7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
Three carts of different masses move along a straight, level track with different velocities. A student wants to analyze the three carts as a single system. Which statement about the velocity of this system is correct?
Answer and reasoning
AIt cannot have a single velocity unless the three carts all move with the same velocity. A student who thinks a group of objects has a velocity only when it moves as one piece picks this. The center-of-mass velocity is defined for any system, however its parts move.
BIts velocity is the average of the three carts' velocities, with each cart counted equally. A student who takes a plain average of the velocities picks this. The system's velocity is the mass-weighted average, (Σ mi v⃗i)/(Σ mi); a plain average gives it only if the carts have equal masses, and these carts do not.
CIts velocity is that of its most massive cart, whose motion dominates the motion of the system. A student who identifies the system with its most massive part picks this. Every cart contributes to v⃗cm in proportion to its mass, so in general v⃗cm differs from the velocity of any one cart.
DIt has one velocity, its center-of-mass velocity, though the carts' velocities differ.Correct Any collection of objects can be described as one system with one center-of-mass velocity, v⃗cm = (Σ mi v⃗i)/(Σ mi). It exists whatever the carts' individual velocities, and the system's total momentum is (Σ mi)v⃗cm.
Each of the systems I, II and III shown in the diagram consists of two carts moving along a straight, level track, with the masses and velocities shown in terms of m and v. Which ranks the magnitudes of the systems' total momenta, pI, pII and pIII?
Answer and reasoning
ApIII > pI = pII A student who adds momentum magnitudes picks this: I, 2mv + mv = 3mv; II, 3mv; III, 2mv + 2mv = 4mv. In I and III the carts move in opposite directions, so their momenta partly or wholly cancel.
BpII > pIII > pI A student who adds the velocities and ignores the masses picks this: I, v − v = 0; II, 3v; III, 2v − v = v. Momentum is mv: in III the 2m cart's momentum, −2mv, cancels the other cart's +2mv, while in I the momenta are +2mv and −mv.
CpII > pI = pIII A student who thinks momenta in opposite directions always cancel completely gives both I and III zero momentum. In I the momenta are +2mv and −mv, which do not cancel; the total is mv.
DpII > pI > pIIICorrect Add each system's momenta with signs, right positive: I, 2mv − mv = mv; II, 2mv + mv = 3mv; III, 2mv − 2mv = 0. So pII > pI > pIII.
Working Right positive. I: (2m)(+v) + (m)(−v) = +mv. II: (m)(+2v) + (m)(+v) = +3mv. III: (m)(+2v) + (2m)(−v) = 0. Ranking of magnitudes: 3mv > mv > 0.
A ball is released from rest and falls toward the ground. Air resistance is negligible. While the ball falls, which system has a total momentum that stays constant?
Answer and reasoning
AThe ball alone, since no other object collides with the ball while it is falling. A student who links changes in momentum only to collisions picks this. The gravitational force is a net external force on the ball alone, so its momentum increases downward at the rate mg.
BNo system, since the gravitational force on the ball is an external force on any system. A student who treats gravity as external in every case picks this. Whether a force is internal or external depends on the system chosen: with the Earth included, the gravitational forces are internal.
CThe ball–Earth system, since the gravitational forces in it are internal.Correct With both objects in the system, the Earth's pull on the ball and the ball's pull on the Earth are a third-law pair of internal forces. Their impulses are equal and opposite, so the ball gains downward momentum exactly as fast as the Earth gains upward momentum, and the total stays constant.
DThe Earth alone, since its very large mass keeps its momentum from changing. A student who thinks the Earth is too massive to be affected picks this. The ball pulls up on the Earth, so the Earth's momentum changes by as much as the ball's, in the opposite direction; its change in velocity is just far too small to notice.
Ball 1, of mass 0.20 kg, moving at 3.0 m/s, collides head-on with ball 2, of mass 0.30 kg, which is at rest on a level surface. Immediately after the collision, ball 1 moves back along its original line at 0.60 m/s. What is the speed of ball 2 immediately after the collision?
Answer and reasoning
A1.6 m/s A student who ignores the direction of ball 1's rebound picks this: 0.60 = (0.20)(0.60) + 0.30v₂ gives v₂ = 1.6 m/s. Ball 1's final momentum is −0.12 kg·m/s, so ball 2 must carry 0.72 kg·m/s.
B3.6 m/s A student who gives both balls equal changes in velocity picks this: ball 1's velocity changes by 3.6 m/s, so ball 2 would go from 0 to 3.6 m/s. The balls have equal and opposite changes in momentum, so the more massive ball 2 has the smaller change in velocity.
C2.4 m/sCorrect Take ball 1's initial direction as positive. Before: (0.20)(3.0) = 0.60 kg·m/s. After: (0.20)(−0.60) + (0.30)v₂. Setting the totals equal, 0.30v₂ = 0.60 + 0.12 = 0.72 kg·m/s, so v₂ = 2.4 m/s.
D2.0 m/s A student who thinks ball 2 receives all of ball 1's initial momentum picks this: 0.60/0.30 = 2.0 m/s. Ball 1 rebounds, so it loses more than its initial momentum: ball 2 gains 0.60 + 0.12 = 0.72 kg·m/s.
