + (2.0 kg) v_{B f} = +4 kg⋅m/s, so -2 + 2 v_{B f} = +4, 2 v_{B f} = +6, v_{B f} = +3.0 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and correctly solves the momentum equation for the unknown velocity. Choice D confuses initial and final velocities, using the initial velocity value of B where the final should be calculated in the momentum conservation equation. 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. Common error: forgetting that momentum is a vector—you cannot simply add the speeds without considering direction. If objects move toward each other (head-on collision), one velocity must be negative when calculating total momentum.
A collision occurs on a low-friction track. Cart A has mass mA=1.5 kg and Cart B has mass mB=0.5 kg. Before the collision, Cart B is at rest (vBi=0). After the collision, the carts stick together and move with common velocity vf=+2.0 m/s (right). This is a perfectly inelastic collision. Taking rightward as positive, what was Cart A's initial velocity vAi?
Explanation: 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. After the perfectly inelastic collision, the carts stick together with combined mass (2.0 kg) and v_f = +2.0 m/s, so p_after = (2.0 kg)(+2.0 m/s) = +4.0 kg⋅m/s, which equals p_before. Before the collision, Cart B has p₂ = (0.5 kg)(0 m/s) = 0 kg⋅m/s, so p_before = (1.5 kg) v_{A i} + 0 = +4.0 kg⋅m/s, solving gives v_{A i} = +4.0 / 1.5 = +2.67 m/s. Choice C is correct because it properly applies momentum conservation with correct signs for directions and correctly 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. 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.
On a low-friction track, Cart A (mass mA=2.0 kg) moves rightward at vAi=+6.0 m/s and collides with Cart B (mass mB=1.0 kg) that is initially at rest (vBi=0 m/s). The collision is perfectly inelastic (the carts stick together). Taking rightward as positive, what is the common final velocity vf immediately after the collision?
Explanation: 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 = (2.0 kg)(+6.0 m/s) = +12 kg⋅m/s, and Cart B has p₂ = (1.0 kg)(0 m/s) = 0 kg⋅m/s, giving total initial momentum p_before = +12 kg⋅m/s. After the perfectly inelastic collision, the carts stick together with combined mass (3.0 kg), so v_f = p_before / (m_A + m_B) = +12 / 3 = +4.0 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 multiply velocities by their respective masses, essentially averaging the velocities ((+6.0 + 0)/2 = +3.0 m/s) 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.
In space (negligible external forces), Astronaut A (mass mA=90 kg) and Astronaut B (mass mB=30 kg) push off from each other. Just before pushing, both are drifting together at +2.0 m/s (right). After the push, Astronaut A moves at vAf=+1.0 m/s. This is an explosion/separation event. Taking rightward as positive, what is Astronaut B's final velocity vBf?
Explanation: 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 separation, both astronauts have initial velocity +2.0 m/s, so total momentum p_before = (90 kg + 30 kg)(+2.0 m/s) = 120 kg * +2.0 m/s = +240 kg⋅m/s. After the push, using momentum conservation: (90 kg)(+1.0 m/s) + (30 kg) v_{B f} = +240 kg⋅m/s, so 90 + 30 v_{B f} = 240, 30 v_{B f} = 150, v_{B f} = +5.0 m/s. Choice C is correct because it properly applies momentum conservation with correct signs for directions and correctly solves the momentum equation for the unknown velocity. Choice B forgets to account for the mass difference, essentially assuming both astronauts end up with the same final velocity as A, but the lighter astronaut recoils faster to conserve 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).
In deep space, two astronauts push off from one another (an explosion/separation event). Initially they are at rest together, so the total initial momentum is 0. Astronaut A has mass 60kg and after pushing off moves at vAf=+2.0m/s (right). Astronaut B has mass 90kg. Taking rightward as positive, what is Astronaut B's velocity vBf after the push-off?
Explanation: 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 push-off, both astronauts are at rest together, so p_before = 0. After the push-off, Astronaut A has momentum p_A = (60 kg)(+2.0 m/s) = +120 kg⋅m/s, and Astronaut B has momentum p_B = (90 kg)(v_{Bf}). Using momentum conservation p_after = p_before: +120 + 90v_{Bf} = 0, solving for v_{Bf} gives 90v_{Bf} = -120, so v_{Bf} = -120/90 = -1.33 m/s ≈ -1.3 m/s. Choice A is correct because it properly applies momentum conservation to this explosion/separation event where the total momentum must remain zero. Choice B (+1.3 m/s) correctly calculates the magnitude but fails to include direction (or gets direction wrong), and momentum is a vector quantity requiring both magnitude and direction—Astronaut B must move left (negative) to conserve 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. Common error: forgetting that momentum is a vector—you cannot simply add the speeds without considering direction; in explosion events starting from rest, the objects must move in opposite directions to maintain zero total momentum.
