Conservation of Linear Momentum - AP Physics 1

1. A $0.10\,\text{kg}$ cart moving right at $10\,\text{m/s}$ collides and sticks to a $0.40\,\text{kg}$ cart initially at rest. Consider both carts as the system and assume external forces are negligible during the collision. What is the speed of the combined carts immediately after?
   Answer: $2.0\,\text{m/s}$ This problem requires conservation of linear momentum in a perfectly inelastic collision where one object is initially at rest. When external forces on the system are negligible, momentum before collision equals momentum after. Initial momentum is (0.10 kg)(10 m/s) + (0.40 kg)(0 m/s) = 1.0 kg·m/s right. After sticking together, the combined mass of 0.50 kg moves with velocity v, so 0.50v = 1.0, yielding v = 2.0 m/s. Choice C (10 m/s) incorrectly assumes the combined mass maintains the initial velocity of the moving cart. Remember that in inelastic collisions, the final velocity is always less than the initial velocity of the faster object.
2. A $3.0\,\text{kg}$ cart moving right at $2.0\,\text{m/s}$ collides and sticks to a $1.0\,\text{kg}$ cart initially at rest on a frictionless track. The system is both carts, and external forces are negligible during the collision. What is the final velocity of the combined carts?
   Answer: $1.5\,\text{m/s}$ right This problem demonstrates conservation of linear momentum in a perfectly inelastic collision with one object initially at rest. When external forces are negligible, the total momentum of the system remains constant before and after the collision. Initially, only the 3.0 kg cart moves: p_initial = (3.0 kg)(2.0 m/s) + (1.0 kg)(0 m/s) = 6.0 kg·m/s rightward. After sticking together, the combined mass is 4.0 kg, so p_final = (4.0 kg)v_final. Choice D (6.0 m/s) incorrectly uses the momentum value as velocity. Setting p_initial = p_final: 6.0 = 4.0v_final, giving v_final = 1.5 m/s rightward. To find the final velocity in sticky collisions, divide the initial total momentum by the combined final mass.
3. A $0.50\,\text{kg}$ cart moving left at $2.0\,\text{m/s}$ collides and sticks to a $0.50\,\text{kg}$ cart moving right at $1.0\,\text{m/s}$ on a frictionless track. Consider the system as both carts and assume external forces are negligible during the collision. What is the final velocity of the combined carts?
   Answer: $-0.50\,\text{m/s}$ (left) This problem requires conservation of linear momentum for equal-mass objects moving in opposite directions. When external forces are negligible, the total momentum of the system remains constant during the collision. Taking rightward as positive: p_initial = (0.50 kg)(-2.0 m/s) + (0.50 kg)(+1.0 m/s) = -1.0 + 0.50 = -0.50 kg·m/s. After sticking together, the combined mass is 1.0 kg, so p_final = (1.0 kg)v_final. Choice C (+0.50 m/s) has the wrong sign. Setting p_initial = p_final: -0.50 = 1.0v_final, giving v_final = -0.50 m/s (leftward). When objects have equal mass but different speeds, the final velocity points in the direction of the initially faster object.
4. On a frictionless horizontal track, a $0.60\,\text{kg}$ cart moving right at $2.0\,\text{m/s}$ collides and sticks to a $0.40\,\text{kg}$ cart initially at rest. Treat the two carts as the system; external forces on the system are negligible during the collision. After the collision, the carts move together. What is the speed of the combined carts immediately after the collision?
   Answer: $1.2\,\text{m/s}$ This question assesses the skill of applying conservation of linear momentum to an inelastic collision between two carts. The total system momentum remains constant because external forces are negligible during the collision on the frictionless track. The initial momentum is (0.60 kg)(2.0 m/s) + (0.40 kg)(0 m/s) = 1.2 kg·m/s to the right. After colliding and sticking, the final momentum is (1.0 kg) v, so v = 1.2 m/s to the right. A common distractor like 2.0 m/s might come from ignoring the mass of the second cart and using only the initial velocity. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
5. Two skaters push off each other on level ice. Skater A has mass $50\,\text{kg}$ and is initially at rest; Skater B has mass $70\,\text{kg}$ and is initially at rest. Define the system as both skaters; external forces are negligible. After pushing off, Skater A moves west at $2.8\,\text{m/s}$. What is Skater B's velocity?
