7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
Two skaters, one more massive than the other, stand at rest on level ice, where friction is negligible. They push off each other and glide apart in opposite directions. What happens to the velocity of the center of mass of the two-skater system?
Answer and reasoning
AIt remains zero before, during and after the skaters push.Correct The skaters' pushes on each other are internal forces, and the external forces (gravity and the normal force from the ice) add to zero. With no net external force the center-of-mass velocity is constant, and it was zero at the start.
BIt points toward the lighter skater, who glides away faster. A student who averages the skaters' velocities without their masses picks this. The lighter skater moves faster, but the momenta are equal and opposite, so the mass-weighted average is zero.
CIt is nonzero during the push, then returns to zero. A student who thinks internal forces change the system's motion while they act picks this. The two pushes are a third-law pair that cancel in the total at every instant, so vcm is zero during the push as well.
DIt increases, as both skaters gain kinetic energy. A student who treats momentum as a form of energy picks this. The skaters do gain kinetic energy, from the work their muscles do, but their momenta are equal and opposite, so the total momentum, and vcm, stay zero.
Two identical carts, each of mass m, move toward each other along a straight track, each with speed v. Which statement about the total momentum of the two-cart system is correct?
Answer and reasoning
AIt is 2mv, as the momenta of the two carts add together. A student who adds the magnitudes without their directions picks this. The momenta do add, but as vectors: +mv + (−mv) = 0.
BIt cannot be zero while both of the carts in the system are moving. A student who thinks a moving system must have momentum picks this. The carts move in opposite directions, so their momenta cancel; a system can have zero total momentum while every part of it moves.
CIt is zero, although each cart has momentum of magnitude mv.Correct Each cart has momentum of magnitude mv, but the momenta point in opposite directions: +mv and −mv. The total momentum is their vector sum, which is zero.
DIt is mv², the sum of (1/2)mv² for each of the two carts. A student who uses the kinetic-energy expression for momentum picks this. The sum of (1/2)mv² for the carts is their total kinetic energy, in J; momentum is mv, in kg·m/s, and has direction.
A heavily loaded truck collides with a small car that is parked on a road. Which statement correctly compares the impulse exerted on the car by the truck with the impulse exerted on the truck by the car during the collision?
Answer and reasoning
AThe truck's impulse on the car is larger, as the truck has more mass. A student who thinks the more massive object exerts the larger force picks this. The forces are a third-law pair, equal at every instant, so the impulses are equal in magnitude. The car's motion changes more because its mass is smaller.
BThey are equal in magnitude and opposite in direction.Correct The truck and the car exert forces on each other that are equal in magnitude and opposite in direction at every instant, and both forces last exactly as long as the contact. So the two impulses are equal and opposite, whatever the masses.
CThe car exerts zero impulse on the truck, as the car was at rest. A student who thinks only the moving, 'active' object exerts a force picks this. While the truck pushes on the car, the car pushes back on the truck just as hard, which is why the truck slows down and its front is dented.
DThey are equal and both in the truck's direction of motion. A student who thinks an impulse is in the direction of motion picks this. The car's impulse on the truck points backward, which is why the truck slows; only the truck's impulse on the car points forward.
On a level track with negligible friction, cart 1, of mass m, moves with velocity +2v toward cart 2, of mass 4m, which moves with velocity −v. The carts collide and stick together. Which expression gives the velocity of the carts immediately after the collision?
Answer and reasoning
A−(2/5)vCorrect Momentum is conserved in the collision: m(+2v) + (4m)(−v) = −2mv before. The stuck-together carts have total mass 5m, so vf = −2mv/(5m) = −(2/5)v, in cart 2's original direction.
B+(6/5)v A student who adds the two momenta as positive amounts, 2mv + 4mv = 6mv, and divides by 5m picks this. Cart 2 moves in the negative direction, so its momentum, −4mv, must be subtracted.
C+(1/2)v A student who averages the two velocities, (2v − v)/2, picks this. That ignores the masses: cart 2 has four times the mass of cart 1, so its velocity carries four times the weight.
