7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
A hand pulls block A to the right across a rough horizontal surface, and a light horizontal string joins A to block B. The diagram shows the horizontal forces exerted on each block; the vertical forces on each block are balanced. What is the magnitude of the rate of change of the momentum of the two-block system?
Answer and reasoning
A10 N A student who takes the pulled block as the system picks this: the net force on A alone is 30 − 15 − 5.0 = 10 N. That is the rate of change of A's momentum; the two-block system also includes B, and the string's pulls cancel within it.
B30 N A student who takes the net force to be the applied pull picks this. The two friction forces are also external forces on the system and act opposite to the pull, so they reduce the net external force to 20 N.
C20 NCorrect For the two-block system the string's pulls on A and on B are internal and cancel. The external horizontal forces are the 30 N pull and the two 5.0 N friction forces, so dp/dt = Fnet,ext = 30 N − 5.0 N − 5.0 N = 20 N.
D70 N A student who adds the sizes of all the forces in the diagram picks this: 30 + 15 + 15 + 5.0 + 5.0 = 70. Forces add as vectors: the friction forces point opposite to the pull, and the two tensions are internal and cancel.
Working System: both blocks. The string tensions, 15 N on each block in opposite directions, are internal and cancel. External horizontal forces: 30 N to the right; friction 5.0 N + 5.0 N to the left. dp/dt = Fnet,ext = 30 − 5.0 − 5.0 = 20 N, to the right.
The net force on an object moving along the x-axis is F(t) = 3.0 + 6.0t², where F is in newtons and t is in seconds. What is the impulse delivered to the object from t = 0 to t = 2.0 s?
Answer and reasoning
A22 N·sCorrect The impulse is the integral of the net force over the interval: J = ∫₀²(3.0 + 6.0t²) dt = [3.0t + 2.0t³] from 0 to 2.0 s = 6.0 + 16 = 22 N·s.
B54 N·s A student who treats the force as constant at its final value picks this: F(2.0 s) = 27 N, and 27 N × 2.0 s = 54 N·s. The force was smaller earlier in the interval, so the impulse is less.
C30 N·s A student who averages the force's end values picks this: (3.0 N + 27 N)/2 = 15 N, and 15 N × 2.0 s = 30 N·s. The force grows as t², not linearly, so its time average is 11 N, not 15 N.
D24 N·s A student who uses the force's rate of change picks this: dF/dt = 12t, which is 24 at t = 2.0 s. That is a slope, in N/s; impulse is the area under the force–time graph, ∫F dt.
The diagram shows the velocity of a ball just before and just after it bounces off a horizontal floor. What is the direction of the impulse delivered to the ball during the bounce?
Answer and reasoning
AAlong the ball's velocity after it bounces A student who takes the change in momentum to be the final momentum picks this. The initial momentum must be subtracted: the horizontal parts cancel, leaving a change that is vertical.
BAlong the ball's velocity before it bounces A student who thinks the force on a moving object points along its motion picks this. The floor pushes the ball up and sends it back up; the impulse points along the change in momentum, not along the incoming velocity.
CPerpendicular to the floor, downward A student who uses the push of the ball on the floor picks this. That force acts on the floor; the floor's force on the ball, which delivers the impulse to the ball, is opposite to it, upward.
DPerpendicular to the floor, upwardCorrect The impulse equals the change in momentum, Δp⃗ = p⃗ − p⃗₀. The horizontal velocity component, 4.0 m/s to the right, is the same before and after, while the vertical component changes from 3.0 m/s down to 3.0 m/s up, so Δp⃗, and the impulse, point straight up.
Working Components (5.0 m/s at 37°): before (4.0, −3.0) m/s; after (4.0, +3.0) m/s. Δp⃗ = m(0, +6.0 m/s): straight up. The impulse equals Δp⃗, so it is perpendicular to the floor, upward.
The graph shows the net force exerted on each of two objects, X and Y, as a function of time during two separate collisions. What is the ratio JX/JY of the impulses delivered to X and to Y?
Answer and reasoning
A1.00Correct Each impulse is the area of a triangle under the force–time graph: JX = (1/2)(400 N)(0.010 s) = 2.0 N·s and JY = (1/2)(200 N)(0.020 s) = 2.0 N·s. X's force is twice as large but acts for half as long, so the impulses are equal.
B2.00 A student who judges impulse by the peak force picks this: 400 N is twice 200 N. The impulse is the area under the graph, and Y's force acts for twice as long, which makes up for its smaller peak.