   Answer: $2.0\,\text{m/s}$ east This question assesses the conservation of linear momentum during an interaction where two objects push off each other. In a system where external forces are negligible, such as on level ice, the total linear momentum of the system remains constant and is zero since both skaters start at rest. When Skater A moves west at 2.8 m/s, their momentum is -140 kg·m/s (taking east as positive). To conserve momentum, Skater B must have +140 kg·m/s, resulting in a velocity of 2.0 m/s east. Option A, 2.8 m/s east, might be selected if one assumes equal speeds regardless of mass differences. A transferable strategy is to define a consistent direction for positive momentum and ensure the vector sum remains constant for the isolated system.
6. A $0.80\,\text{kg}$ cart at rest explodes into two pieces on a frictionless track. Piece A has mass $0.30\,\text{kg}$ and moves east at $4.0\,\text{m/s}$. Treat both pieces as the system; external forces are negligible. What is Piece B's velocity?
   Answer: $2.4\,\text{m/s}$ west This question assesses the conservation of linear momentum in an explosion of a cart into pieces. In a system where external forces are negligible, such as on a frictionless track, the total linear momentum remains constant and is zero since it starts at rest. Piece A's momentum is 0.30 kg × 4.0 m/s east = +1.2 kg·m/s. Piece B, with 0.50 kg mass, must have -1.2 kg·m/s to conserve zero, yielding 2.4 m/s west. Option A, 1.5 m/s east, could be from mistakenly adding velocities instead of conserving momentum. A transferable strategy is to set the vector sum of final momenta equal to the initial total momentum, often zero for explosions from rest.
7. Two identical carts on a frictionless track interact via a compressed spring. Initially, both carts are at rest and the spring is compressed between them; the system is the two carts (ignore the spring's mass). External forces on the system are negligible. After release, cart 1 moves left at $3.0\,\text{m/s}$. What is the velocity of cart 2?
   Answer: $3.0\,\text{m/s}$ right This problem demonstrates conservation of linear momentum in an explosion-type interaction. When external forces on a system are negligible, the total momentum remains constant throughout any internal interaction. Initially, both carts are at rest, so p_initial = 0. After the spring releases, the system momentum must still be zero: m₁v₁ + m₂v₂ = 0. Since the carts are identical and cart 1 moves left at 3.0 m/s, we have m(-3.0 m/s) + m(v₂) = 0. Choice A (3.0 m/s left) would give the same direction for both carts, violating momentum conservation. Solving: v₂ = +3.0 m/s (rightward). For explosion problems starting from rest, remember that objects must move in opposite directions with momenta that cancel out.
8. A $0.060\,\text{kg}$ ball moving east at $10\,\text{m/s}$ hits a $0.20\,\text{kg}$ cart initially at rest on a frictionless track and sticks inside it. Treat ball + cart as the system; external forces are negligible. What is the cart's speed immediately after impact?
   Answer: $2.3\,\text{m/s}$ This question assesses the conservation of linear momentum in an inelastic collision where a ball sticks to a cart. In a system where external forces are negligible, such as on a frictionless track, the total linear momentum remains constant before and after impact. The initial momentum is solely from the ball, 0.060 kg × 10 m/s east, equaling 0.60 kg·m/s east. After sticking, the combined mass of 0.26 kg shares this momentum, giving a velocity of approximately 2.3 m/s east. Option A, 1.9 m/s, might be chosen if one underestimates the combined mass in the calculation. A transferable strategy is to treat the system as isolated and equate initial momentum to final momentum times total mass for stuck-together objects.