D−(1/2)v A student who divides the total momentum, −2mv, by the mass of cart 2 alone, 4m, picks this. After the collision both carts move together, so the momentum is shared by the total mass, 5m.
Working Momentum before = m(+2v) + (4m)(−v) = 2mv − 4mv = −2mv. After, the carts move together: (m + 4m)vf = −2mv, so vf = −(2/5)v.
A ball of clay is thrown horizontally at a wall, hits it and sticks, so the ball's momentum decreases to zero. Which statement correctly describes momentum in this interaction?
Answer and reasoning
AThe ball's momentum was converted into internal energy of the clay. A student who treats momentum as a form of energy picks this. The ball's kinetic energy does become internal energy, but momentum is a different quantity; it cannot be converted into energy, only passed to other objects.
BTotal momentum decreased, since the ball stuck to the wall instead of bouncing. A student who thinks momentum is lost when objects stick together picks this. Momentum is conserved in all interactions, including those in which objects stick or deform; here the ball's momentum passes to the wall and Earth.
CTotal momentum was not conserved, since the wall exerted a force on the ball. A student who thinks any force spoils conservation picks this. Every interaction involves forces; the momentum the wall's force removes from the ball is exactly the momentum the ball's force gives to the wall and Earth.
DThe momentum the ball lost was transferred to the wall and Earth.Correct Momentum is conserved in every interaction. The ball pushes on the wall while the wall stops the ball, so the wall and Earth, to which it is fixed, gain exactly the momentum the ball loses. Their huge mass makes the resulting change in velocity far too small to see.
Two carts collide on a level track. Friction and air resistance are negligible. During the collision, Earth exerts a gravitational force and the track exerts a normal force on each cart. Is the total momentum of the two-cart system constant during the collision?
Answer and reasoning
AYes, because Earth's and the track's forces are internal. A student who counts every force on a cart as internal to the carts' system picks this. Earth and the track are not part of the system, so their forces are external; the momentum is constant because those forces add to zero.
BNo, because the carts exert large forces on each other. A student who thinks collision forces change the system's momentum picks this. The carts' forces on each other are internal, a third-law pair, so one cart gains exactly the momentum the other loses.
CYes, because the external forces on the system add up to zero.Correct Earth and the track are outside the system, so their forces are external, but on the level track the normal forces balance the gravitational forces. The net external force is zero, so the total momentum is constant, even though external forces act.
DNo, because each cart's velocity changes in the collision. A student who thinks conservation means each object keeps its own momentum picks this. The carts' individual momenta change, but in equal and opposite amounts, so the total stays the same.
A cart of mass m moves with speed v along a level track toward a wall. It hits the wall and rebounds along the same line with speed v/3. Taking the cart's initial direction of motion as positive and the cart as the system, which expression gives the momentum transferred to the system from its surroundings during the collision?
Answer and reasoning
A+(2/3)mv A student who treats momentum as an amount without direction sees the cart's momentum fall from mv to (1/3)mv and takes the (2/3)mv it lost as the momentum transferred. The cart's velocity reverses, so its final momentum is −(1/3)mv, and the momentum transferred is −(1/3)mv − mv = −(4/3)mv.
B−(4/3)mvCorrect The momentum transferred to the cart equals its change in momentum, final minus initial: m(−v/3) − m(+v) = −(4/3)mv. The wall first removes the cart's forward momentum, mv, and then gives it backward momentum, (1/3)mv.
C−(1/3)mv A student who takes the momentum transferred to be the cart's final momentum picks this. The transfer is the change in momentum, which includes removing the cart's initial momentum, +mv.
D+(4/3)mv A student who thinks the impulse is in the direction the cart was moving picks this. The wall pushes the cart back, in the negative direction, so the momentum transferred to the cart is negative.
Working p0 = +mv; p = m(−v/3) = −(1/3)mv. Momentum transferred = Δp = p − p0 = −(1/3)mv − mv = −(4/3)mv.
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 (of objects) Fix
System (of objects)
A chosen collection of objects that is analyzed together. When the internal details do not matter, the collection can be described as one object with one mass (the total mass) and one velocity, the velocity of its center of mass.