Two hockey pucks slide on nearly frictionless ice. Puck A has mass 0.20kg and initial velocity vAi=+8.0m/s (to the right). Puck B has mass 0.10kg and is initially at rest (vBi=0). After an inelastic collision (they bounce apart), puck A is observed moving at vAf=+2.0m/s. Taking rightward as positive, what is puck B's final velocity vBf?
Explanation: 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} = (0.20 kg)(+8.0 m/s) + (0.10 kg)(0) = +1.6 + 0 = +1.6 kg⋅m/s. After the collision, using momentum conservation p_after = p_before: (0.20 kg)(+2.0 m/s) + (0.10 kg)(v_{Bf}) = +1.6 kg⋅m/s, which gives +0.40 + 0.10v_{Bf} = +1.6, solving for v_{Bf} = (+1.6 - 0.40)/0.10 = +1.2/0.10 = +12 m/s. Choice A is correct because it properly applies momentum conservation and correctly solves the momentum equation for the unknown velocity of puck B. Choice C (+3.0 m/s) 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. 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). 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.
On a low-friction track, Cart A (mass mA=2.0kg) moves rightward at vAi=+6.0m/s and catches up to Cart B (mass mB=1.0kg) moving rightward at vBi=+2.0m/s. They collide and stick together (perfectly inelastic). Taking rightward as positive, what is their common final velocity vf immediately after the collision?
Explanation: 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 = m_A v_{Ai} = (2.0 kg)(+6.0 m/s) = +12 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(+2.0 m/s) = +2.0 kg⋅m/s, giving total initial momentum p_before = +12 + 2.0 = +14 kg⋅m/s. After the perfectly inelastic collision, the objects stick together with combined mass (2.0 + 1.0 = 3.0 kg), so v_f = p_before / (m_A + m_B) = +14 / 3.0 = +4.67 m/s ≈ +4.7 m/s. Choice B is correct because it properly applies momentum conservation with the correct calculation of the common final velocity for the combined mass. Choice D (+3.0 m/s) 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.
On a frictionless track (rightward positive), Cart A has mass mA=3.0kg and moves at vAi=+4.0m/s. Cart B has mass mB=1.0kg and moves toward it at vBi=−2.0m/s. They collide and stick together (perfectly inelastic). What is the common final velocity vf of the combined carts?
Explanation: 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 = m_A v_{Ai} = (3.0 kg)(+4.0 m/s) = +12 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(-2.0 m/s) = -2.0 kg⋅m/s, giving total initial momentum p_before = +12 + (-2.0) = +10 kg⋅m/s. After the perfectly inelastic collision, the objects stick together with combined mass (3.0 + 1.0 = 4.0 kg), so v_f = p_before / (m_A + m_B) = +10 / 4.0 = +2.5 m/s. Choice B (+2.5 m/s) 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 (+4.0 m/s) incorrectly assumes the heavier cart's velocity dominates without properly accounting for the opposing momentum of Cart B, essentially ignoring the leftward momentum contribution. 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. Common error: forgetting that momentum is a vector—you cannot simply add the speeds without considering direction; if objects move toward each other (head-on collision), one velocity must be negative when calculating total momentum.
A 1500 kg car (Object A) and a 1500 kg car (Object B) collide on an icy straight road (rightward positive). Before the collision, Object A moves at vAi=+6.0m/s and Object B moves at vBi=−4.0m/s. After the collision (elastic), Object A moves at vAf=−4.0m/s. Using conservation of momentum, what is Object B's final velocity vBf?
Explanation: 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} = (1500 kg)(+6.0 m/s) + (1500 kg)(-4.0 m/s) = +9000 + (-6000) = +3000 kg⋅m/s. After the collision, using momentum conservation p_after = p_before: (1500 kg)(-4.0 m/s) + (1500 kg)(v_{Bf}) = +3000 kg⋅m/s, which gives -6000 + 1500v_{Bf} = +3000, solving for the unknown velocity gives v_{Bf} = (+3000 + 6000)/1500 = +9000/1500 = +6.0 m/s. Choice B (+6.0 m/s) is correct because it accurately shows p_before equals p_after and properly solves the momentum equation for the unknown velocity. Choice A (-6.0 m/s) has the correct magnitude but wrong sign—it fails to properly account for the direction when solving the momentum equation. 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. In elastic collisions between objects of equal mass, the velocities are exchanged—here we see Car A went from +6.0 to -4.0 m/s while Car B goes from -4.0 to +6.0 m/s, which is characteristic of elastic collisions between equal masses.