Working Positive = ball 1's initial direction. p before = (0.20 kg)(3.0 m/s) = 0.60 kg·m/s. p after = (0.20 kg)(−0.60 m/s) + (0.30 kg)v₂ = −0.12 + 0.30v₂. Conservation: v₂ = 0.72/0.30 = 2.4 m/s.
A box slides across a level floor and comes to rest because of friction. Which statement about the box's momentum is correct?
Answer and reasoning
AIt was transferred to the Earth by friction, so the box–Earth total did not change.Correct Friction from the floor gave the box a backward impulse, and the box gave the floor, and the Earth it is attached to, an equal and opposite forward impulse. The momentum the box lost was transferred to the Earth, so the momentum of the box–Earth system is unchanged.
BIt was destroyed by friction, which turned it into thermal energy in the box and floor. A student who thinks momentum can be dissipated like kinetic energy picks this. Friction transforms the box's kinetic energy into thermal energy, but momentum is only transferred: here, to the Earth.
CIt was not conserved, since momentum is conserved only in collisions and explosions. A student who links conservation of momentum only to collisions picks this. Momentum is conserved in all interactions, including friction: the box's loss is the Earth's gain.
DIt was transferred to the Earth, which gained the speed that the box lost. A student who gives interacting objects equal and opposite changes in velocity picks this. The interaction gives the box and the Earth equal and opposite changes in momentum, not in velocity; with its enormous mass, the Earth's change in velocity is immeasurably small.
An open cart of mass 3m rolls at speed v₀ along a level track with negligible friction. A ball of mass m falls vertically into the cart and stays in it; just before it lands, the ball's speed is 4v₀. What is the speed of the cart, with the ball in it, just after the ball lands?
Answer and reasoning
A1.00v₀ A student who thinks a vertically falling ball cannot change the cart's horizontal velocity picks this. Friction from the cart must bring the ball up to the cart's horizontal velocity, and the ball pushes back on the cart, slowing it.
B1.25v₀ A student who includes the ball's vertical momentum, 4mv₀, as if it were conserved too, picks this: √((3mv₀)² + (4mv₀)²)/(4m) = 5v₀/4. The track exerts a large vertical external impulse while the ball lands, so only the horizontal momentum is conserved.
C0.75v₀Correct While the ball lands, the track pushes up on the cart with a large normal force, so the system's vertical momentum is not conserved; no horizontal external force acts, so its horizontal momentum is. Horizontally, (3m)v₀ + m(0) = (4m)v, so v = 3v₀/4 = 0.75v₀.
D0.50v₀ A student who takes the final velocity to be the plain average of the cart's horizontal velocity, v₀, and the ball's, 0, picks this. The velocities must be weighted by mass: (3m·v₀ + m·0)/(4m) = 0.75v₀.
Working System: cart + ball. Horizontal: no external force, so (3m)v₀ + m(0) = (4m)v and v = 3v₀/4 = 0.75v₀. Vertical: the track's normal force gives the system an upward impulse of 4mv₀ (the ball's vertical momentum is transferred to the Earth), so vertical momentum is not conserved.
A ball moving horizontally at speed v₀ strikes a wall that is fixed to the ground. In trial 1 the ball sticks to the wall. In trial 2 an identical ball with the same initial speed bounces straight back at speed v₀/2. What is the ratio of the momentum transferred from the ball to the wall and Earth in trial 2 to that transferred in trial 1?
Answer and reasoning
A1.0 A student who thinks the wall receives the ball's incoming momentum, mv₀, in both trials picks this. In trial 2 the ball's momentum reverses, so it loses more than it brought: 3mv₀/2.
B1.5Correct Toward the wall positive. Trial 1: the ball's momentum changes from mv₀ to 0, so mv₀ is transferred. Trial 2: it changes from mv₀ to −mv₀/2, a change of magnitude 3mv₀/2, which is transferred. The ratio is 1.5.
C0.5 A student who subtracts the sizes of the ball's momenta picks this: mv₀ − mv₀/2 = mv₀/2 in trial 2. The rebound is in the opposite direction, so the change is mv₀ − (−mv₀/2) = 3mv₀/2.
D2.0 A student who has learned that a bounce transfers twice the momentum of a stick picks this. That holds only for a rebound at the original speed; this ball rebounds at v₀/2, so the transfer is 3mv₀/2, not 2mv₀.
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.3.A.1 System Fix
System
A collection of objects chosen for analysis and treated together. Forces exerted by one part of the system on another are internal forces; forces exerted on the system's parts by objects outside it are external forces. A system can be described by its total mass and a single center-of-mass velocity.
Center-of-mass velocity, v⃗cm
The velocity of a system's center of mass, v⃗cm = (Σ p⃗i)/(Σ mi) = (Σ mi v⃗i)/(Σ mi): the mass-weighted average of the parts' velocities, equal to the system's total momentum divided by its total mass. It is the time derivative of the center-of-mass position. Unit: m/s.
Constant center-of-mass velocity
If the net external force on a system is zero, its center-of-mass velocity is constant, however its parts move, collide or push on one another. A system that starts at rest then keeps its center of mass at rest.