9. Two ice skaters initially at rest push off each other on level ice. Skater A has mass $50\,\text{kg}$ and skater B has mass $75\,\text{kg}$. Define the system as both skaters together; external forces on the system are negligible during the push. Afterward, skater A moves left at $3.0\,\text{m/s}$. What is skater B's velocity?
   Answer: $2.0\,\text{m/s}$ right This question assesses the skill of applying conservation of linear momentum to two skaters pushing off each other. The total system momentum remains constant because external forces are negligible during the push on level ice. The initial momentum is zero since both skaters are at rest. After pushing, the momentum of skater A is (50 kg)(-3.0 m/s) = -150 kg·m/s, so skater B's momentum must be +150 kg·m/s, giving v = 2.0 m/s to the right. A common distractor like 3.0 m/s right might come from assuming equal velocities without considering the mass difference. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
10. A $0.50\,\text{kg}$ cart moving east at $2.0\,\text{m/s}$ collides and sticks to a $1.5\,\text{kg}$ cart moving east at $1.0\,\text{m/s}$. Define the system as both carts; external forces are negligible. What is the final speed of the combined carts?
    Answer: $1.25\,\text{m/s}$ This question assesses the conservation of linear momentum in an inelastic collision where both objects move in the same direction. In a system where external forces are negligible, the total linear momentum remains constant, so initial total equals final total. The initial momentum is 0.50 kg × 2.0 m/s plus 1.5 kg × 1.0 m/s, both east, totaling 2.5 kg·m/s east. After sticking, the combined 2.0 kg mass has a velocity of 2.5 / 2.0 = 1.25 m/s east. Option A, 0.75 m/s, could result from incorrectly subtracting the velocities instead of conserving momentum. A transferable strategy is to sum the individual momenta initially and divide by the total mass for inelastic collisions where objects stick together.
11. A person of mass $60\,\text{kg}$ stands on a $20\,\text{kg}$ skateboard at rest on smooth level ground. The person jumps forward so that just after leaving the board, the person's speed is $3.0\,\text{m/s}$ relative to the ground. Take the person+skateboard as the system; external horizontal forces are negligible during the jump. What is the skateboard's speed just after the jump?
    Answer: $9.0\,\text{m/s}$ backward This question assesses the skill of applying conservation of linear momentum to a person jumping off a skateboard. The total system momentum remains constant because external horizontal forces are negligible during the jump on smooth ground. The initial momentum is zero since the system is at rest. After jumping, the person's momentum is (60 kg)(3.0 m/s) = 180 kg·m/s forward, so the skateboard's momentum must be -180 kg·m/s, giving v = -9.0 m/s (9.0 m/s backward). A common distractor like 3.0 m/s backward might come from using equal velocities without masses. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
12. Two students stand on frictionless ice. Student A ($60\,\text{kg}$) is initially at rest and pushes Student B ($40\,\text{kg}$). Define the system as both students; external forces are negligible during the push. Immediately after, Student B moves east at $3.0\,\text{m/s}$. What is Student A's velocity?
    Answer: $2.0\,\text{m/s}$ west This problem demonstrates conservation of linear momentum during an internal push between two students. Since external forces are negligible and the system starts at rest, the initial total momentum is zero, which means the final total momentum must also be zero. After the push, Student B (40 kg) moves east at 3.0 m/s, giving momentum of +120 kg⋅m/s. To maintain zero total momentum, Student A (60 kg) must have momentum of -120 kg⋅m/s, so vA = -120/60 = -2.0 m/s (west). Choice C incorrectly calculates using a 2:3 mass ratio instead of proper momentum conservation. When analyzing explosions or pushes from rest, remember that the total momentum remains zero throughout.