On a low-friction track (rightward positive), Cart A has mass mA=4.0kg and is initially at rest (vAi=0). Cart B has mass mB=1.0kg and moves right at vBi=+12m/s. They collide and stick together (perfectly inelastic). What is the common final velocity vf?
Explanation: 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 = m_A v_{Ai} = (4.0 kg)(0) = 0 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(+12 m/s) = +12 kg⋅m/s, giving total initial momentum p_before = 0 + 12 = +12 kg⋅m/s. After the perfectly inelastic collision, the objects stick together with combined mass (4.0 + 1.0 = 5.0 kg), so v_f = p_before / (m_A + m_B) = +12 / 5.0 = +2.4 m/s. Choice A (+2.4 m/s) is correct because it properly applies momentum conservation and correctly calculates the common final velocity for the combined mass. Choice C (+12 m/s) incorrectly assumes the moving cart maintains its velocity after collision, ignoring that momentum must be shared among the larger combined mass—this would violate conservation of 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.
On a low-friction track, Cart A (mass 2.0kg) moves rightward at vAi=+4.0m/s. Cart B (mass 1.0kg) moves leftward at vBi=−2.0m/s. After an inelastic collision (they do not stick), the measured final velocities are vAf=+1.0m/s and vBf=+4.0m/s. Which statement correctly verifies whether momentum is conserved (take rightward as positive)?
Explanation: 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. Calculating momentum before: p_before = (2.0 kg)(+4.0 m/s) + (1.0 kg)(-2.0 m/s) = +8.0 + (-2.0) = +6.0 kg⋅m/s. Calculating momentum after: p_after = (2.0 kg)(+1.0 m/s) + (1.0 kg)(+4.0 m/s) = +2.0 + 4.0 = +6.0 kg⋅m/s. Since p_before = p_after = +6.0 kg⋅m/s, momentum is conserved. Choice B is correct because it accurately shows p_before equals p_after with the correct calculation of +6.0 kg⋅m/s for both. Choice A incorrectly calculates p_after as +4.0 kg⋅m/s, likely by making an arithmetic error or forgetting to include one of the terms in the final momentum calculation. 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. Common error: forgetting that momentum is a vector—you cannot simply add the speeds without considering direction; if objects move toward each other (head-on collision), one velocity must be negative when calculating total momentum.
A rear-end collision occurs on a straight, icy road (treat external forces as negligible during the short collision). Car A has mass mA=1000 kg and is moving at vAi=+20 m/s (east). Car B has mass mB=1000 kg and is moving at vBi=+10 m/s (east). The cars lock together in a perfectly inelastic collision. Taking east as positive, what is their common final velocity vf?
Explanation: 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 = (1000 kg)(+20 m/s) = +20000 kg⋅m/s, and Car B has p₂ = (1000 kg)(+10 m/s) = +10000 kg⋅m/s, giving total initial momentum p_before = +20000 + 10000 = +30000 kg⋅m/s. After the perfectly inelastic collision, the cars stick together with combined mass (2000 kg), so v_f = p_before / (m_A + m_B) = +30000 / 2000 = +15 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 D forgets to multiply velocities by their respective masses, essentially taking half the difference ((20 - 10)/2 = 5 m/s) instead of the weighted average—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.
Two carts collide head-on on a frictionless track. Cart A has mass mA=1.0 kg and initial velocity vAi=+8.0 m/s (right). Cart B has mass mB=3.0 kg and initial velocity vBi=−2.0 m/s (left). The collision is perfectly inelastic (they stick together). Taking rightward as positive, what is the common final velocity vf?
Explanation: 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).
A 0.20 kg hockey puck (Puck A) slides right at vAi=+10m/s and collides head-on with a 0.40 kg puck (Puck B) sliding left at vBi=−2.0m/s on nearly frictionless ice. After an inelastic collision, Puck A rebounds left at vAf=−4.0m/s. Taking rightward as positive, what is Puck B's final velocity vBf (in m/s)?
Explanation: 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).
In a 1D elastic collision on a frictionless track, two equal-mass carts collide head-on. Cart A has mass mA=1.0kg and initial velocity vAi=+4.0m/s. Cart B has mass mB=1.0kg and initial velocity vBi=−1.0m/s. After the collision, Cart A is observed moving at vAf=−1.0m/s. Using conservation of momentum, what is Cart B's final velocity vBf (in m/s)?
Explanation: 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).