C0.50 A student who thinks the longer interaction delivers the larger impulse picks this: X's force lasts half as long. Duration alone does not decide the impulse; X's larger force makes the two areas equal.
D4.00 A student who divides the force by the time picks this: (400/0.010)/(200/0.020) = 4. Impulse is the force multiplied by time, the area under the graph, not the force divided by time.
The graph shows the momentum p of an object moving along a straight line as a function of time t, in three intervals labeled I, II and III. Rank the magnitudes of the net force on the object in the three intervals, from largest to smallest.
Answer and reasoning
AII > III > I A student who reads the force from the height of the momentum–time graph, comparing the average momentum in each interval (12, 11 and 5 kg·m/s for II, III and I), picks this. The force is the slope; II has the largest momentum but the gentlest slope.
BIII > II > I A student who takes the force to be the change in momentum alone picks this: |Δp| is 10, 8 and 6 kg·m/s. The force is the change in momentum per unit time, and II's change is spread over 4 s.
CIII > I > IICorrect The net force is the slope of the momentum–time graph. Interval I: 6 kg·m/s in 2 s, 3.0 N. Interval II: 8 kg·m/s in 4 s, 2.0 N. Interval III: −10 kg·m/s in 2 s, magnitude 5.0 N. So III > I > II.
DI > II > III A student who treats a negative force as smaller than a positive one picks this: the slopes are +3.0, +2.0 and −5.0 N. The sign gives only the direction; the magnitude in III, 5.0 N, is the largest.
Working |Fnet| = |slope|. I: (8 − 2)/2 = 3.0 N. II: (16 − 8)/4 = 2.0 N. III: (6 − 16)/2 = −5.0 N, magnitude 5.0 N. Ranking III > I > II. (Heights: mean p 5, 12, 11; |Δp|: 6, 8, 10; signed slopes 3, 2, −5.)
A clay ball and a rubber ball have equal masses. Each is thrown horizontally at a wall with the same speed. The clay ball sticks to the wall; the rubber ball rebounds with the same speed it had before the collision. How does the magnitude of the impulse delivered by the wall to the rubber ball compare with that delivered to the clay ball?
Answer and reasoning
AThe clay ball's is larger, since the wall must absorb all its motion. A student who thinks the object stopped dead receives the larger impulse picks this. The wall must stop both balls; for the rubber ball it must also send it back at the same speed, so the rubber ball's change in momentum is larger.
BThe rubber ball's is twice as large, since its momentum reverses.Correct Taking the initial direction as positive, the clay ball's change in momentum is 0 − mv = −mv, and the rubber ball's is (−mv) − mv = −2mv. The impulse equals the change in momentum, so the rubber ball's is twice as large.
CThey are equal, since both balls reach the wall with equal momenta. A student who thinks the wall only stops each ball picks this. Equal initial momenta do not give equal changes; the wall also pushes the rubber ball back out, which doubles its change in momentum.
DThe rubber ball's is zero, since its speed after the collision is unchanged. A student who takes the change in momentum to be the change in its size picks this. The speed is unchanged but the direction reverses, so the change in momentum, and the impulse, has magnitude 2mv.
An egg dropped onto a pillow does not break, but an identical egg dropped from the same height onto a hard floor does break. In both cases the egg comes to rest. Which reasoning best explains the difference?
Answer and reasoning
AThe pillow is soft and gives way, so the egg's change in momentum on the pillow is smaller than on the floor. A student who thinks a soft landing reduces the change in momentum picks this. Both eggs go from the same momentum to rest, so the change is the same; the pillow changes only how long it takes.
BThe hard floor pushes up on the egg, but the soft pillow only absorbs the egg's motion without pushing. A student who thinks surfaces such as pillows are passive picks this. The pillow does push up on the egg; that push is what stops it, with a smaller force acting for a longer time.
CThe pillow acts on the egg for a longer time, so the egg's change in momentum is larger but gentler. A student who thinks a longer contact time means a larger effect picks this. The egg's change in momentum is fixed by its initial and final states and is the same on both; a longer time means a smaller average force.
DThe egg's momentum change is the same on both, but the pillow takes longer, so the average force is smaller.Correct Both eggs arrive with the same momentum and end at rest, so their changes in momentum, and the impulses of the net forces on them, are equal. The pillow spreads that change over a longer time, so the average net force on the egg, Δp/Δt, is smaller.
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.2.A.1 Net external force as rate of change of momentum Fix
Net external force as rate of change of momentum
The net external force on a system equals the rate of change of the system's momentum, F⃗net = dp⃗/dt. Forces that objects inside the system exert on each other do not change the system's momentum. Unit: N = kg·m/s².