13. A $0.25\,\text{kg}$ glider moving right at $6.0\,\text{m/s}$ catches and sticks to a $0.75\,\text{kg}$ glider moving left at $2.0\,\text{m/s}$ on an air track. Define the system as both gliders; external forces are negligible. What is the final velocity of the stuck gliders?
    Answer: $0\,\text{m/s}$ This question assesses the conservation of linear momentum in an inelastic collision on an air track. In a system where external forces are negligible, the total linear momentum remains constant, equating initial to final. The initial momentum is 0.25 kg × 6.0 m/s right plus 0.75 kg × (-2.0 m/s) left, totaling 0 kg·m/s. After sticking, the 1.0 kg combined mass has velocity 0 / 1.0 = 0 m/s. Option B, 1.0 m/s right, could result from averaging speeds without momentum conservation. A transferable strategy is to check if initial momenta cancel out, leading to zero final velocity for equal but opposite contributions adjusted by mass.
14. On frictionless ice, a $60\,\text{kg}$ skater initially at rest pushes a $20\,\text{kg}$ sled so the sled moves east at $3.0\,\text{m/s}$. Define the system as skater + sled, and assume external forces on the system are negligible during the push. What is the skater's velocity after the push?
    Answer: $1.0\,\text{m/s}$ west This problem involves conservation of linear momentum during a push between two objects initially at rest. When external forces on a system are negligible, the total momentum before and after an interaction must be equal. The initial momentum is zero since both the skater and sled start at rest. After the push, the sled has momentum (20 kg)(3.0 m/s) = 60 kg·m/s east, so the skater must have momentum 60 kg·m/s west to maintain zero total momentum. With the skater's mass of 60 kg, the velocity is 60 kg·m/s ÷ 60 kg = 1.0 m/s west. Choice B (3.0 m/s west) incorrectly assumes equal speeds rather than equal magnitudes of momentum. Remember that in zero initial momentum problems, the momenta of the parts must be equal and opposite after separation.
15. A $0.10\,\text{kg}$ glider moving left at $5.0\,\text{m/s}$ collides with a $0.40\,\text{kg}$ glider initially at rest on an air track. After the collision, the $0.10\,\text{kg}$ glider moves right at $1.0\,\text{m/s}$. Define the system as both gliders; external forces are negligible. What is the $0.40\,\text{kg}$ glider's velocity after the collision?
    Answer: $-1.5\,\text{m/s}$ This question assesses the skill of applying conservation of linear momentum to a collision between two gliders. The total system momentum remains constant because external forces are negligible on the air track. The initial momentum is (0.10 kg)(-5.0 m/s) + (0.40 kg)(0 m/s) = -0.50 kg·m/s. After collision, with the 0.10 kg glider at +1.0 m/s, its momentum is (0.10 kg)(1.0 m/s) = 0.10 kg·m/s, so the 0.40 kg glider's momentum must be -0.60 kg·m/s, giving v = -1.5 m/s. A common distractor like +1.0 m/s might come from assuming symmetric velocities. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
16. Two ice skaters push off each other on nearly frictionless ice. Skater A has mass $60\,\text{kg}$ and skater B has mass $40\,\text{kg}$. Initially, both are at rest; take the system as both skaters and assume external forces are negligible during the push. If skater A moves left at $2.0\,\text{m/s}$ afterward, what is skater B's velocity?
    Answer: $3.0\,\text{m/s}$ right This problem demonstrates conservation of linear momentum when two objects push apart from rest. When external forces are negligible, the total momentum of an isolated system remains constant. Initially, both skaters are at rest, so p_initial = 0. After pushing apart, the total momentum must still be zero: (60 kg)(-2.0 m/s) + (40 kg)$v_B$ = 0. This gives -120 + 40v_B = 0. Choice B (1.33 m/s) incorrectly calculates the velocity ratio. Solving: $v_B$ = 120/40 = 3.0 m/s rightward. In push-off problems from rest, the momentum magnitudes must be equal and opposite, so lighter objects move faster than heavier ones.