Students often think A system can be given a single velocity only when all of its parts move together with the same velocity. In fact Yes. Any system has one center-of-mass velocity, v⃗cm = (Σ mi v⃗i)/(Σ mi), whatever the motions of its parts, and the system's total momentum equals its total mass times v⃗cm.
Students often think The velocity of a system, or the common velocity of objects that end up moving together, is the plain average of the objects' velocities, each counted equally. In fact Not in general; for two objects with different velocities it is only if their masses are equal. In general it is the mass-weighted average, (Σ mi v⃗i)/(Σ mi), so more massive objects count for more.
4.3.A.2 Total momentum of a system, p⃗sysFix
Total momentum of a system, p⃗sys
The vector sum of the momenta of the system's parts, p⃗sys = Σ mi v⃗i = (Σ mi)v⃗cm. Along a line, momenta in opposite directions are added with opposite signs; in two dimensions they are added by components. Unit: kg·m/s.
Students often think The total momentum of a system is the sum of the magnitudes of its parts' momenta, whatever their directions. In fact No. Momentum is a vector. Along a line, momenta in opposite directions have opposite signs and partly or wholly cancel; in two dimensions momenta are added by components. The magnitude of the total is less than the sum of the magnitudes unless all the moving parts move in the same direction.
Students often think When two objects move in opposite directions, their momenta cancel and the system's total momentum is zero, whatever their masses and speeds. In fact Only if their momenta have equal magnitudes. Opposite momenta partly cancel, and the total is in the direction of the larger momentum magnitude, mv, which depends on mass as well as speed.
4.3.A.3 Internal and external forces Fix
Internal and external forces
Internal forces act between parts of the system and come in Newton's third-law pairs inside it, so their impulses cancel in the system's total momentum. Only external forces can change the total momentum of a system; a change in one part's momentum caused by internal forces is balanced by an equal and opposite change in another part's.
Equal and opposite impulses
During an interaction, the impulse exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the impulse exerted on object 2 by object 1, because the two forces form a Newton's third-law pair and act over the same time interval. The objects' momentum changes are therefore equal and opposite; their velocity changes are in inverse ratio to their masses. Unit of impulse: N·s.
Choice of system
Where the system boundary is drawn is a choice. Including both interacting objects makes the forces between them internal, so a system can be chosen whose total momentum is constant; a system containing only one of the objects has its momentum changed by the force from the other.
Impulse on a system
The impulse exerted on a system by the net external force, J⃗ = ∫F⃗net dt, equals the change in the system's total momentum, J⃗ = Δp⃗. On a graph of net external force against time it is the signed area between the curve and the time axis. Unit: N·s.
Students often think Internal forces, such as a spring launching one part of a system or a person pushing on another part, add momentum to the system and change its center-of-mass velocity. In fact No. Internal forces form Newton's third-law pairs inside the system, so their impulses are equal and opposite and cancel in the total. Only a net external force changes the total momentum and v⃗cm.
Students often think In a collision, the more massive object exerts the larger force, and therefore the larger impulse, on the other object. In fact No. The forces the two objects exert on each other are a Newton's third-law pair, equal in magnitude and opposite in direction at every instant, so the impulses are equal and opposite. The less massive object has the larger change in velocity.
4.3.A.4 Conservation of momentum in collisions and explosions Fix
Conservation of momentum in collisions and explosions
When the net external impulse during a collision or explosion is negligible, the system's total momentum immediately after equals its total momentum immediately before: Σ mi v⃗i (before) = Σ mi v⃗i (after). In two dimensions this holds separately for the x- and y-components.
Students often think Two objects that push apart from rest move off with equal speeds, or move equal distances, whatever their masses. In fact Only if their masses are equal. Their momenta are equal and opposite, so their speeds are in inverse ratio to their masses: the less massive object moves faster and, in the same time, farther.
Students often think Equal and opposite forces from a spring or an explosion give the two objects equal kinetic energies. In fact No. The spring exerts equal and opposite forces on the objects for the same time, so they receive equal and opposite impulses. The less massive object moves farther while the forces act, so more work is done on it and it receives more kinetic energy: for momenta of equal magnitude p, K = p²/(2m).
4.3.B.1 Momentum transfer Fix
Momentum transfer
Momentum is conserved in all interactions: the momentum one object loses is gained by the object it interacts with. When friction slows an object, the momentum it loses goes to the object exerting the friction, often the Earth through a floor.
Students often think Momentum can be lost or used up, for example dissipated as thermal energy by friction or in a collision, in the way kinetic energy can be. In fact No. Every interaction conserves momentum. When friction slows an object, the momentum it loses is transferred to the object exerting the friction, often the Earth through a floor; in a collision, the momentum one object loses is gained by the other.
Students often think Momentum conservation applies only to collisions and explosions; in other interactions, such as friction slowing an object, momentum is not conserved. In fact No. Momentum is conserved in all interactions, including friction, gravitational attraction and pushes between people. A chosen system's total is constant when the system includes all the interacting objects and no net external force acts on it.
4.3.B.2 Constant total momentum Fix
Constant total momentum
If the net external force on the selected system is zero, dp⃗sys/dt = 0 and the system's total momentum is constant. If only one component of the net external force is zero, the momentum component in that direction is constant.
