8 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 8
A hockey puck slides east across level ice at constant velocity, so its momentum stays constant. Friction and air resistance are negligible. Which statement about the forces exerted on the puck is correct?
Answer and reasoning
AThe forces on it add to zero, since its momentum does not change.Correct The net external force equals the rate of change of momentum, F⃗net = Δp⃗/Δt. The puck's momentum is constant, so Δp⃗ = 0 and the net force is zero. Forces are still exerted on the puck (its weight and the normal force from the ice), but they add to zero.
BA net force directed east is exerted on it, keeping it moving east. A student who believes that motion needs a force in its direction picks this. A net force would change the puck's momentum at a rate equal to that force; because the momentum is constant, the net force is zero, and the puck keeps moving east with no push.
CNo forces are exerted on the puck, as its velocity is constant. A student who thinks zero net force means no forces picks this. Earth exerts a downward gravitational force on the puck and the ice exerts an upward normal force; they balance, so the net force is zero even though two forces act.
DIts momentum gives no clue to its net force, as momentum is conserved. A student who thinks momentum stays constant whatever forces act, and so says nothing about them, picks this. Momentum is constant only when the net external force is zero, so constant momentum is exactly the evidence that the net force is zero.
A cart is at rest on a level track with negligible friction. A student pushes it with a constant horizontal force for a time interval Δt. The cart is returned to rest, and the student pushes it again with the same force for a time interval 2Δt. How has the impulse exerted on the cart by the student changed from the first push to the second?
Answer and reasoning
AUnchanged A student who thinks impulse depends only on the size of the force picks this. The force is the same, but it is exerted for twice as long, and J = Favg Δt doubles with Δt.
BHalved A student who has learned that a longer time interval goes with a smaller effect, as when a cushion lengthens a collision, picks this. A cushion lowers the average force for a fixed impulse; here the force is fixed, so the longer push delivers the larger impulse.
CDoubledCorrect Impulse is the average force multiplied by the time interval, J = Favg Δt. With the same force exerted for twice as long, the impulse is twice as great.
DQuadrupled A student who treats impulse as work picks this: from rest under a constant force, twice the time gives four times the distance, so four times the work. Impulse is force × time, not force × distance, so it only doubles.
Working J = Favg Δt. The force is the same and the time interval is doubled, so J2/J1 = (F)(2Δt)/[(F)(Δt)] = 2: the impulse is doubled.
A cart moves in the +x direction along a level track during the whole time shown. The graph shows the net force F exerted on the cart as a function of time t, with +x as the positive direction. Which statement describes the impulse delivered to the cart from t = 0 to t = 2.0 s?
Answer and reasoning
AIt is in the +x direction, the cart's direction of motion. A student who thinks impulse points in the direction of motion picks this. The direction of an impulse is that of the net force, here −x; an impulse opposite to the velocity slows the cart without reversing it.
BIt has no direction, since it is an area on a graph. A student who thinks an area cannot have a direction picks this. Impulse is a vector; on a force–time graph for one axis, area below the time axis is impulse in the negative direction, here −6.0 N·s, in −x.
CIt is in the −x direction, opposite to the cart's velocity.Correct Impulse has the direction of the net force. The net force is −3.0 N throughout, so the impulse, (−3.0 N)(2.0 s) = −6.0 N·s, is in the −x direction. The cart keeps moving in +x but slows down, so its momentum decreases.
DIt is zero, since the force stays constant from t = 0 to 2.0 s. A student who takes the impulse from the slope of the graph picks this: a horizontal line has zero slope. The impulse is the area between the graph and the time axis, (−3.0 N)(2.0 s) = −6.0 N·s, which is not zero.
Working J = area between the graph and the time axis = (−3.0 N)(2.0 s) = −6.0 N·s. The negative sign means the impulse is in the −x direction, the direction of the net force, even though the cart keeps moving in +x (it slows down).
Each graph shows the net force F exerted on a different cart, P, Q or R, as a function of time t. Which ranking of the magnitudes of the impulses delivered to the carts is correct?
Answer and reasoning
AP > R > QCorrect The impulse is the area under each graph. P: (4 N)(3 s) = 12 N·s. R: ½(4 s)(5 N) = 10 N·s. Q: ½(2 s)(8 N) = 8 N·s. So P > R > Q, even though Q has the largest peak force and R lasts longest.
