Apply Momentum Conservation to Collisions - Physics

1. Two hockey pucks slide on nearly frictionless ice and collide in 1D. Puck A has mass $m_A=0.20\ \text{kg}$ and initial velocity $v_{A i}=+10\ \text{m/s}$ (right). Puck B has mass $m_B=0.10\ \text{kg}$ and is initially at rest ($v_{B i}=0$). After an elastic collision, Puck A is observed to move at $v_{A f}=+3.3\ \text{m/s}$ (right). Taking rightward as positive, what is Puck B's final velocity $v_{B f}$ (from momentum conservation)?
   Answer: $+13.4\ \text{m/s}$ 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, Puck A has momentum p₁ = (0.20 kg)(+10 m/s) = +2.0 kg⋅m/s, and Puck B has p₂ = (0.10 kg)(0 m/s) = 0 kg⋅m/s, giving total initial momentum p_before = +2.0 kg⋅m/s. After the elastic collision, using momentum conservation: (0.20 kg)(+3.3 m/s) + (0.10 kg) $v_{B f}$ = +2.0 kg⋅m/s, so +0.66 + 0.10 $v_{B f}$ = +2.0, 0.10 $v_{B f}$ = +1.34, $v_{B f}$ = +13.4 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 B forgets to multiply velocities by their respective masses, essentially taking the difference in velocities of A (10 - 3.3 = 6.7 m/s) without weighting by mass—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. 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).
2. A collision occurs on a low-friction track. Cart A has mass $m_A=1.5\ \text{kg}$ and Cart B has mass $m_B=0.5\ \text{kg}$. Before the collision, Cart B is at rest ($v_{B i}=0$). After the collision, the carts stick together and move with common velocity $v_f=+2.0\ \text{m/s}$ (right). This is a perfectly inelastic collision. Taking rightward as positive, what was Cart A's initial velocity $v_{A i}$?
   Answer: $+2.67\ \text{m/s}$ 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.
3. Two carts collide on a frictionless track. Cart A has mass $m_A=2.0\ \text{kg}$ and Cart B has mass $m_B=2.0\ \text{kg}$. Before the collision, Cart A moves at $v_{A i}=+5.0\ \text{m/s}$ and Cart B moves at $v_{B i}=-3.0\ \text{m/s}$. After an inelastic collision (they bounce apart), Cart A is measured to have $v_{A f}=-1.0\ \text{m/s}$. Taking rightward as positive, what is Cart B's final velocity $v_{B f}$ from momentum conservation?
   Answer: $+3.0\ \text{m/s}$ 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.
4. Two carts collide on a low-friction track and stick together (perfectly inelastic). Cart A has mass $3.0\,\text{kg}$ and is initially at rest ($v_{A i}=0$). Cart B has mass $1.0\,\text{kg}$ and initial velocity $v_{B i} = +8.0\,\text{m/s}$ (right). What is the common final velocity $v_f$ of the stuck-together carts?
   Answer: $+2.0\,\text{m/s}$ 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)(0) = 0 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(+8.0 m/s) = +8.0 kg⋅m/s, giving total initial momentum p_before = 0 + 8.0 = +8.0 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 / ($m_A$ + $m_B$) = +8.0 / 4.0 = +2.0 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 (+4.0 m/s) incorrectly assumes the final velocity is simply half of the initial velocity of the moving cart, failing to account for the different masses—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.
5. On a low-friction track, Cart A (mass $m_A = 2.0\,\text{kg}$) moves rightward at $v_{A i}=+6.0\,\text{m/s}$ and collides with Cart B (mass $m_B = 1.0\,\text{kg}$) initially at rest ($v_{B i}=0\,\text{m/s}$). The collision is perfectly inelastic (the carts stick together). Taking rightward as positive, what is the common final velocity $v_f$ immediately after the collision?