Students often think An object that falls vertically into a moving cart does not change the cart's horizontal velocity, because it exerts no horizontal force on the cart. In fact Yes. The dropped object has no horizontal momentum, so friction between it and the cart must bring it up to the cart's horizontal velocity; the cart receives an equal and opposite horizontal impulse and slows, keeping the system's horizontal momentum constant.
Students often think Whenever momentum is conserved in a collision, the whole momentum vector is conserved, including components along which external forces act. In fact No. A momentum component is constant only if the net external force has no component in that direction. When a falling object lands in a cart on a track, the track's normal force gives the system a large upward impulse, so vertical momentum is not conserved, while horizontal momentum is.
4.3.B.3 Momentum exchange with the environment Fix
Momentum exchange with the environment
If the net external force on a system is nonzero, the system's momentum changes at the rate F⃗net = dp⃗/dt, and the objects in the environment that exert that force undergo an equal and opposite change in momentum.
Students often think A ball that bounces back transfers twice as much momentum as one that sticks, whatever its rebound speed. In fact Only if it rebounds with its original speed. The momentum transferred equals the ball's change in momentum, m(v₀ + v) for a rebound at speed v, which lies between mv₀ (sticking) and 2mv₀ (rebound at the original speed).
Students often think An object's change in momentum is the difference between the magnitudes of its final and initial momenta, whatever their directions. In fact No. Δp⃗ = p⃗ − p⃗₀ is a vector difference. If the object reverses direction the magnitudes add: a ball that arrives at 4.0 kg·m/s and rebounds at 3.0 kg·m/s has a change in momentum of magnitude 7.0 kg·m/s, not 1.0 kg·m/s.
15 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 15
The diagram shows three carts moving along a straight, level track. What is the speed of the center of mass of the three-cart system?
Answer and reasoning
A4.0 m/s A student who adds the momentum magnitudes picks this: (6.0 + 6.0 + 20)/8.0 = 4.0 m/s. The 3.0 kg cart moves left, so its momentum, −6.0 kg·m/s, cancels part of the others'.
B3.0 m/s A student who averages the three velocities without weighting them by mass picks this: (6.0 − 2.0 + 5.0)/3 = 3.0 m/s. The carts' masses differ, so each velocity must be weighted by its mass.
C2.5 m/sCorrect Take right as positive and add the momenta with their signs: (1.0)(6.0) + (3.0)(−2.0) + (4.0)(5.0) = 20 kg·m/s. Dividing by the total mass, 8.0 kg, gives vcm = 2.5 m/s, to the right.
D5.0 m/s A student who gives the system the velocity of its most massive cart, the 4.0 kg cart, picks this. The other carts contribute too: with every momentum included, vcm = 20/8.0 = 2.5 m/s.
Working Right positive. Σp = (1.0 kg)(+6.0 m/s) + (3.0 kg)(−2.0 m/s) + (4.0 kg)(+5.0 m/s) = 6.0 − 6.0 + 20 = +20 kg·m/s. Σm = 8.0 kg. vcm = (20 kg·m/s)/(8.0 kg) = +2.5 m/s, so the speed is 2.5 m/s (to the right).
Two carts move on two parallel, straight, level tracks that run in the x-direction. Cart P, of mass 2m, has x-coordinate xP(t) = bt², and cart Q, of mass 3m, has x-coordinate xQ(t) = −ct³, where b and c are positive constants. What is the velocity of the center of mass of the two-cart system at time t?
Answer and reasoning
A(2bt − 3ct²)/5 A student who finds each velocity as position divided by time picks this: xP/t = bt and xQ/t = −ct², which give [2m(bt) + 3m(−ct²)]/(5m). The carts' velocities are changing, so the velocity at time t is the derivative dx/dt: 2bt and −3ct².
B(4bt − 9ct²)/5Correct Differentiate each position: vP = dxP/dt = 2bt and vQ = dxQ/dt = −3ct². Then vcm = [2m(2bt) + 3m(−3ct²)]/(2m + 3m) = (4bt − 9ct²)/5.
C(2bt − 3ct²)/2 A student who averages the two velocities without weighting them by mass picks this: (2bt − 3ct²)/2. The carts have masses 2m and 3m, so vcm = (2m vP + 3m vQ)/(5m).
D(4bt + 9ct²)/5 A student who adds the momentum magnitudes picks this. Cart Q's velocity, −3ct², is in the −x-direction while P's is in the +x-direction, so Q's momentum, −9mct², must be added with its negative sign.
Working vP = d(bt²)/dt = 2bt; vQ = d(−ct³)/dt = −3ct². vcm = (mP vP + mQ vQ)/(mP + mQ) = [2m(2bt) + 3m(−3ct²)]/(5m) = (4bt − 9ct²)/5. (Checked with sympy; each distractor reproduced from its stated error.)
A student of mass m stands at one end of a plank of mass M that rests on frictionless ice; both are at rest. The student walks a distance L along the plank, measured relative to the plank, and then stops. How far does the plank move relative to the ice?
Answer and reasoning
AmL/(m + M)Correct No horizontal external force acts on the student–plank system, which starts at rest, so its center of mass stays at rest: m ds = M dp, where ds and dp are the distances the student and plank move relative to the ice, in opposite directions. The student's displacement relative to the plank is ds + dp = L, so dp = mL/(m + M).