17. A $0.20\,\text{kg}$ puck slides east at $4.0\,\text{m/s}$ and makes a head-on elastic collision with a $0.30\,\text{kg}$ puck sliding west at $2.0\,\text{m/s}$. Consider the system of both pucks; external forces are negligible. What is the system's total momentum after the collision?
    Answer: $+0.20\,\text{kg}\cdot\text{m/s}$ This question assesses the conservation of linear momentum in an elastic collision. In a system where external forces are negligible, the total linear momentum of the system remains constant, equal before and after the collision. The initial momentum is 0.20 kg × 4.0 m/s east plus 0.30 kg × (-2.0 m/s) west, totaling +0.20 kg·m/s. Since momentum is conserved regardless of the collision type, the total momentum after is also +0.20 kg·m/s. Option B, 0 kg·m/s, might arise from mistakenly assuming the momenta cancel out completely without calculation. A transferable strategy is to compute the initial total momentum vectorially and recognize it must equal the final total for isolated systems.
18. On a frictionless horizontal track, a $0.50\,\text{kg}$ cart moving right at $2.0\,\text{m/s}$ collides and sticks to a $1.0\,\text{kg}$ cart initially at rest. Define the system as both carts; external forces on the system are negligible during the collision. After they stick, what is their common velocity?
    Answer: $0.67\,\text{m/s}$ to the right This problem tests conservation of linear momentum in a perfectly inelastic collision. When external forces are negligible (as stated), the total momentum of the system before collision equals the total momentum after collision. Initially, the 0.50 kg cart has momentum (0.50 kg)(2.0 m/s) = 1.0 kg⋅m/s to the right, while the 1.0 kg cart has zero momentum. After sticking together, the combined mass of 1.5 kg moves with velocity v, so (1.5 kg)v = 1.0 kg⋅m/s, giving v = 0.67 m/s to the right. Choice C (2.0 m/s) incorrectly assumes the stuck carts maintain the original cart's speed. To solve momentum conservation problems: identify the system, confirm external forces are negligible, then set initial total momentum equal to final total momentum.
19. A $1.5\,\text{kg}$ cart moving right at $1.6\,\text{m/s}$ collides with a $0.50\,\text{kg}$ cart moving right at $0.40\,\text{m/s}$. The carts stick together. Define the system as both carts; external forces on the system are negligible during the collision. What is the final velocity of the stuck-together carts?
    Answer: $1.3\,\text{m/s}$ right This question assesses the skill of applying conservation of linear momentum to an inelastic collision between two carts. The total system momentum remains constant because external forces are negligible during the collision. The initial momentum is (1.5 kg)(1.6 m/s) + (0.50 kg)(0.40 m/s) = 2.6 kg·m/s to the right. After sticking together, the final momentum is (2.0 kg) v, so v = 1.3 m/s to the right. A common distractor like 1.6 m/s might come from using only the velocity of the faster cart. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
20. A $1.5\,\text{kg}$ cart moving right at $2.0\,\text{m/s}$ collides and sticks to a $0.50\,\text{kg}$ cart moving left at $4.0\,\text{m/s}$ on a low-friction track. Treat both carts as the system and assume external forces are negligible during the collision. What is the velocity of the combined carts immediately after?
    Answer: $0.50\,\text{m/s}$ right This problem requires conservation of linear momentum for a perfectly inelastic collision between objects moving in opposite directions. When external forces on the system are negligible during collision, total momentum is conserved. Taking right as positive, initial momentum is (1.5 kg)(2.0 m/s) + (0.50 kg)(-4.0 m/s) = 3.0 - 2.0 = 1.0 kg·m/s. After sticking together, the combined mass of 2.0 kg moves with velocity v, so 2.0v = 1.0, yielding v = 0.50 m/s right. Choice C (0.50 m/s left) has the correct magnitude but wrong direction. Always check your sign conventions and ensure the final direction matches the net initial momentum direction.

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