   Answer: $+4.0\,\text{m/s}$ 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) = +12.0 kg⋅m/s rightward, and Cart B has momentum p₂ = (1.0 kg)(0 m/s) = 0 kg⋅m/s, giving total initial momentum p_before = +12.0 + 0 = +12.0 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 / ($m_A$ + $m_B$) = +12.0 / 3.0 = +4.0 m/s rightward. Choice A is correct because it properly applies momentum conservation with the correct calculation of the common final velocity for the combined mass. Choice B (+3.0 m/s) incorrectly calculates by averaging the initial velocities (6.0 + 0)/2 = 3.0 instead of using momentum conservation—momentum depends on both mass and velocity, not just 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. 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.
6. In a 1D elastic collision on a frictionless track, two equal-mass carts collide head-on. Cart A has mass $m_A = 1.0\,\text{kg}$ and initial velocity $v_{A i} = +4.0\,\text{m/s}$. Cart B has mass $m_B = 1.0\,\text{kg}$ and initial velocity $v_{B i} = -1.0\,\text{m/s}$. After the collision, Cart A is observed moving at $v_{A f} = -1.0\,\text{m/s}$. Using conservation of momentum, what is Cart B's final velocity $v_{B f}$ (in m/s)?
   Answer: $+4.0\,\text{m/s}$ 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).
7. In space (negligible external forces), Astronaut A (mass $m_A = 90\,\text{kg}$) is moving right at $v_{A i} = +1.0\,\text{m/s}$ and grabs Astronaut B (mass $m_B = 60\,\text{kg}$) who is moving left at $v_{B i} = -2.0\,\text{m/s}$. They move together afterward in a perfectly inelastic collision. What is their common final velocity $v_f$ (in m/s)?
   Answer: $-0.20\,\text{m/s}$ 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.
8. A 0.20 kg hockey puck (Puck A) slides right at $v_{A i} = +10\,\text{m/s}$ and collides head-on with a 0.40 kg puck (Puck B) sliding left at $v_{B i} = -2.0\,\text{m/s}$ on nearly frictionless ice. After an inelastic collision, Puck A rebounds left at $v_{A f} = -4.0\,\text{m/s}$. Taking rightward as positive, what is Puck B's final velocity $v_{B f}$ (in m/s)?
   Answer: $+5.0\,\text{m/s}$ 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).
9. A head-on collision occurs on a low-friction track. Cart A has mass $1.0\,\text{kg}$ and initial velocity $v_{A i} = +5.0\,\text{m/s}$ (right). Cart B has mass $1.0\,\text{kg}$ and initial velocity $v_{B i} = -3.0\,\text{m/s}$ (left). The collision is perfectly elastic. What are the final velocities $(v_{A f}, v_{B f})$ immediately after the collision?
   Answer: $(v_{A f}, v_{B f}) = (-3.0\,\text{m/s},\,+5.0\,\text{m/s})$ 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 = $m_A$ $v_{Ai}$ = (1.0 kg)(+5.0 m/s) = +5.0 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(-3.0 m/s) = -3.0 kg⋅m/s, giving total initial momentum p_before = +5.0 + (-3.0) = +2.0 kg⋅m/s. For a perfectly elastic collision between equal masses, the velocities are exchanged: Cart A takes on Cart B's initial velocity and Cart B takes on Cart A's initial velocity, so $v_{Af}$ = -3.0 m/s and $v_{Bf}$ = +5.0 m/s. Choice B is correct because it properly applies the velocity exchange rule for elastic collisions between equal masses, resulting in the correct final velocities that conserve both momentum and kinetic energy. Choice A makes a sign error by treating the leftward-moving cart as having positive momentum after the collision, when leftward motion should be negative in the chosen coordinate system. 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.
10. Two hockey pucks slide on nearly frictionless ice. Puck A has mass $0.20\,\text{kg}$ and initial velocity $v_{A i} = +8.0\,\text{m/s}$ (to the right). Puck B has mass $0.10\,\text{kg}$ and is initially at rest ($v_{B i}=0$). After an inelastic collision (they bounce apart), puck A is observed moving at $v_{A f} = +2.0\,\text{m/s}$. Taking rightward as positive, what is puck B's final velocity $v_{B f}$?