BmL/M A student who treats L as the walker's displacement relative to the ice picks this: mL = M dp gives dp = mL/M. L is measured relative to the plank, which itself moves backward; relative to the ice the walker moves only L − dp.
CL/2 A student who thinks the walker and the plank move equal distances in opposite directions picks this. Their momenta are equal and opposite at every instant, so their displacements are in inverse ratio to their masses and are equal only if m = M.
D0 A student who thinks the plank stays put while the walker moves picks this. To walk forward, the walker pushes backward on the plank, and on frictionless ice the plank moves backward, keeping the system's center of mass at rest.
Working System: student + plank. No horizontal external force, vcm = 0 throughout, so xcm is fixed: m ds − M dp = 0 (ds forward, dp backward, both relative to the ice). Relative displacement: ds + dp = L. Solving: dp = mL/(m + M), ds = ML/(m + M). (Checked with sympy.)
A cart moves at constant velocity along a level track with negligible friction. A spring inside the cart then launches a small block forward, and the block slides ahead of the cart along the same track. How does the velocity of the center of mass of the cart–block system after the launch compare with its velocity before the launch?
Answer and reasoning
AIt is greater, since the spring gives the block extra forward momentum during the launch. A student who thinks an internal push adds momentum to the system picks this. The spring gives the block forward momentum but gives the cart an equal amount of backward momentum, so the total is unchanged.
BIt is unchanged, since the spring's forces on the cart and the block are internal.Correct The spring pushes the block forward and the cart backward with equal and opposite forces, which are internal to the cart–block system. With no net external force, the system's total momentum, and so v⃗cm, is unchanged.
CIt is smaller, since the cart, which is the most massive part, slows down in the launch. A student who takes the system's velocity to be that of its most massive part picks this. The cart does slow down, but the block speeds up by an equal and opposite momentum change, so v⃗cm is unchanged.
DIt is not defined, since the cart and block now move with different velocities. A student who thinks a system has a velocity only when its parts move together picks this. v⃗cm = (Σ mi v⃗i)/(Σ mi) is defined for any system; here it equals the velocity the cart had before the launch.
Carts A and B collide on a level track. The bar chart shows the momentum p of each cart just before and just after the collision, with rightward taken as positive. Which claim do these data support?
Answer and reasoning
AThe system lost momentum, since the bar lengths add to 8 before and to 6 after. A student who adds momentum magnitudes picks this: 6 + 2 = 8 before and 1 + 5 = 6 after. Momenta in opposite directions must be added with signs: 6 − 2 = 4 before and −1 + 5 = 4 after.
BMomentum was not conserved, since each cart's momentum changed in the collision. A student who thinks conservation requires each object to keep its momentum picks this. Conservation concerns the total: the carts' momenta changed by equal and opposite amounts, and the total stayed at 4 kg·m/s.
CThe momentum gained by B equals the momentum lost by A, so the total is unchanged.Correct From the chart, A's momentum changes from +6 to −1 kg·m/s, a change of −7 kg·m/s, and B's from −2 to +5 kg·m/s, a change of +7 kg·m/s. The changes are equal and opposite, so the total is +4 kg·m/s both before and after.
DNo external forces acted on either cart, since their total momentum did not change. A student who thinks constant momentum means there are no external forces picks this. Gravity and the track's normal force act on both carts; the data show only that the net external impulse during the collision was negligible.
A small block A of mass 0.50 kg slides at 4.0 m/s onto one end of a long board B of mass 1.5 kg that is at rest on frictionless ice. The coefficient of kinetic friction between A and B is 0.20. For how long does A slide along B before the two move with the same velocity? Use g = 10 m/s².
Answer and reasoning
A1.5 sCorrect The ice exerts no horizontal force, so the A–B system's momentum is constant and the common velocity is (0.50)(4.0)/2.0 = 1.0 m/s. Friction μk mA g = 1.0 N acts on A, so A's change in momentum, (0.50)(1.0 − 4.0) = −1.5 kg·m/s, takes 1.5 s, while B gains +1.5 kg·m/s.
B2.0 s A student who treats the board as fixed, like a floor, picks this: A would slide until it stopped, after (4.0 m/s)/(2.0 m/s²) = 2.0 s. The board is free on the ice; it gains the momentum A loses and moves forward, so A stops slipping at the common velocity, 1.0 m/s.
C1.0 s A student who takes the common velocity to be the plain average of 4.0 m/s and 0, 2.0 m/s, picks this: t = (4.0 − 2.0)/2.0 = 1.0 s. The common velocity weights each velocity by mass: (0.50)(4.0)/2.0 = 1.0 m/s.
D6.0 s A student who divides the friction force on A by the total mass, 2.0 kg, finds A's acceleration as 0.50 m/s² and picks this: t = (3.0 m/s)/(0.50 m/s²) = 6.0 s. The friction force acts on A alone, so A's acceleration is (1.0 N)/(0.50 kg) = 2.0 m/s².