Velocity of the center of mass, v⃗cm
The velocity of a system treated as a single object: the total momentum divided by the total mass, v⃗cm = (Σ mi v⃗i)/(Σ mi). It is a mass-weighted average of the objects' velocities, so it lies closer to the velocity of the more massive objects. Unit: m/s.
Net external force
The vector sum of the forces exerted on a system by objects outside the system. Unit: N. When it is zero, the velocity of the system's center of mass is constant, even if the objects in the system push, pull or collide with each other.
Students often think The velocity of a system's center of mass (and the common velocity of objects that stick together) is the simple average of the objects' velocities, whatever their masses. In fact Only when the objects have equal masses. The center-of-mass velocity is a mass-weighted average, v⃗cm = (Σ mi v⃗i)/(Σ mi), so it lies closer to the velocity of the more massive object.
Students often think The total momentum of a system is divided by the mass of just one object (usually the moving one or the most massive one) to find the system's velocity. In fact The total mass of all the objects in the system: v⃗cm = (Σ p⃗i)/(Σ mi).
4.3.A.2 Total momentum of a system Fix
Total momentum of a system
The vector sum of the momenta of all the objects in the system, Σ p⃗i = Σ mi v⃗i. Along a line, momenta in opposite directions have opposite signs. It also equals (Σ mi)v⃗cm. Unit: kg·m/s.
Students often think Momentum and impulse are amounts without direction, so momenta (and impulses) are combined as positive numbers whatever their directions. In fact No. Momentum is a vector. Along a line, choose a positive direction; a momentum in the opposite direction is negative and reduces the total.
Students often think Momentum is calculated with the kinetic-energy expression (1/2)mv², since both describe 'how much motion' an object has. In fact No. Momentum is p⃗ = mv⃗, a vector in kg·m/s; kinetic energy is K = (1/2)mv², a scalar in J. They are different quantities with different units.
4.3.A.3 Internal and external forces Fix
Internal and external forces
An internal force is exerted by one object in the system on another object in the system; an external force is exerted on an object in the system by an object outside it. Whether a force is internal depends on the system chosen.
Equal and opposite impulses
In an interaction, the impulse exerted on object 2 by object 1 has the same magnitude as, and the opposite direction to, the impulse exerted on object 1 by object 2, because the third-law forces are equal and opposite and act for the same time interval. Impulse unit: N·s (= kg·m/s).
Choosing a system
Deciding which objects to include in the system. Including both objects of every interaction that matters makes those forces internal, so a system can be chosen whose total momentum is constant (for example, a falling ball together with Earth).
Impulse on a system
The product of the average net external force on a system and the time interval over which it acts, J⃗ = F⃗avg Δt, equal to the area under a graph of net external force against time. It equals the change in the system's total momentum, J⃗ = Δp⃗. Unit: N·s.
Students often think A system whose objects are moving must have nonzero total momentum, so a push-off or explosion from rest creates momentum. In fact No. Total momentum is a vector sum. Objects moving in opposite directions can have momenta that cancel, so a system can have zero total momentum while its parts move, as after a push-off from rest.
Students often think A negative momentum (a bar below the axis) is smaller than a positive one, so the object with the positive value has more momentum. In fact Not in magnitude. Along a line, the sign of a momentum gives only its direction: −4 kg·m/s and +4 kg·m/s have the same magnitude.
4.3.A.4 Collision Fix
Collision
An interaction in which the forces the objects exert on each other are much larger than the net external force during the interaction, so the total momentum of the objects just before the collision equals their total momentum just after it.
Explosion
An interaction in which forces internal to a system push its objects apart, such as a compressed spring released between two carts or a person throwing a ball. A system at rest before an explosion has zero total momentum after it.
Students often think Equal and opposite forces make two objects that push apart (such as a thrower and the object thrown) move off with equal speeds. In fact No. Equal and opposite impulses give momenta of equal magnitude, so m1v1 = m2v2, and the less massive object moves faster.