Students often think The system whose momentum changes is the object that is pushed or pulled, so the net force on that object alone is the rate of change of the whole system's momentum. In fact No. The two-block system's momentum changes at a rate equal to the net external force on both blocks together. The pulled block alone also feels the string's backward pull, which is internal to the two-block system and cancels against the string's forward pull on the other block.
Students often think The net force on a system is the force applied by whoever pushes or pulls it; other forces, such as friction, do not count toward it. In fact No. The net force is the vector sum of all the external forces, including friction, gravity and normal forces, not only the force applied by the agent.
4.2.A.2 Impulse, J⃗ Fix
Impulse, J⃗
The integral of the net force over a time interval, J⃗ = ∫F⃗net dt from t₁ to t₂. For a constant net force, J⃗ = F⃗net Δt. Unit: N·s, equivalent to kg·m/s.
Students often think A force that varies during an interval can be replaced by one of its values (its value at the start, its peak or its value at the end) acting for the whole interval. In fact No. When the force varies, the impulse is the integral J = ∫Fnet dt, the area under the force–time graph. Using the force's initial, peak or final value for the whole interval gives the wrong impulse unless the force is actually constant.
Students often think The average of any varying force over an interval is the mean of its values at the start and at the end of the interval. In fact Not in general. The time-averaged force is (1/Δt)∫F dt. The mean of the end values is guaranteed to equal it only when the force changes linearly with time; for a curved or segmented force–time graph it usually gives a different value.
4.2.A.3 Direction of impulse Fix
Direction of impulse
Impulse is a vector with the direction of the net force over the interval (for a varying direction, the direction of the vector integral). It need not point along the object's velocity.
Students often think A moving object carries a force in its direction of motion that keeps it moving: the net force, and so the impulse, points along the motion, and the object slows as that force is used up. In fact No. The net force and the impulse point along the change in momentum, which can be in any direction relative to the motion. An object moving forward while slowing down has a net force pointing backward, and an object moving at constant velocity with constant mass has zero net force.
Students often think The force or impulse exerted by an object (a ball pushing on a floor, say) can be used as the force or impulse exerted on it. In fact No. The impulse on an object comes from the forces exerted on it. The force an object exerts on something else acts on that other thing, and by Newton's third law it is opposite in direction to the force that thing exerts on the object.
4.2.A.4 Impulse from a force–time graph Fix
Impulse from a force–time graph
The impulse delivered by a net force is the area between the graph of Fnet against t and the time axis; area below the axis counts as negative impulse.
Students often think The impulse delivered in an interaction can be judged by the peak force alone: a larger maximum force means a larger impulse. In fact No. The impulse is the area under the force–time graph, which depends both on how large the force is and on how long it acts. A tall, narrow pulse and a low, wide pulse can have equal areas.
4.2.A.5 Net force from a momentum–time graph Fix
Net force from a momentum–time graph
The net force on a system at an instant is the slope of its momentum–time graph at that instant; a steeper graph means a larger net force, and a negative slope means a net force in the negative direction.
Students often think The net force on an object can be read from the height of its momentum–time graph: the larger the momentum, the larger the net force. In fact No. The net force is the slope of the momentum–time graph, Fnet = dp/dt. A large momentum can be changing slowly (small net force), and a momentum of zero can be changing quickly (large net force).
Students often think A negative force, or a negative slope, is smaller than a positive one or than zero, so it ranks lowest when magnitudes are compared. In fact No. The sign of a force component gives its direction; its magnitude is the absolute value. A force component of −5 N has a larger magnitude than one of +2 N.
4.2.B.1 Change in momentum, Δp⃗ Fix
Change in momentum, Δp⃗
The final momentum minus the initial momentum, Δp⃗ = p⃗ − p⃗₀, found by vector subtraction: with signs in one dimension, by components in two. Unit: kg·m/s.
Students often think An object's change in momentum, or the impulse on it, is its final momentum; the initial momentum can be left out. In fact Only if the object starts at rest. The change in momentum is Δp⃗ = p⃗ − p⃗₀, so the initial momentum must be subtracted, as a vector.
Students often think The change in momentum is the change in its size, |p| − |p₀|, so an object whose speed is unchanged has no change in momentum. In fact Not necessarily. Momentum is a vector, so a change of direction is a change in momentum. A ball that rebounds with unchanged speed has a change in momentum of magnitude 2mv, and in general Δp⃗ = p⃗ − p⃗₀ must be found by vector subtraction.