BQ > R > P A student who ranks by the height of each graph picks this (peaks 8 N, 5 N, 4 N). The impulse is the area, which depends on how long each force acts as well as on how large it is.
CR > Q > P A student who multiplies each peak force by the whole duration picks this (R 20, Q 16, P 12). That is right only for the rectangle P; each triangle's area is half of peak × duration, which puts Q last.
DR > P > Q A student who ranks by how long each force lasts picks this (4 s, 3 s, 2 s). The duration is only one factor; R's smaller forces give it an area of 10 N·s, less than P's 12 N·s.
Working Impulse = area under each graph. P: rectangle (4 N)(3 s) = 12 N·s. Q: triangle ½(2 s)(8 N) = 8 N·s. R: triangle ½(4 s)(5 N) = 10 N·s. Ranking P > R > Q. (Peaks: Q > R > P. Peak × duration: R 20 > Q 16 > P 12. Durations: R 4 s > P 3 s > Q 2 s.)
The graph shows the momentum p of a cart moving along a straight track as a function of time t. What is the magnitude of the net force exerted on the cart?
Answer and reasoning
A4.0 N A student who reads the height of the graph picks this, taking the final momentum, 4.0 kg·m/s, as the force. The height is the momentum at one instant; the net force is how fast the momentum changes, the slope.
B1.5 NCorrect The net force is the slope of the momentum–time graph: (4.0 − 1.0) kg·m/s ÷ 2.0 s = 1.5 N. The graph is a straight line, so the net force is constant.
C5.0 N A student who uses the area under the graph picks this: ½(1.0 + 4.0)(2.0) = 5.0. The area under a momentum–time graph has units kg·m, not N; the net force is the slope.
D2.0 N A student who divides the final momentum by the time, 4.0 ÷ 2.0, picks this. The cart already had 1.0 kg·m/s at t = 0, so the change in momentum is 3.0 kg·m/s, not 4.0 kg·m/s.
Working Fnet = slope of the p–t graph = Δp/Δt = (4.0 − 1.0) kg·m/s ÷ (2.0 − 0) s = 1.5 N.
A baseball of mass m moving horizontally with speed v is struck by a bat. It leaves the bat moving horizontally with speed 3v/2 in the opposite direction. Taking the ball's initial direction of motion as positive, what is the change in the ball's momentum?
Answer and reasoning
A+(5/2)mv A student who subtracts final from initial, p0 − p, picks this: mv − (−(3/2)mv) = +(5/2)mv. The magnitude is right, but Δp = p − p0, and the change points in the negative direction, the way the bat pushes the ball.
B−(5/2)mvCorrect Δp = p − p0 = (−(3/2)mv) − (+mv) = −(5/2)mv. The ball reverses direction, so the magnitudes of the two momenta add, and the change points in the negative direction, the direction of the bat's force on the ball.
C+(1/2)mv A student who subtracts the speeds, m(3v/2 − v), picks this. The ball reverses direction, so its final momentum is negative: Δp = −(3/2)mv − mv = −(5/2)mv.
D−(3/2)mv A student who takes the final momentum, −(3/2)mv, as the change picks this. The ball arrived with momentum +mv, which must be subtracted: Δp = −(3/2)mv − mv = −(5/2)mv.
Working p0 = +mv; p = −(3/2)mv. Δp = p − p0 = −(3/2)mv − mv = −(5/2)mv. (Errors: p0 − p = +(5/2)mv; subtracting speeds m(3v/2 − v) = +(1/2)mv; final momentum only −(3/2)mv.)
Two identical carts roll at the same speed along a level track with negligible friction. Cart 1 hits a rigid wall and cart 2 hits a block of soft foam; both carts come to rest without bouncing. Which statement correctly compares the foam's effect on cart 2 with the wall's effect on cart 1?
Answer and reasoning
AThe foam exerts a smaller impulse, since it stops the cart more gently. A student who thinks cushioning reduces the change in momentum picks this. Both carts start with the same momentum and end at rest, so the impulses are equal; what the foam reduces is the average force.
BThe foam exerts a larger impulse, as it touches the cart for a longer time. A student who ranks impulses by duration alone picks this. The foam acts for longer but with a smaller force; since both carts' momenta change by the same amount, the impulses are equal.
CThe foam exerts the same impulse over a longer time, so a smaller average force.Correct Both carts go from the same momentum to zero, so their changes in momentum, and the impulses exerted on them, are equal. The foam takes longer to stop cart 2, so by Favg = Δp/Δt it exerts a smaller average force.