Students often think After an explosion or a throw, the momentum of one piece is divided by the total mass of the system, as if the pieces still moved together. In fact No. After the objects separate, each moves with its own velocity, so each piece's momentum is divided by that piece's own mass. The total mass is used only for objects that move together.
4.3.B.1 Conservation of momentum Fix
Conservation of momentum
In every interaction, the momentum gained by one object is lost by the other objects it interacts with, so momentum is never created or destroyed; it can only be transferred between objects.
Students often think Momentum is a form of energy: a system gains momentum when it gains kinetic energy, and momentum can be converted into heat or internal energy. In fact No. Momentum (kg·m/s, a vector) and energy (J, a scalar) are different quantities. Kinetic energy can be transformed into internal energy; momentum cannot be transformed into anything, only transferred between objects.
Students often think Momentum is conserved only in collisions in which the objects bounce apart; when objects stick together or deform, momentum is lost. In fact No. Momentum is conserved in all interactions: collisions in which the objects bounce, stick together or crumple, and explosions.
4.3.B.2 Condition for constant total momentum Fix
Condition for constant total momentum
The total momentum of a selected system is constant if, and only if, the net external force on the system is zero. External forces may act on the system as long as they add to zero.
Students often think Momentum is conserved only when no forces of any kind act; where forces act momentum is not conserved, and a constant total momentum means that no forces were exerted. In fact No. The total momentum of a system is constant when the net external force on it is zero. Internal forces, including the large forces of a collision, do not change the total.
Students often think The forces between objects in a system, such as collision forces, change the system's total momentum. In fact No. They are internal forces, a third-law pair that gives equal and opposite impulses. One object gains exactly the momentum the other loses, so the total is unchanged.
4.3.B.3 Momentum transfer between a system and its environment Fix
Momentum transfer between a system and its environment
When the net external force on a system is not zero, momentum passes between the system and the objects outside it. The momentum the system gains or loses equals the impulse of the net external force, and the environment changes by the same amount in the opposite direction.
Students often think Friction destroys momentum: when a system slows down, the momentum it loses ceases to exist instead of going anywhere. In fact No. Every interaction conserves momentum. Friction is exerted by the ground, part of Earth, and the momentum the system loses is transferred to Earth.
Students often think The impulse exerted on an object, and its change in momentum, are in the direction the object is moving. In fact Not necessarily. The impulse is in the direction of the net force. A cart that rebounds from a wall receives an impulse directed away from the wall, opposite to its initial motion.
12 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 12
Cart A (mass 2.0 kg) and cart B (mass 1.0 kg) move toward each other along a straight, level track. The graph shows the position x of each cart as a function of time t until they meet at t = 2.0 s. What is the velocity of the center of mass of the two-cart system during this time?
Answer and reasoning
A+1.0 m/s A student who averages the two velocities, (+3.0 − 1.0) m/s ÷ 2, picks this. That ignores the masses: cart A has twice the mass of cart B, so the center of mass moves at a velocity closer to A's.
B+2.3 m/s A student who adds the momenta as positive amounts, (6.0 + 1.0) kg·m/s ÷ 3.0 kg, picks this. Cart B moves in the negative direction (its line slopes down), so its momentum, −1.0 kg·m/s, reduces the total.
C+1.7 m/sCorrect Each cart's velocity is the slope of its line: A, +3.0 m/s; B, −1.0 m/s. The total momentum is (2.0 kg)(+3.0 m/s) + (1.0 kg)(−1.0 m/s) = +5.0 kg·m/s, and dividing by the total mass, 3.0 kg, gives vcm = +1.7 m/s.
D+2.5 m/s A student who divides the total momentum, +5.0 kg·m/s, by the mass of cart A alone picks this. The center of mass belongs to the whole system, so the total momentum is divided by the total mass, 3.0 kg.
Working Slopes: vA = (6.0 − 0) m ÷ 2.0 s = +3.0 m/s; vB = (6.0 − 8.0) m ÷ 2.0 s = −1.0 m/s. Σp = (2.0 kg)(+3.0 m/s) + (1.0 kg)(−1.0 m/s) = +5.0 kg·m/s. vcm = Σp/Σm = 5.0 kg·m/s ÷ 3.0 kg = +1.7 m/s.
