This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions where both momentum and kinetic energy are conserved. In elastic collisions, knowing the masses, initial velocities, and one final velocity allows us to calculate the other final velocity using momentum conservation. In this problem, a 50 kg skater moving at +4.0 m/s collides elastically with a stationary 70 kg skater, and after collision the second skater moves at +3.0 m/s. Choice A (-0.20 m/s) is correct because applying momentum conservation: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f gives us (50)(4.0) + (70)(0) = (50)v₁f + (70)(3.0), which simplifies to 200 = 50v₁f + 210, yielding v₁f = -10/50 = -0.20 m/s. Choice C (+1.0 m/s) is incorrect because it would not conserve momentum, as the total final momentum would exceed the initial momentum. To help students: In elastic collisions, always verify both momentum and energy conservation. The negative final velocity indicates the first skater rebounds backward, which is physically reasonable when a lighter object collides with a heavier stationary object.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where fragments move in opposite directions. When a stationary object explodes, the total momentum must remain zero, requiring the momentum vectors of all fragments to sum to zero. In this problem, a stationary 10 kg device explodes into 6.0 kg and 4.0 kg fragments, with the 6.0 kg fragment moving at -5.0 m/s. Choice A (+7.5 m/s) is correct because applying momentum conservation: 0 = m₁v₁ + m₂v₂ gives us 0 = (6.0)(-5.0) + (4.0)v₂, which simplifies to 0 = -30 + 4v₂, yielding v₂ = +7.5 m/s. Choice B (-7.5 m/s) is incorrect because it has the wrong sign, which would result in both fragments moving in the same direction, creating net momentum from an initially stationary system. To help students: In explosion problems from rest, fragments must move in opposite directions. The lighter fragment moves faster to compensate for its smaller mass, maintaining zero total momentum.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil problems where an initially stationary system separates into moving parts. In recoil situations, the total momentum must remain zero if the system starts at rest, meaning the momentum of the projectile and the recoiling object must be equal in magnitude but opposite in direction. In this problem, a 250 kg cannon fires a 5.0 kg projectile at +180 m/s relative to the ground. Choice A (-3.6 m/s) is correct because applying momentum conservation: 0 = mcvc + mpvp gives us 0 = (250)vc + (5.0)(180), which simplifies to 0 = 250vc + 900, yielding vc = -900/250 = -3.6 m/s. Choice B (+3.6 m/s) is incorrect because it has the wrong sign, which would create net momentum in the positive direction instead of maintaining zero total momentum. To help students: Emphasize that in recoil problems from rest, the objects must move in opposite directions. The lighter object moves faster and the heavier object moves slower, with their momenta canceling exactly.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically using the impulse-momentum theorem to find velocity changes when forces act over time. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum, allowing us to calculate final velocities when initial conditions and applied forces are known. In this problem, a 0.20 kg tennis ball initially moving at -15 m/s experiences a +120 N force for 0.050 s, and we must find the final velocity. Choice A (+15 m/s) is correct because impulse = FΔt = (120 N)(0.050 s) = 6.0 N·s = Δp = m(vf - vi), so 6.0 = 0.20(vf - (-15)), giving 6.0 = 0.20vf + 3.0, thus vf = 3.0/0.20 = +15 m/s. Choice B (-15 m/s) is incorrect because it suggests no change in speed, ignoring the substantial impulse applied to the ball. To help students: Emphasize the vector nature of impulse and momentum, ensuring proper sign conventions are maintained. Practice problems involving direction changes and use impulse-momentum bar charts to visualize the momentum change process.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically finding the second object's velocity in an elastic collision where the first object hits a stationary target. In elastic collisions between two objects where one is initially at rest, specific formulas relate the final velocities to the initial conditions and mass ratio. In this problem, a 50 kg skater at +4.0 m/s collides elastically with a 70 kg skater at rest, and we need the final velocity of the second skater. Choice A (+3.33 m/s) is correct because for elastic collisions with the second object initially at rest, v2f = (2m1/(m1+m2))v1i = (2(50)/(50+70))(4.0) = (100/120)(4.0) = 3.33 m/s. Choice B (+2.86 m/s) is incorrect because it might result from using the wrong formula or making calculation errors with the mass ratio. To help students: Emphasize memorizing or deriving the elastic collision formulas for the special case of one object at rest. Practice problems with various mass ratios and use energy bar charts alongside momentum diagrams to verify both conservation laws are satisfied.