DThe two exert equal average forces, as the momentum changes are equal. A student who treats force and change in momentum as the same quantity picks this. The changes in momentum are equal, but the foam spreads it over a longer time, so Favg = Δp/Δt is smaller for the foam.
Working Each cart's momentum changes from mv to 0, so Δp, and the impulse exerted by the wall or the foam, is the same. Favg = Δp/Δt, and the foam takes longer to stop the cart, so its average force is smaller.
A cart of constant mass 0.50 kg moves along a straight, level track. While a constant net force is exerted on it for 0.40 s, the cart's momentum increases from 0.20 kg·m/s to 0.80 kg·m/s. What is the magnitude of the cart's acceleration during this time?
Answer and reasoning
A1.5 m/s² A student who treats momentum as if it were velocity picks this, reporting the rate of change of momentum, 0.60 ÷ 0.40 = 1.5, as the acceleration. That is the net force, 1.5 N; dividing by the 0.50 kg mass gives the acceleration.
B3.0 m/s²Correct The net force is the rate of change of momentum: (0.80 − 0.20) kg·m/s ÷ 0.40 s = 1.5 N. For a constant mass, Fnet = Δp/Δt = mΔv/Δt = ma, so a = 1.5 N ÷ 0.50 kg = 3.0 m/s².
C1.2 m/s² A student who treats the change in momentum as the force picks this, dividing it by the mass, 0.60 ÷ 0.50 = 1.2, as in a = F/m. That is the change in velocity, 1.2 m/s; the acceleration is that change divided by the 0.40 s it took.
D4.0 m/s² A student who uses the final momentum as the change picks this: 0.80 ÷ (0.50 × 0.40) = 4.0. The cart already had 0.20 kg·m/s, so only 0.60 kg·m/s was gained during the 0.40 s.
Working Fnet = Δp/Δt = (0.80 − 0.20) kg·m/s ÷ 0.40 s = 1.5 N. The mass is constant, so Fnet = ma and a = 1.5 N ÷ 0.50 kg = 3.0 m/s². (Equivalently Δv = Δp/m = 0.60 ÷ 0.50 = 1.2 m/s and a = 1.2 ÷ 0.40 = 3.0 m/s².)
In preparation: 0 of 8 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.2.A.1 Net external force Fix
Net external force
The vector sum of the forces exerted on an object or system by objects outside it, in newtons (N). It equals the rate of change of the momentum of the object or system: F⃗net = Δp⃗/Δt.
Rate of change of momentum
The change in momentum divided by the time interval over which it happens, Δp⃗/Δt, in kg·m/s² (equivalent to N). It is zero when the momentum is constant, whatever the speed.
Internal and external forces
Internal forces are exerted by objects inside the chosen system on each other; they come in third-law pairs and do not change the system's total momentum. External forces are exerted on the system by objects outside it; only their sum changes the system's momentum.
Students often think An object that keeps moving with constant momentum must have a net force exerted on it in its direction of motion. In fact No. When the net external force on an object is zero, its momentum does not change. A net force is needed only to change the momentum, not to keep it.
Students often think An object with zero net force exerted on it has no forces exerted on it at all. In fact No. Zero net force means that the forces exerted on the object add to zero. A puck gliding on ice still has a downward gravitational force and an upward normal force exerted on it; they balance.
4.2.A.2 Impulse Fix
Impulse
The product of the average force exerted on a system and the time interval during which it is exerted, J⃗ = F⃗avg Δt. SI unit: newton-second (N·s), equivalent to kg·m/s.
Average force
The constant force that would deliver the same impulse over the same time interval as the actual, possibly changing, force: Favg = J/Δt, in N. It is not in general halfway between the largest and smallest forces.
Students often think Force and impulse (or change in momentum) are the same quantity, so a value of one can be used as the other, and equal changes in momentum mean equal forces. In fact No. Force is measured in newtons; impulse is the average force multiplied by the time interval during which it is exerted, measured in N·s, and it equals the change in momentum. The same impulse can come from a large force for a short time or a small force for a long time.
Students often think The impulse delivered by a force, and the change in momentum it causes, depend only on the size of the force, not on how long it is exerted. In fact No. Impulse is the average force multiplied by the time interval during which it is exerted, so the same force exerted for twice as long delivers twice the impulse.