    Answer: $+12\,\text{m/s}$ 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.
11. On a low-friction track, Cart A (mass $m_A = 2.0\,\text{kg}$) moves rightward at $v_{A i} = +6.0\,\text{m/s}$ and catches up to Cart B (mass $m_B = 1.0\,\text{kg}$) moving rightward at $v_{B i} = +2.0\,\text{m/s}$. They collide and stick together (perfectly inelastic). Taking rightward as positive, what is their common final velocity $v_f$ immediately after the collision?
    Answer: $+4.7\,\text{m/s}$ 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.
12. In space (negligible external forces), Astronaut A (mass $m_A=90\ \text{kg}$) and Astronaut B (mass $m_B=30\ \text{kg}$) push off from each other. Just before pushing, both are drifting together at $+2.0\ \text{m/s}$ (right). After the push, Astronaut A moves at $v_{A f}=+1.0\ \text{m/s}$. This is an explosion/separation event. Taking rightward as positive, what is Astronaut B's final velocity $v_{B f}$?
    Answer: $+5.0\ \text{m/s}$ 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).
13. A head-on elastic collision occurs between two carts on a low-friction track. Cart A has mass $m_A = 1.0\ \text{kg}$ with initial velocity $v_{A i} = +4.0\ \text{m/s}$ (right). Cart B has mass $m_B = 1.0\ \text{kg}$ with initial velocity $v_{B i} = -2.0\ \text{m/s}$ (left). After the collision, Cart B is observed moving right at $v_{B f} = +4.0\ \text{m/s}$. Using conservation of momentum only, what must Cart A's final velocity $v_{A f}$ be (in m/s) for momentum to be conserved?
    Answer: $-2.0\ \text{m/s}$ 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).
14. On an icy (nearly frictionless) straight road, Car A (mass $m_A = 1000\ \text{kg}$) travels east at $v_{A i} = +12\ \text{m/s}$ and rear-ends Car B (mass $m_B = 1000\ \text{kg}$) traveling east at $v_{B i} = +6\ \text{m/s}$. They lock together in a perfectly inelastic collision. Taking east as positive, what is the common final velocity $v_f$ (in m/s)?
    Answer: $+9.0\ \text{m/s}$ 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.
15. Two carts on a low-friction track collide head-on (rightward positive). Cart A: $m_A = 1.0\,\text{kg}$, $v_{A i} = +5.0\,\text{m/s}$. Cart B: $m_B = 2.0\,\text{kg}$, $v_{B i} = -1.0\,\text{m/s}$. After the collision (inelastic; they do not stick), Cart A has velocity $v_{A f} = -1.0\,\text{m/s}$. Using conservation of momentum, what is Cart B's final velocity $v_{B f}$?
    Answer: $+2.0\,\text{m/s}$ 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)(+5.0 m/s) + (2.0 kg)(-1.0 m/s) = +5.0 + (-2.0) = +3.0 kg⋅m/s. After the collision, using momentum conservation p_after = p_before: (1.0 kg)(-1.0 m/s) + (2.0 kg)($v_{Bf}$) = +3.0 kg⋅m/s, which gives -1.0 + $2.0v_{Bf}$ = +3.0, solving for the unknown velocity gives $v_{Bf}$ = (+3.0 + 1.0)/2.0 = +4.0/2.0 = +2.0 m/s. Choice B (+2.0 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 (+1.0 m/s) makes a sign error or arithmetic mistake in solving the momentum equation, leading to an incorrect final velocity for Cart B. 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).