Working No horizontal external force on A + B: (0.50 kg)(4.0 m/s) = (2.0 kg)vf, vf = 1.0 m/s. Friction on A: μk mA g = (0.20)(0.50 kg)(10 m/s²) = 1.0 N, backward. Impulse–momentum for A: −(1.0 N)t = (0.50 kg)(1.0 − 4.0) m/s = −1.5 N·s, so t = 1.5 s. (B: (1.0 N)(1.5 s) = (1.5 kg)(1.0 m/s), consistent.)
Ball A, of mass m, moving at speed v₀, collides head-on with ball B, of mass 3m, which is at rest. During the collision, B exerts an impulse of magnitude JA on A and A exerts an impulse of magnitude JB on B, and the magnitudes of the balls' changes in velocity are ΔvA and ΔvB. Which correctly compares these magnitudes?
Answer and reasoning
AJA < JB and ΔvA = ΔvB A student who gives both balls equal changes in velocity, and then finds each impulse as mΔv, picks this: B, with three times the mass, would receive three times the impulse. The impulses are a third-law pair and are equal; with masses m and 3m, ΔvA = J/m is three times ΔvB = J/(3m).
BJA < JB and ΔvA < ΔvB A student who thinks only the moving ball exerts a force, so that ball B, at rest, pushes back weakly or not at all, picks this. B pushes on A exactly as hard as A pushes on B, so JA = JB, and the lighter ball A has the larger change in velocity.
CJA > JB and ΔvA > ΔvB A student who thinks the more massive ball exerts the larger force picks this. The forces are a third-law pair, so the impulses are equal; A's larger change in velocity comes from its smaller mass, not from a larger impulse.
DJA = JB and ΔvA > ΔvBCorrect The forces between the balls are a third-law pair, equal and opposite and acting for the same time, so JA = JB. Each change in velocity is J/m, so A, with one third of B's mass, has three times B's change in velocity.
Cart X, of mass 1.5 kg, collides with cart Y, of mass 2.0 kg, on a level track with negligible friction. The graph shows the force exerted on Y by X as a function of time t during the collision. What is the magnitude of the change in X's velocity during the collision?
Answer and reasoning
A4.0 m/sCorrect The impulse on Y is the area under the graph, ½(0.040 s)(300 N) = 6.0 N·s. Y exerts an equal and opposite impulse on X, so X's momentum changes by 6.0 kg·m/s in magnitude and its velocity by (6.0 N·s)/(1.5 kg) = 4.0 m/s.
B8.0 m/s A student who multiplies the peak force by the duration picks this: (300 N)(0.040 s)/(1.5 kg) = 8.0 m/s. The force rises and falls, so the impulse is the triangle's area, half that product.
C3.0 m/s A student who gives both carts the same change in velocity picks this: Y's change is (6.0 N·s)/(2.0 kg) = 3.0 m/s. The carts receive equal impulses, not equal changes in velocity; the less massive cart X changes velocity more.
D0.0 m/s A student who thinks the cart doing the pushing receives no force picks this. Y pushes back on X with a force equal and opposite to the one graphed, so X also receives an impulse of 6.0 N·s.
Working Impulse on Y by X = area of triangle = ½(0.040 s)(300 N) = 6.0 N·s. By Newton's third law the impulse on X by Y has magnitude 6.0 N·s. |ΔvX| = (6.0 N·s)/(1.5 kg) = 4.0 m/s.
A system of two carts joined by a spring moves along a level track with negligible friction, and at t = 0 its total momentum is +2.0 kg·m/s. A student's hand then pushes and pulls on one of the carts. The graph shows the force F exerted by the hand, the only horizontal external force on the system, as a function of time t. What is the total momentum of the system at t = 0.50 s?
Answer and reasoning
A7.0 kg·m/s A student who adds the area below the axis as positive picks this: 2.0 + 3.0 + 2.0 = 7.0 kg·m/s. The hand pulls in the negative direction from 0.30 s to 0.50 s, so that area, −2.0 N·s, reduces the momentum.
B3.0 kg·m/sCorrect The impulse is the signed area: the triangle above the axis gives ½(0.30 s)(20 N) = +3.0 N·s and the rectangle below it gives (0.20 s)(−10 N) = −2.0 N·s, so J = +1.0 N·s. The total momentum is 2.0 + 1.0 = 3.0 kg·m/s.
C1.0 kg·m/s A student who takes the impulse, +1.0 N·s, to be the final momentum picks this. The impulse is the change in momentum, so it must be added to the initial +2.0 kg·m/s.
D6.0 kg·m/s A student who multiplies the peak force of the push by its duration picks this: 2.0 + (20 N)(0.30 s) − 2.0 = 6.0 kg·m/s. The push rises and falls, so its impulse is the triangle's area, ½(0.30 s)(20 N) = 3.0 N·s.
Working J = ∫F dt from 0 to 0.50 s = ½(0.30 s)(20 N) + (0.20 s)(−10 N) = 3.0 − 2.0 = +1.0 N·s. The spring forces are internal. p(0.50 s) = p0 + J = 2.0 + 1.0 = +3.0 kg·m/s.
Two blocks, of masses m and 3m, are at rest on a frictionless horizontal surface with a compressed light spring between them; the spring is not attached to either block and stores potential energy U. The blocks are released and the spring pushes them apart. What is the speed of the block of mass m after it leaves the spring?