4.2.A.3 Direction of impulse Fix
Direction of impulse
Impulse is a vector with the direction of the net force exerted on the system. It need not point in the direction of the velocity: a braking force delivers an impulse opposite to the motion. For motion along one axis, the sign of the impulse gives its direction.
Students often think The impulse delivered to an object, like the net force on it, points in the direction in which the object is moving. In fact No. Impulse has the direction of the net force, which is the direction of the change in momentum. An object that slows down while moving in the +x direction receives an impulse in the −x direction.
Students often think Impulse has no direction, because it is calculated as an area under a graph (or as a force multiplied by a time). In fact No. Impulse is a vector with the direction of the net force. For motion along one axis, the sign shows the direction: area above the time axis of a force–time graph is impulse in the positive direction, and area below it is impulse in the negative direction.
4.2.A.4 Area under a force–time graph Fix
Area under a force–time graph
The area between a graph of net external force against time and the time axis equals the impulse delivered, in N·s. Area above the axis counts as positive impulse and area below the axis as negative impulse.
Students often think The height of a graph gives the quantity asked for: the peak of a force–time graph is the impulse or the average force, and the height of a momentum–time graph is the net force. In fact No. On a force–time graph the impulse is the area under the curve, and the average force is that area divided by the time interval; on a momentum–time graph the net force is the slope. The height of either graph gives only the force or the momentum at one instant.
Students often think The area under a force–time graph is the peak force multiplied by the whole time interval, whatever the shape of the graph. In fact Only if the force stays at its peak for the whole interval. For a triangular pulse the area is ½ × base × height; in general, the parts of the interval where the force is below its peak contribute less.
4.2.A.5 Slope of a momentum–time graph Fix
Slope of a momentum–time graph
The slope of a graph of momentum against time equals the net external force, in (kg·m/s)/s = N. A horizontal segment means zero net force; a downward slope means a net force in the negative direction.
Students often think The net force on an object is the area under its momentum–time graph. In fact No. The net force is the slope of the momentum–time graph, Fnet = Δp/Δt. The area under a momentum–time graph has units of kg·m, not newtons.
Students often think A negative value, such as the downward slope of a momentum–time graph, means a smaller force than any positive value. In fact Not necessarily. A downward slope means that the net force is in the negative direction; its magnitude is the steepness of the graph. A steep downward slope means a larger net force than a gentle upward slope.
4.2.B.1 Change in momentum Fix
Change in momentum
The final momentum minus the initial momentum, Δp⃗ = p⃗ − p⃗0, in kg·m/s. Because momentum is a vector, a reversal of direction makes the magnitudes add: a ball that rebounds has a larger change in momentum than an identical ball, arriving at the same speed, that stops.
Students often think The change in momentum of a rebounding object is its mass multiplied by the difference between its final and initial speeds, so an object that bounces back at the same speed has no change in momentum. In fact No. Momentum is a vector, and a rebound reverses its direction. With the initial direction positive, Δp = p − p0 = (−mv) − (mv0) = −m(v + v0), so the magnitudes add.
Students often think The change in momentum is the initial momentum minus the final momentum. In fact No. By definition Δp⃗ = p⃗ − p⃗0, final minus initial. Reversing the order gives the right magnitude but the opposite direction.
4.2.B.2 Impulse–momentum theorem Fix
Impulse–momentum theorem
The impulse exerted on a system equals the system's change in momentum: J⃗ = F⃗avg Δt = Δp⃗. For a given change in momentum, a longer time interval means a smaller average force.
Newton-second and kg·m/s
The units of impulse (N·s) and of momentum (kg·m/s) are equivalent, since 1 N = 1 kg·m/s². This is why an impulse can be set equal to a change in momentum.
Students often think A longer time interval, such as a cushion provides, reduces the impulse (the change in momentum) exerted on an object. In fact No. For an object brought to rest from a given velocity, the change in momentum, and therefore the impulse, is fixed. The cushion makes the stopping time longer, so the average force Favg = Δp/Δt is smaller.
Students often think Momentum and kinetic energy are the same quantity, so a change in momentum can be found from ½mv² or from a change in kinetic energy. In fact No. Momentum p⃗ = mv⃗ is proportional to the velocity and is a vector; kinetic energy K = ½mv² depends on the square of the speed and has no direction. The change in momentum equals the impulse; the change in kinetic energy equals the work done.