16. On a low-friction track, Cart A (mass $2.0\,\text{kg}$) moves rightward at $v_{A i} = +4.0\,\text{m/s}$. Cart B (mass $1.0\,\text{kg}$) moves leftward at $v_{B i} = -2.0\,\text{m/s}$. After an inelastic collision (they do not stick), the measured final velocities are $v_{A f} = +1.0\,\text{m/s}$ and $v_{B f} = +4.0\,\text{m/s}$. Which statement correctly verifies whether momentum is conserved (take rightward as positive)?
    Answer: Momentum is conserved: $p_{\text{before}}=+6\,\text{kg}\cdot\text{m/s}$ and $p_{\text{after}}=+6\,\text{kg}\cdot\text{m/s}$. 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.
17. In deep space (negligible external forces), two astronauts push off from each other starting from rest (an explosion/separation). Astronaut A has mass $m_A = 60\,\text{kg}$ and moves left at $v_{A f} = -0.50\,\text{m/s}$ (rightward is positive). Astronaut B has mass $m_B = 90\,\text{kg}$. What is Astronaut B's velocity $v_{B f}$ after they push off?
    Answer: $+0.33\,\text{m/s}$ 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, so total momentum is p_before = $m_A$ $v_{Ai}$ + $m_B$ $v_{Bi}$ = (60 kg)(0) + (90 kg)(0) = 0 kg⋅m/s. After the push-off, using momentum conservation p_after = p_before: (60 kg)(-0.50 m/s) + (90 kg)($v_{Bf}$) = 0 kg⋅m/s, which gives -30 + $90v_{Bf}$ = 0, solving for the unknown velocity gives $v_{Bf}$ = 30/90 = +0.33 m/s. Choice C (+0.33 m/s) is correct because it properly applies momentum conservation, recognizing that the total momentum must remain zero, so the astronauts must move in opposite directions with momenta that cancel. Choice A (-0.33 m/s) has the correct magnitude but wrong direction—it fails to recognize that if Astronaut A moves left (negative), Astronaut B must move right (positive) 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. A key principle in explosion problems: when objects start from rest, the total momentum before and after must be zero, meaning the objects move in opposite directions with momenta that exactly cancel each other out.
18. On a low-friction track, Cart A (mass $m_A=3.0\ \text{kg}$) moves rightward at $v_{A i}=+4.0\ \text{m/s}$ and Cart B (mass $m_B=1.0\ \text{kg}$) moves rightward at $v_{B i}=+1.0\ \text{m/s}$. They collide and stick together in a perfectly inelastic collision. Taking rightward as positive, what is the common final velocity $v_f$?
    Answer: $+3.25\ \text{m/s}$ 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 = (3.0 kg)(+4.0 m/s) = +12 kg⋅m/s, and Cart B has p₂ = (1.0 kg)(+1.0 m/s) = +1 kg⋅m/s, giving total initial momentum p_before = +12 + 1 = +13 kg⋅m/s. After the perfectly inelastic collision, the carts stick together with combined mass (4.0 kg), so $v_f$ = p_before / ($m_A$ + $m_B$) = +13 / 4 = +3.25 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 forgets to multiply velocities by their respective masses, essentially averaging the velocities ((4.0 + 1.0)/2 = 2.50 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. 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).
19. 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 $v_{A i} = +6.0\,\text{m/s}$ and Object B moves at $v_{B i} = -4.0\,\text{m/s}$. After the collision (elastic), Object A moves at $v_{A f} = -4.0\,\text{m/s}$. Using conservation of momentum, what is Object B's final velocity $v_{B f}$?
    Answer: $+6.0\,\text{m/s}$ 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.
20. A 1D collision occurs in space (negligible external forces). Satellite A has mass $m_A = 300\,\text{kg}$ and velocity $v_{A i} = +2.0\,\text{m/s}$ (right). Satellite B has mass $m_B = 100\,\text{kg}$ and velocity $v_{B i} = -4.0\,\text{m/s}$ (left). They collide and stick together in a perfectly inelastic collision. What is the common final velocity $v_f$ (in m/s)?
    Answer: $+0.50\,\text{m/s}$ 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.

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