Answer and reasoning
A√(U/(2m)) A student who thinks the blocks move off with equal speeds picks this: U = ½(4m)v² gives v = √(U/(2m)). Equal speeds would give the system a total momentum of 2mv, not zero; the less massive block moves three times as fast.
B√(2U/m) A student who treats the heavier block as a fixed wall picks this: U = ½mv₁² gives v₁ = √(2U/m). The 3m block is free, so it recoils and takes a quarter of U.
C√(U/m) A student who thinks equal and opposite spring forces give the blocks equal kinetic energies picks this: ½mv₁² = U/2. The forces give equal impulses, not equal work; the lighter block moves farther while the spring pushes, so it receives three quarters of U.
D√(3U/(2m))Correct The system starts at rest with no net external force, so the momenta stay equal and opposite: mv₁ = 3mv₂, v₂ = v₁/3. All of U becomes kinetic energy: U = ½mv₁² + ½(3m)(v₁/3)² = (2/3)mv₁², so v₁ = √(3U/(2m)).
Working p: 0 = mv₁ − 3mv₂ → v₂ = v₁/3. Energy: U = ½mv₁² + ½(3m)(v₁/3)² = ½mv₁² + (1/6)mv₁² = (2/3)mv₁² → v₁ = √(3U/(2m)). K of block m = (3/4)U; K of block 3m = (1/4)U. (Checked with sympy, with each distractor.)
An object of mass 4m at rest on a frictionless horizontal surface explodes into three pieces that slide along the surface. Immediately after the explosion, a piece of mass m moves at speed v in the +x-direction and a piece of mass 2m moves at speed 2v in the +y-direction. What is the speed of the third piece, of mass m?
Answer and reasoning
A√17 vCorrect The total momentum stays zero, so the third piece's momentum cancels the other two: px = −mv and py = −(2m)(2v) = −4mv. Its magnitude is √((mv)² + (4mv)²) = √17 mv, so its speed is √17 mv/m = √17 v.
B5v A student who adds the magnitudes of the two known momenta picks this: mv + 4mv = 5mv, giving 5v. The two momenta are perpendicular, so they combine by components: the magnitude of their sum is √17 mv.
C√5 v A student who balances the pieces' velocities instead of their momenta picks this: the third piece would need velocity components −v and −2v. Momentum, not velocity, must balance: the 2m piece's momentum is (2m)(2v) = 4mv, so the components to cancel are mv and 4mv.
D(√17/4)v A student who divides the third piece's momentum, √17 mv, by the total mass 4m picks this. The speed of one piece is its own momentum divided by its own mass, m.
Working Initial momentum 0; no horizontal external force. After: (mv, 0) + (0, 4mv) + p⃗₃ = 0 → p⃗₃ = (−mv, −4mv), |p⃗₃| = √17 mv, v₃ = √17 v ≈ 4.1v. (Checked with sympy.)
Cart X, of mass m, moving at speed v₀ along a level track with negligible friction, collides with and sticks to cart Y, which is at rest. In trial 1, Y has mass m. In trial 2, Y has mass 3m, and X has the same initial speed. What is the ratio of Y's momentum just after the collision in trial 2 to Y's momentum just after the collision in trial 1?
Answer and reasoning
A1.00 A student who thinks the struck cart receives all of X's momentum, mv₀, in both trials picks this. X keeps part of the momentum because it moves on with Y: Y receives mv₀/2 in trial 1 and 3mv₀/4 in trial 2.
B1.50Correct The carts share momentum mv₀. Trial 1: common speed v₀/2, so Y's momentum is m(v₀/2) = mv₀/2. Trial 2: common speed v₀/4, so Y's momentum is 3m(v₀/4) = 3mv₀/4. The ratio is (3/4)/(1/2) = 1.50.
C3.00 A student who thinks the stuck carts move off at X's original speed in both trials picks this: Y's momentum would be mv₀ and then 3mv₀. The pair shares X's momentum, so its speed falls from v₀/2 to v₀/4.
D0.50 A student who compares velocities instead of momenta picks this: the common speed halves, from v₀/2 to v₀/4. Y's mass triples at the same time, so its momentum changes by a factor of 3 × ½ = 1.5.
Two carts roll along a track and collide with each other at t = 1.0 s. Motion sensors record both carts, and the graph shows the total momentum p of the two-cart system as a function of time t. Which claim do the data support?
Answer and reasoning
AThe collision destroyed some of the system's momentum, turning it into thermal energy. A student who thinks momentum can be dissipated like energy picks this. The momentum falls at the same steady rate before and after the collision, with no drop at t = 1.0 s; the decrease is a transfer to the surroundings, not a loss.
BDuring the collision, one cart exerted a larger force on the other than it received. A student who thinks one object in a collision can push harder than it is pushed picks this. The forces between the carts are a third-law pair, so they cancel in the total; the graph also shows the total falling when the carts are not in contact.
CThe sensors must be faulty, since the total momentum of any system is constant. A student who thinks every system's momentum is constant picks this. The total is constant only if the net external force on the system is zero; the steady decrease shows that it was not.
DA net external force, such as friction from the track, acted on the carts throughout.Correct The total momentum falls steadily, at the same rate before, during and after the collision: dp/dt = (3.0 − 4.0)/2.0 = −0.50 N. A constant net external force, for example friction from the track, transferred momentum from the system to the surroundings; the collision itself did not change the total.