4.2.B.3 Constant-mass system Fix
Constant-mass system
A system whose mass does not change during the interval considered. For such a system Δp⃗ = mΔv⃗, so F⃗net = Δp⃗/Δt = mΔv⃗/Δt = ma⃗: Newton's second law follows from the impulse–momentum theorem.
8 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 8
Carts A and B are joined by a spring and rest on a level track with negligible friction. A student then pulls cart A horizontally with a constant force, and the spring stretches and compresses as the carts move. Consider the system of both carts and the spring. Which statement about the momentum of this system is correct?
Answer and reasoning
AIt changes at a rate that depends on the spring's forces. A student who thinks internal forces change a system's momentum picks this. The spring pushes or pulls on cart A and on cart B with forces of equal magnitude in opposite directions, so the spring's forces add to zero for the system and cannot change its momentum.
BIt changes at a rate equal to the student's pull alone.Correct The rate of change of the system's momentum equals the net external force on the system. The spring's forces on the carts are internal and cancel in pairs; the vertical forces (weights and normal forces) balance; friction is negligible. The only unbalanced external force is the student's pull.
CIt changes at a rate equal to the net force on cart A. A student who takes the net force on the pulled object as the net force on the system picks this. The net force on cart A includes the spring's force, which is internal to the system; the system's momentum changes at a rate equal to the net external force, the student's pull.
DIt stays constant, as the momentum of a system is conserved. A student who thinks a system's momentum is conserved whatever acts on it picks this. The student's pull is an external force that is not balanced, so the system's momentum increases at a rate equal to the pull.
A student pushes a cart along a level track. The graph shows the horizontal force F that the student exerts on the cart as a function of time t. What is the average force exerted on the cart by the student from t = 0 to t = 0.40 s?
Answer and reasoning
A4.0 N A student who reads the height of the graph picks this, the peak value. The force is 4.0 N only from 0.10 s to 0.30 s; during the rise and fall it is less, so the average over the whole 0.40 s is less than the peak.
B2.0 N A student who averages the largest and smallest forces, (4.0 N + 0)/2, picks this. That works only for a force that changes linearly; this force stays at its peak for half the interval, so the average is J/Δt = 3.0 N.
C1.2 N A student who treats impulse and force as the same quantity picks this, reporting the area, 1.2 N·s, as the force. The area is the impulse; dividing it by the 0.40 s time interval gives the average force.
D3.0 NCorrect The average force is the impulse divided by the time interval. The impulse is the area under the graph: two triangles of ½(0.10 s)(4.0 N) = 0.20 N·s each and a rectangle of (0.20 s)(4.0 N) = 0.80 N·s, total 1.2 N·s. Favg = 1.2 N·s ÷ 0.40 s = 3.0 N.
Working Impulse = area under the graph = ½(0.10 s)(4.0 N) + (0.20 s)(4.0 N) + ½(0.10 s)(4.0 N) = 0.20 + 0.80 + 0.20 = 1.2 N·s. Favg = J/Δt = 1.2 N·s ÷ 0.40 s = 3.0 N.
A cart of mass 1.5 kg is at rest on a level track at t = 0. The graph shows the net force F exerted on the cart as a function of time t. What is the speed of the cart at t = 4.0 s?
Answer and reasoning
A6.0 m/s A student who adds the area below the axis picks this: 6.0 + 3.0 = 9.0 N·s, and 9.0 ÷ 1.5 = 6.0 m/s. The force after 2.0 s is negative, so its impulse of −3.0 N·s reduces the cart's momentum.
B4.0 m/s A student who multiplies each peak force by its whole duration picks this: (6.0)(2.0) − (3.0)(2.0) = 6.0 N·s, and 6.0 ÷ 1.5 = 4.0 m/s. Each pulse is a triangle, so its area is half of peak × duration.
C3.0 m/s A student who uses the net impulse, 3.0 N·s, as the speed picks this. The impulse equals the change in momentum; the speed is that momentum divided by the mass: 3.0 kg·m/s ÷ 1.5 kg = 2.0 m/s.
D2.0 m/sCorrect The net impulse is the area above the axis minus the area below: ½(2.0 s)(6.0 N) − ½(2.0 s)(3.0 N) = 6.0 N·s − 3.0 N·s = 3.0 N·s. Starting from rest, the cart's momentum is 3.0 kg·m/s, so v = 3.0 kg·m/s ÷ 1.5 kg = 2.0 m/s.