Cart P, of mass m, moves along a level track with velocity v⃗. Cart Q, of mass 2m, is at rest on the track. Cart Q is replaced by a cart of mass 4m, also at rest, while cart P's velocity stays the same. By what factor does the velocity of the center of mass of the two-cart system change?
Answer and reasoning
A1.00 A student who takes vcm to be the average of the two velocities, (v + 0)/2, picks this, because that average does not depend on the masses. Adding mass at rest to the system lowers the center-of-mass velocity.
B0.60Correct The only momentum is P's, mv, which does not change. Before, vcm = mv/(3m) = v/3; after, vcm = mv/(5m) = v/5. The factor is (1/5) ÷ (1/3) = 0.60: the total mass, not Q's mass alone, is in the denominator.
C0.50 A student who treats vcm as inversely proportional to Q's mass alone picks this: Q's mass doubles, so vcm halves. The denominator is the total mass, which grows from 3m to 5m, not from 2m to 4m.
D1.20 A student who weights each velocity by the other cart's mass, vcm = (mQ v)/(m + mQ), gets 2v/3 and then 4v/5, and picks this. Each velocity is weighted by its own mass; Q is at rest, so only P's momentum, mv, is in the numerator.
Cart A (mass 2.0 kg) and cart B (mass 1.0 kg) collide on a level track with negligible friction. The graph shows the velocity v of each cart as a function of time t; the collision lasts from t = 0.40 s to t = 0.60 s. Which statement correctly describes the velocity of the center of mass of the two-cart system?
Answer and reasoning
AIt rises from 0 before the collision to +2 m/s after it. A student who averages the two velocities without the masses, (3 − 3)/2 and (−1 + 5)/2, picks this. Cart A has twice the mass of cart B, so each velocity must be weighted by its cart's mass.
BIt falls from +3 m/s before the collision to −1 m/s after. A student who thinks the center of mass moves with the more massive cart, A, picks this. The center of mass depends on both carts: B's momentum counts too, and the total, +3 kg·m/s, does not change.
CIt is +1.5 m/s before and after the collision. A student who divides the total momentum, +3 kg·m/s, by the mass of cart A alone, 2.0 kg, picks this. The center-of-mass velocity is the total momentum divided by the total mass, 3.0 kg.
DIt is +1 m/s, both before and after the collision.Correct Before: [(2.0)(+3) + (1.0)(−3)] kg·m/s ÷ 3.0 kg = +1 m/s. After: [(2.0)(−1) + (1.0)(+5)] kg·m/s ÷ 3.0 kg = +1 m/s. The carts exert forces only on each other (friction is negligible, and gravity and the normal force balance), so there is no net external force and vcm cannot change.
Working Before: Σp = (2.0 kg)(+3 m/s) + (1.0 kg)(−3 m/s) = +3 kg·m/s; vcm = 3 kg·m/s ÷ 3.0 kg = +1 m/s. After: Σp = (2.0 kg)(−1 m/s) + (1.0 kg)(+5 m/s) = +3 kg·m/s; vcm = +1 m/s. (During the collision the slopes are −20 m/s² and +40 m/s², so the forces are −40 N and +40 N: equal and opposite, and Σp stays +3 kg·m/s throughout.)
Two carts move along a straight track. Cart 1, of mass 3.0 kg, moves at +2.0 m/s, and cart 2, of mass 2.0 kg, moves at −1.0 m/s. What is the total momentum of the two-cart system?
Answer and reasoning
A+8.0 kg·m/s A student who adds the two momenta as positive amounts, 6.0 + 2.0, picks this. Cart 2 moves in the negative direction, so its momentum is −2.0 kg·m/s and reduces the total.
B+4.0 kg·m/sCorrect The total momentum is the sum of the carts' momenta, with signs: (3.0 kg)(+2.0 m/s) + (2.0 kg)(−1.0 m/s) = +6.0 kg·m/s − 2.0 kg·m/s = +4.0 kg·m/s.