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where fragments move in opposite directions. When an object at rest explodes, the total momentum remains zero, requiring the momentum vectors of all fragments to sum to zero. In this problem, a stationary 8.0 kg object explodes into 5.0 kg and 3.0 kg fragments, with the 5.0 kg fragment moving at -4.0 m/s. Choice A (+6.7 m/s) is correct because initial momentum = 0, so (5.0 kg)(-4.0 m/s) + (3.0 kg)(v) = 0, giving -20.0 + 3.0v = 0, thus v = +20.0/3.0 = +6.67 m/s ≈ +6.7 m/s. Choice B (-6.7 m/s) is incorrect because it has the wrong sign, failing to recognize that the fragments must move in opposite directions. To help students: Use vector diagrams to visualize how explosion fragments must have opposite momentum vectors. Practice with different mass splits to show how the lighter fragment always moves faster to conserve momentum.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in perfectly inelastic collisions where objects stick together. Linear momentum, the product of mass and velocity, remains constant in isolated systems, meaning the total momentum before collision equals the total momentum after. In this problem, a 2.0 kg cart moving at +3.0 m/s collides with and sticks to a 1.0 kg cart initially at rest, and we must find their common final velocity. Choice A (+2.0 m/s) is correct because applying conservation of momentum: initial momentum = (2.0 kg)(+3.0 m/s) + (1.0 kg)(0 m/s) = 6.0 kg·m/s, and final momentum = (2.0 + 1.0 kg)(v_f) = 3.0v_f, so v_f = 6.0/3.0 = +2.0 m/s. Choice B (+1.0 m/s) is incorrect because it might result from dividing the initial velocity by the total mass without considering the initial momentum. To help students: Use the systematic approach of calculating total initial momentum, then dividing by total final mass. Visual representations like momentum bar charts can reinforce the conservation concept and help students avoid common calculation errors.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil situations where a projectile is fired from an initially stationary system. Linear momentum conservation requires that the total momentum before firing (zero for a system at rest) equals the total momentum after firing, resulting in the cannon recoiling in the opposite direction to the projectile. In this problem, a 10 kg cannon fires a 0.50 kg projectile at +80 m/s, and we must find the cannon's recoil velocity. Choice A (-4.0 m/s) is correct because applying conservation of momentum: initial momentum = 0, final momentum = (0.50 kg)(+80 m/s) + (10 kg)(v_c) = 0, which gives 40 + 10v_c = 0, so v_c = -40/10 = -4.0 m/s. Choice B (+4.0 m/s) is incorrect because it has the wrong sign, suggesting the cannon moves in the same direction as the projectile, which violates momentum conservation. To help students: Stress that recoil always occurs in the opposite direction to the projectile motion. Practice with various mass ratios and use free-body diagrams to reinforce the action-reaction principle underlying momentum conservation.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions where both momentum and kinetic energy are conserved. In elastic collisions, objects bounce off each other without losing mechanical energy, requiring simultaneous solution of momentum and energy conservation equations. In this problem, a 6.0 kg cart at +2.0 m/s collides elastically with a 2.0 kg cart at rest. Choice B (+3.0 m/s) is correct because for elastic collisions with one object initially at rest, the final velocity of the initially stationary object is v_2f = (2m_1/(m_1+m_2))v_1i = (2×6.0/(6.0+2.0))×2.0 = (12.0/8.0)×2.0 = 3.0 m/s. Choice C (+1.5 m/s) is incorrect because it might result from treating this as an inelastic collision or using incorrect elastic collision formulas. To help students: Teach the specific formulas for elastic collisions, especially the special case where one object is initially at rest. Use energy bar charts alongside momentum calculations to verify that kinetic energy is conserved in elastic collisions.

This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil problems where momentum is conserved in firing projectiles. When a cannon fires a projectile, the system's total momentum remains zero if initially at rest, causing the cannon to recoil in the opposite direction. In this problem, a 40 kg cannon fires a 2.0 kg projectile at +120 m/s from rest. Choice A (-6.0 m/s) is correct because initial momentum = 0, so (2.0 kg)(+120 m/s) + (40 kg)(v) = 0, giving v = -240/40 = -6.0 m/s. Choice D (-60 m/s) is incorrect because it might result from using the wrong mass ratio or confusing the projectile and cannon masses. To help students: Draw clear before/after diagrams showing the recoil effect, emphasize that the more massive object moves slower to conserve momentum. Practice with various mass ratios to show how recoil velocity depends inversely on mass.