Working Area 0–2.0 s: ½(2.0 s)(6.0 N) = +6.0 N·s. Area 2.0–4.0 s: ½(2.0 s)(−3.0 N) = −3.0 N·s. Net impulse J = +3.0 N·s = Δp. The cart starts at rest, so p = 3.0 kg·m/s and v = p/m = 3.0 ÷ 1.5 = 2.0 m/s.
The graph shows the momentum p of a cart on a straight track as a function of time t. The cart starts from rest at t = 0, and the direction in which it then moves is positive. Which claim about the net force exerted on the cart is supported by the graph?
Answer and reasoning
AIt is greatest from 2.0 s to 5.0 s, where the momentum is greatest. A student who reads the height of the graph as the force picks this. The net force is the slope, and the graph is level from 2.0 s to 5.0 s, so there the net force is zero, not greatest.
BIt is in the positive direction from 5.0 s to 6.0 s, as p is positive. A student who thinks the net force points the way the cart moves picks this. The momentum is still positive, but it is decreasing: the slope is (2.0 − 6.0) ÷ 1.0 = −4.0 N, so the net force is in the negative direction.
CIts magnitude is greatest from 0 to 2.0 s, where the graph rises. A student who treats a downward slope as a smaller force picks this. From 0 to 2.0 s the slope is +3.0 N; from 5.0 s to 6.0 s it is −4.0 N, a force of greater magnitude in the negative direction.
DIt is zero from 2.0 s to 5.0 s, although the cart is moving.Correct The net force is the slope of the momentum–time graph. From 2.0 s to 5.0 s the graph is level, so the slope and the net force are zero, even though the momentum, 6.0 kg·m/s, is at its largest and the cart is moving.
Working Net force = slope of p–t. 0–2.0 s: (6.0 − 0)/2.0 = +3.0 N. 2.0–5.0 s: slope 0, so Fnet = 0. 5.0–6.0 s: (2.0 − 6.0)/1.0 = −4.0 N, magnitude 4.0 N, in the negative direction. Greatest magnitude: 5.0–6.0 s. Zero: 2.0–5.0 s, while p = 6.0 kg·m/s.
Two balls of equal mass move horizontally toward a wall at the same speed v. The rubber ball bounces straight back with speed v, and the clay ball sticks to the wall. Which statement correctly compares the impulses exerted on the two balls by the wall during their collisions with it?
Answer and reasoning
AThe rubber ball's is greater, as the wall must also send it back.Correct The wall must stop the rubber ball and then send it back. Taking the direction toward the wall as positive, its momentum changes from +mv to −mv, a change of magnitude 2mv; the clay ball's changes from +mv to 0, a change of mv. The impulse equals the change in momentum, so the rubber ball's is twice as great.
BThe clay ball's is greater, since it loses all of its momentum. A student who thinks a sticking object 'absorbs' more of the blow picks this. The clay ball's momentum changes by mv; the rubber ball's must be stopped and then sent back, a change of 2mv.
CThey are equal, as each ball reaches the wall with equal momentum. A student who thinks the wall only has to stop each ball picks this. The wall must also push the rubber ball back, so its impulse on that ball is mv to stop it plus mv to send it back.
DThe rubber ball's is zero, as its speed does not change. A student who subtracts speeds, v − v = 0, picks this. Momentum is a vector: the rubber ball's changes from +mv to −mv, a change of magnitude 2mv.
Working Toward the wall positive. Rubber: Δp = (−mv) − (mv) = −2mv. Clay: Δp = 0 − mv = −mv. The impulse equals Δp, so the rubber ball's impulse is twice the clay ball's.
Pucks A and B are at rest on level ice with negligible friction. Puck A has mass m, and puck B has mass 2m. Each puck is pushed by the same constant horizontal force F for the same time interval Δt. After the push, puck A has momentum of magnitude pA. What is the magnitude of puck B's momentum after the push?
Answer and reasoning
ApB = pA/2 A student who follows the speed instead of the momentum picks this: puck B reaches half of A's speed. Momentum is mass × velocity, and B's doubled mass makes up for its halved speed.
BpB = √2 pA A student who treats the impulse as work, so that both pucks get the same kinetic energy, finds p = √(2mK) and picks this. The same force for the same time gives the same momentum, not the same kinetic energy.