C+2.5 kg·m/s A student who multiplies the total mass, 5.0 kg, by the simple average of the velocities, (2.0 − 1.0)/2 = 0.50 m/s, picks this. That average ignores the masses; the momenta must be added cart by cart.
D+7.0 kg·m/s A student who adds the kinetic energies, (1/2)(3.0)(2.0)² + (1/2)(2.0)(1.0)² = 6.0 + 1.0, picks this. That total, 7.0 J, is energy, not momentum; momentum is mv, with the sign of v.
Cart A (mass 1.0 kg) and cart B (mass 2.0 kg) are held at rest on a level track with negligible friction, with a compressed spring between them. When released, the carts move apart. The bar chart shows the momentum p of cart A, cart B and the two-cart system before and after the release. Which claim does the bar chart support?
Answer and reasoning
AB's momentum changed by as much as A's did, in the opposite direction.Correct A's momentum changed from 0 to −4 kg·m/s and B's from 0 to +4 kg·m/s: equal changes in opposite directions. The spring's forces on the carts are internal, so the system bar stays at zero.
BThe spring gave the system momentum, since both carts moved off. A student who thinks a system in motion must have momentum picks this. The chart shows the system bar at zero after the release: the carts' momenta, −4 and +4 kg·m/s, cancel.
CB ended with more momentum than A, since B's bar is above the axis. A student who reads a negative momentum as a smaller one picks this. The bars have the same length, 4 kg·m/s; the signs show only that the carts move in opposite directions.
DThe carts moved apart with equal speeds, since their bars are equal. A student who thinks equal and opposite forces give equal speeds picks this. Equal momenta of 4 kg·m/s give A a speed of 4.0 m/s and B, with twice the mass, a speed of 2.0 m/s.
Cart 1, of mass 1.0 kg, moves in the positive direction along a level track with negligible friction and collides with cart 2, of mass 2.0 kg, which is moving in the negative direction at 1.0 m/s. The graph shows the force exerted on cart 1 by cart 2 as a function of time t during the collision. What is the momentum of cart 2 just after the collision?
Answer and reasoning
A−5.0 kg·m/s A student who gives cart 2 the impulse read from the graph, sign included, picks this: −2.0 kg·m/s + (−3.0 N·s) = −5.0 kg·m/s. That is the impulse on cart 1. The impulse on cart 2 is equal in magnitude and opposite in direction, +3.0 N·s: cart 1 pushes cart 2 in the positive direction.
B+4.0 kg·m/s A student who multiplies the peak force, 60 N, by the whole contact time, 0.10 s, gives cart 2 an impulse of +6.0 N·s and picks this. The size of the force rises and then falls, so the area is a triangle: (1/2)(0.10 s)(60 N) = 3.0 N·s, and cart 2's final momentum is −2.0 kg·m/s + 3.0 N·s = +1.0 kg·m/s.
C−0.5 kg·m/s A student who judges each impulse by the change in velocity it produces halves the 3.0 N·s for cart 2, which has twice cart 1's mass, and gets −2.0 kg·m/s + 1.5 N·s = −0.5 kg·m/s. The impulses are equal in magnitude; the larger mass only makes cart 2's change in velocity smaller.
D+1.0 kg·m/sCorrect The impulse on cart 1 is the area under the graph: (1/2)(0.10 s)(−60 N) = −3.0 N·s. The carts exert equal and opposite forces on each other at every instant, so the impulse on cart 2 is +3.0 N·s, whatever the carts' masses. Cart 2's momentum changes from (2.0 kg)(−1.0 m/s) = −2.0 kg·m/s to −2.0 kg·m/s + 3.0 N·s = +1.0 kg·m/s.
Working Impulse on cart 1 by cart 2 = area = (1/2)(0.10 s)(−60 N) = −3.0 N·s. By Newton's third law the impulse on cart 2 by cart 1 is equal and opposite: +3.0 N·s. Cart 2's initial momentum is (2.0 kg)(−1.0 m/s) = −2.0 kg·m/s, so its final momentum is −2.0 kg·m/s + 3.0 N·s = +1.0 kg·m/s.