Block A, of mass m, slides at speed v₀ along a frictionless horizontal surface toward block B, of mass 2m, which is at rest. A light spring of spring constant k is attached to the side of B that faces A. A strikes the spring and compresses it. What is the maximum compression of the spring?
Answer and reasoning
A1.00 v₀√(m/k) A student who treats B as fixed, like a wall, picks this: all of A's kinetic energy goes into the spring, ½mv₀² = ½k(Δx)². B is free, so the spring pushes it forward; at maximum compression both blocks move at v₀/3 and still carry one third of the initial kinetic energy.
B0.71 v₀√(m/k) A student who thinks B receives all of A's momentum, leaving A at rest at maximum compression, picks this: vB = v₀/2 and ½k(Δx)² = ½mv₀² − ½(2m)(v₀/2)² = ¼mv₀². With A at rest and B moving away, the spring would already be lengthening; at maximum compression the two velocities are equal.
C0.82 v₀√(m/k)Correct The spring is shortest when the blocks move with the same velocity. The spring forces are internal, so momentum is conserved: mv₀ = 3mv, v = v₀/3. The kinetic energy the blocks no longer carry is stored in the spring: ½k(Δx)² = ½mv₀² − ½(3m)(v₀/3)² = (1/3)mv₀², so Δx = √(2/3) v₀√(m/k) ≈ 0.82 v₀√(m/k).
D0.50 v₀√(m/k) A student who takes the common velocity at maximum compression to be the plain average of v₀ and 0 picks this: v = v₀/2 gives ½k(Δx)² = ½mv₀² − ½(3m)(v₀/2)² = (1/8)mv₀². The common velocity is the mass-weighted average, mv₀/(3m) = v₀/3.
Working The spring is shortest when A and B have the same velocity. The spring forces are internal and the surface is frictionless, so the system's momentum is constant: mv₀ = (m + 2m)v, v = v₀/3. The surface is frictionless and the spring ideal, so mechanical energy is conserved: ½mv₀² = ½(3m)(v₀/3)² + ½k(Δx)², so ½k(Δx)² = ½mv₀² − (1/6)mv₀² = (1/3)mv₀² and Δx = v₀√(2m/(3k)) = √(2/3) v₀√(m/k) ≈ 0.82 v₀√(m/k). Distractors: B treated as fixed, ½k(Δx)² = ½mv₀² → 1.00 v₀√(m/k); A at rest with B carrying all the momentum, vB = v₀/2, ½k(Δx)² = ½mv₀² − ½(2m)(v₀/2)² = ¼mv₀² → √(1/2) ≈ 0.71 v₀√(m/k); common velocity as the plain average v₀/2, ½k(Δx)² = ½mv₀² − ½(3m)(v₀/2)² = (1/8)mv₀² → 0.50 v₀√(m/k). Key and distractors checked with sympy.
A block of mass 4m and a block of mass m are joined by a light spring and are at rest on a frictionless horizontal surface. Starting at t = 0, a horizontal force of magnitude F = bt², where b is a positive constant, is exerted on the block of mass 4m along the line of the spring; it is the only horizontal external force on the two-block system. What is the speed of the system's center of mass at time t = T?
Answer and reasoning
A0.200 bT³/m A student who takes the impulse to be the peak force multiplied by the duration picks this: bT² × T = bT³, and bT³/(5m) = 0.200 bT³/m. The impulse is the area under the force–time graph, ∫₀ᵀ bt² dt = bT³/3, one third of that product.
B0.067 bT³/mCorrect Taking both blocks as the system, the spring forces are internal, so the system's momentum changes only by the impulse of F: J = ∫₀ᵀ bt² dt = bT³/3. Dividing the total momentum by the total mass, 5m, gives vcm = bT³/(15m) ≈ 0.067 bT³/m, whatever the spring is doing.
C0.100 bT³/m A student who multiplies the average of the initial and final forces by the time picks this: (0 + bT²)/2 × T = bT³/2, giving bT³/(10m). That average gives the area only when F changes linearly with t; F = bt² rises slowly at first, and the impulse is bT³/3.
D0.083 bT³/m A student who treats the system as moving with its most massive block, ignoring the lighter one, picks this: (bT³/3)/(4m) = bT³/(12m) ≈ 0.083 bT³/m. The center-of-mass velocity is the total momentum divided by the total mass, 5m, which includes the block of mass m.
Working System: both blocks and the spring. The spring forces are internal, so only F changes the system's momentum: Δp = J = ∫₀ᵀ bt² dt = bT³/3. The system starts at rest and its total mass is 5m, so vcm = (Σp)/(Σm) = (bT³/3)/(5m) = bT³/(15m) ≈ 0.067 bT³/m. Distractors: peak force times duration, bT² × T = bT³, gives bT³/(5m) = 0.200 bT³/m; average of initial and final force, (0 + bT²)/2 × T = bT³/2, gives bT³/(10m) = 0.100 bT³/m; the correct impulse divided by the 4m block's mass alone gives bT³/(12m) ≈ 0.083 bT³/m. Key and distractors checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