CpB = 2pA A student who thinks the heavier puck gains more momentum because momentum is proportional to mass picks this. B's speed is only half of A's, so mass × speed is the same for both.
DpB = pACorrect Each puck receives the same impulse, FΔt, and starts at rest, so by the impulse–momentum theorem each ends with momentum FΔt: pB = pA. Puck B reaches half the speed of puck A, and its mass is twice as great.
Working Impulse on each puck = FΔt, the same. Each starts at rest, so p = Δp = FΔt for both: pB = pA. (B's speed is half of A's, but its mass is twice as great.)
A ball of mass m is released from rest at a height h above a hard floor. It bounces and rises to a maximum height h/4. Air resistance is negligible, and the bounce can be modeled as a collision in which the force exerted by the floor is much larger than the ball's weight. What is the magnitude of the impulse exerted on the ball by the floor?
Answer and reasoning
Am√(gh/2) A student who subtracts the speeds, √(2gh) − ½√(2gh) = √(gh/2), picks this. The ball reverses direction, so the two momenta have opposite signs and their magnitudes add: m√(2gh) + ½m√(2gh) = (3/2)m√(2gh) = m√(9gh/2).
Bm√(2gh) A student who thinks the floor only has to stop the ball picks this, the momentum just before the bounce. The floor also sends the ball back up at ½√(2gh), which adds ½m√(2gh) to the impulse, giving (3/2)m√(2gh) = m√(9gh/2).
Cm√(9gh/2)Correct Just before the bounce the ball moves down at √(2gh); just after, it moves up at √(2g·h/4) = √(gh/2) = ½√(2gh). With up positive, J = Δp = m√(gh/2) − (−m√(2gh)) = (3/2)m√(2gh) = m√(9gh/2). The floor must stop the ball and send it back up.
D3mgh/4 A student who uses the loss of kinetic energy, mgh − mgh/4, as the impulse picks this. That is an energy, in joules; the impulse equals the change in momentum, in N·s, which depends on the velocities, not on their squares.
Working Energy: speed just before the bounce v0 = √(2gh) (downward); speed just after v = √(2g·h/4) = ½√(2gh) (upward). Upward positive: J = Δp = m(½√(2gh)) − m(−√(2gh)) = (3/2)m√(2gh) = m√(9gh/2), since (3/2)√(2gh) = √((9/4)(2gh)) = √(9gh/2). Distractors: m(√(2gh) − ½√(2gh)) = m√(gh/2); m√(2gh); mgh − mgh/4 = 3mgh/4.
A ball of mass m falls vertically and hits a level floor with speed v. It rebounds vertically with speed v/2. The ball is in contact with the floor for a time interval Δt, and air resistance is negligible. Taking upward as positive, which expression gives the average force exerted on the ball by the floor during the contact?
Answer and reasoning
Am(g + 1.5v/Δt)Correct The ball's momentum changes from −mv to +mv/2, so Δp = +1.5mv. The net force, Ffloor − mg, equals Δp/Δt, so the floor pushes with mg + 1.5mv/Δt: enough to balance the ball's weight and to reverse its momentum.
Bm(g + 1.0v/Δt) A student who counts only the impulse needed to stop the ball, mv, picks this. The floor must also send the ball back up at v/2, so Δp = mv/2 − (−mv) = +1.5mv.
Cm(1.5v/Δt − g) A student who adds the weight to the floor's force as a magnitude, Ffloor + mg = Δp/Δt, picks this. The gravitational force points down, so with upward positive it enters as −mg: Ffloor − mg = 1.5mv/Δt, and Ffloor = m(g + 1.5v/Δt).
Dm(g − 1.5v/Δt) A student who takes the change in momentum as initial minus final, (−mv) − (+mv/2) = −1.5mv, picks this. Δp⃗ = p⃗ − p⃗0 is final minus initial, +1.5mv: upward, the direction in which the floor pushes.
Working Upward positive. Momentum just before contact: −mv; just after: +mv/2. Δp = mv/2 − (−mv) = +1.5mv. During the contact two external forces are exerted on the ball: the floor's force Ffloor (upward) and the gravitational force, −mg. The net external force equals the rate of change of momentum: Ffloor − mg = Δp/Δt = 1.5mv/Δt, so Ffloor = mg + 1.5mv/Δt = m(g + 1.5v/Δt).
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