A ball is dropped from rest and falls toward the ground. Air resistance is negligible. For which choice of system is the total momentum constant while the ball falls?
Answer and reasoning
AThe ball alone, since air resistance on it is negligible A student who thinks each object's momentum is constant when air resistance is negligible picks this. Earth's gravitational force is external to the ball, so the ball's downward momentum increases throughout the fall.
BEarth alone, as a small ball cannot change Earth's motion A student who thinks a small object cannot change the momentum of a massive one picks this. The ball pulls Earth upward as hard as Earth pulls the ball down, so Earth's momentum changes by the same amount as the ball's; its velocity change is tiny only because its mass is enormous.
CNo system, since a gravitational force acts during the fall A student who thinks momentum is conserved only when no forces act picks this. A force changes a system's total momentum only if it is external; choosing the ball–Earth system makes both gravitational forces internal.
DThe ball–Earth system, as the gravitational forces are internal to itCorrect Earth pulls the ball down and the ball pulls Earth up with a force of equal magnitude. With both objects in the system these forces are internal, and no significant external force acts, so the ball gains downward momentum exactly as fast as Earth gains upward momentum.
Two carts joined together form a system that moves along a level track with negligible friction; the system's total momentum is +3.0 kg·m/s. Starting at t = 0, a student's hand pushes on the system in the negative direction. The graph shows the force F exerted on the system by the hand as a function of time t. What is the total momentum of the system at t = 0.40 s?
Answer and reasoning
A+1.0 kg·m/s A student who takes the impulse as the peak force times the whole time, (−5.0 N)(0.40 s) = −2.0 N·s, picks this. The force builds up and dies away, so the area is a trapezoid of −1.5 N·s, not a rectangle.
B+3.0 kg·m/s A student who thinks a system's total momentum cannot change picks this. The hand is outside the system, so its push is a net external force; the system's momentum changes by the impulse, −1.5 N·s.
C+1.5 kg·m/sCorrect The hand is the only object outside the system exerting an unbalanced force, so the change in the system's momentum equals the hand's impulse, the area under the graph: two triangles of (1/2)(0.10 s)(−5.0 N) = −0.25 N·s and a rectangle of (0.20 s)(−5.0 N) = −1.0 N·s, a total of −1.5 N·s. So p = +3.0 − 1.5 = +1.5 kg·m/s.
D+4.5 kg·m/s A student who adds the size of the impulse without its direction picks this. The hand pushes in the negative direction, so its impulse, −1.5 N·s, reduces the system's momentum.
A student of mass M stands at rest on level ice, where friction is negligible, and throws a ball of mass m horizontally with speed v relative to the ice. Which expression gives the student's velocity immediately after the throw?
Answer and reasoning
Av, in the direction opposite to the ball A student who thinks the equal and opposite forces give equal speeds picks this. The impulses are equal in magnitude, so the momenta are equal in magnitude; the student's larger mass gives a smaller speed, mv/M.
Bmv/M, in the direction opposite to the ballCorrect The student and the ball start at rest, so their total momentum is zero, and the forces between them during the throw are internal. After the throw mv + Mvs = 0, so vs = −mv/M: the student slides backward at mv/M, more slowly than the ball because M > m.
CZero, since the student remains at rest A student who thinks only the thrower exerts a force picks this. While the student pushes the ball forward, the ball pushes the student backward with a force of equal magnitude, so the student recoils.
Dmv/(M + m), in the direction opposite to the ball A student who divides the momentum by the total mass, as if the student and the ball moved together, picks this. After the throw they move separately, so the student's momentum, of magnitude mv, is divided by the student's own mass, M.
Working Momentum of student + ball before = 0. Taking the ball's direction as positive, after the throw mv + Mvs = 0, so vs = −mv/M: speed mv/M, opposite to the ball.
On a level air table, where friction is negligible, puck 1 of mass m and puck 2 of mass 2m slide toward the collision point shown in the diagram, each with speed v. The pucks collide and stick together. How does the y-component of the combined pucks' momentum immediately after the collision compare with its x-component?
Answer and reasoning
AThe two components have the same magnitude. A student who averages the two velocities without the masses picks this, expecting the pucks to move off at 45°. Puck 2 has twice the mass, so it brings twice as much momentum along y as puck 1 brings along x.
BThe x-component is zero; the y-component is not. A student who thinks the combined object moves with the more massive puck picks this. Puck 1's x-momentum, mv, does not disappear: the combined pucks keep it and move off between the +x and +y directions.
CThe y-component is twice the x-component.Correct Momentum is conserved in each direction separately. Only puck 1 has x-momentum, mv, and only puck 2 has y-momentum, (2m)v = 2mv. After the collision the combined pucks keep both, so the y-component is twice the x-component.
DThe x-component is double the y-component. A student who weights each puck's velocity by the other puck's mass picks this. Each momentum is the puck's own mass times its own velocity: mv along x and 2mv along y.
Working Before: px = m·v = mv (puck 1 only); py = 2m·v = 2mv (puck 2 only). Momentum is conserved in each direction, so after the collision px = mv and py = 2mv: the y-component is twice the x-component. (No value of the final speed or angle is needed.)
A spring launcher of mass M rests on a level track with negligible friction. It fires a ball of mass m horizontally, and the launcher recoils. In a second trial, starting again from rest, the launcher fires a ball of mass 2m, which leaves at half the speed of the first ball. Both speeds are measured relative to the track. By what factor does the launcher's recoil speed change?
Answer and reasoning
A1.00Correct The launcher and ball start at rest, so after the launch their momenta are equal and opposite: MvL = mb vb. The ball's momentum is mv in the first trial and (2m)(v/2) = mv in the second, so the launcher's recoil speed does not change.
B0.50 A student who thinks the launcher and ball move off with equal speeds picks this: the ball's speed halves, so the recoil speed halves. It is the momenta, not the speeds, that are equal in magnitude, and the ball's momentum is unchanged.
C0.71 A student who thinks the launcher and ball receive equal kinetic energies picks this: the ball's kinetic energy halves, so the recoil speed changes by 1/√2. Conservation of momentum, not an equal share of energy, sets the recoil.
D2.00 A student who tracks only the ball's mass, which doubles, picks this. The ball's speed halves at the same time, so the product mb vb, and with it the recoil momentum, is unchanged.
Working From rest, total momentum is zero: MvL = mb vb. Trial 1: vL = mv/M. Trial 2: vL = (2m)(v/2)/M = mv/M. Factor = 1.00.
Carts A and B move in the same direction along a straight track. The graph shows the momentum p of cart A, of cart B and of the two-cart system as functions of time t. Which claim is supported by the graph?
Answer and reasoning
AA net external force was exerted on the system between 2 s and 3 s.Correct The system's momentum is constant (6 kg·m/s) until 2 s and then falls to 3 kg·m/s at 3 s. A system's total momentum changes only when a net external force transfers momentum between the system and its surroundings, so one acted between 2 s and 3 s.
BThe forces between the carts at about 1 s changed the system's momentum. A student who thinks collision forces change a system's momentum picks this. At about 1 s cart A loses 3 kg·m/s and cart B gains 3 kg·m/s, and the system line stays flat at 6 kg·m/s.
CNo forces were exerted on either cart between 0 and 2 s. A student who reads a constant total momentum as 'no forces at all' picks this. At about 1 s both carts' momenta change sharply, so the carts exerted large forces on each other; those forces are internal and cancel in the total.
DThe momentum lost between 2 s and 3 s was destroyed, not transferred. A student who thinks friction destroys momentum picks this. Momentum is conserved in all interactions: the 3 kg·m/s the system lost went to the objects outside it that exerted the external force.
Working System line: constant at 6 kg·m/s from 0 to 2 s (at about 1 s, A drops from 5 to 2 kg·m/s while B rises from 1 to 4 kg·m/s, so the internal collision leaves the total unchanged), then falling to 3 kg·m/s at 3 s. A change in total momentum (−3 kg·m/s) requires a net external impulse, so a net external force acted between 2 s and 3 s.
Compiled from the AP Physics 1 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