4.2.B.2 Impulse–momentum theorem Fix
Impulse–momentum theorem
The impulse of the net force on an object over an interval equals the object's change in momentum over that interval: J⃗ = ∫F⃗net dt = Δp⃗.
Average net force
For an interaction lasting Δt, the average net force is Δp⃗/Δt: the constant force that would deliver the same impulse in the same time. For a given change in momentum, a longer interaction means a smaller average net force.
Newton's second law as a special case
For a system of constant mass, F⃗net = dp⃗/dt = m dv⃗/dt = ma⃗. F⃗net = ma⃗ therefore holds only when the system's mass does not change.
Constant-velocity system of changing mass
For a system whose velocity v⃗ is constant while its mass changes at the rate dm/dt, such as a conveyor belt onto which sand drops, F⃗net = dp⃗/dt = (dm/dt)v⃗, so a net external force is needed although the acceleration is zero.
Students often think An object's velocity, or its change in velocity, is set by the net force acting at that moment: where the force is zero the object is at rest, and the change in velocity is F/m. In fact No. The net force sets how fast the momentum is changing, not the momentum itself. The change in velocity over an interval depends on the impulse, ∫Fnet dt, so an object can be moving fast at an instant when the net force on it is zero.
Students often think Impulse and work, and momentum and kinetic energy, are interchangeable, so an impulse or a force can be found from changes in kinetic energy, or taken to grow as v². In fact No. Impulse, ∫Fnet dt, changes momentum, a vector proportional to v; work, ∫F⃗·dr⃗, changes kinetic energy, a scalar proportional to v². Doubling an object's speed doubles its momentum but quadruples its kinetic energy.
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
A cart moving in the +x direction is slowed by a resistive force, and its momentum is p(t) = p₀e−t/τ, where p₀ and τ are positive constants and t is time. What is the magnitude of the net force on the cart at the instant its momentum is p₀/2?
Answer and reasoning
A1.00p₀/τ A student who treats the net force as constant at its initial value picks this: at t = 0, |dp/dt| = p₀/τ. The force is proportional to the momentum, so it has halved by the time the momentum has halved.
B0.72p₀/τ A student who divides the change in momentum since t = 0 by the elapsed time picks this: (p₀/2)/(τ ln 2) ≈ 0.72p₀/τ. That is the average net force over the interval; the force at the instant is the slope dp/dt there, p₀/(2τ).
C0.37p₀/τ A student who thinks the momentum halves at t = τ picks this: the force magnitude at t = τ is (p₀/τ)e⁻¹ ≈ 0.37p₀/τ. At t = τ the momentum is p₀/e, not p₀/2; it halves at t = τ ln 2, where |dp/dt| = p₀/(2τ).
D0.50p₀/τCorrect The net force is the rate of change of momentum: Fnet = dp/dt = −(p₀/τ)e−t/τ = −p/τ. At the instant p = p₀/2 its magnitude is p₀/(2τ) = 0.50p₀/τ, directed opposite to the motion.
Working Fnet = dp/dt = −(p₀/τ)e−t/τ = −p(t)/τ. When p = p₀/2, |Fnet| = p₀/(2τ); this happens at t = τ ln 2. Distractors: force at t = 0, p₀/τ; average force from t = 0 to τ ln 2, (p₀/2)/(τ ln 2) = p₀/(2τ ln 2) ≈ 0.72p₀/τ; force at t = τ, (p₀/τ)e⁻¹ ≈ 0.37p₀/τ. The options give the coefficients to two decimal places.
A net force F(t) = F₀(1 − t²/T²), where F₀ and T are positive constants, acts in the +x direction on an object of mass m from t = 0 to t = T. The object is at rest at t = 0. What is the object's speed at t = T?
Answer and reasoning
AF₀T/(2m) A student who takes the average force to be the mean of its end values, (F₀ + 0)/2, picks this. That works only for a force that changes linearly; this force falls slowly at first, so its time average, 2F₀/3, is larger than F₀/2.
B2F₀T/(3m)Correct The impulse is the integral of the force: J = ∫₀ᵀ F₀(1 − t²/T²) dt = F₀(T − T/3) = 2F₀T/3. The object starts at rest, so mv = J and v = 2F₀T/(3m).
CF₀T/m A student who treats the force as constant at its initial value, F₀, picks this. The force falls to zero by t = T, so the impulse is less than F₀T.
D0 A student who thinks an object is at rest wherever the net force on it is zero picks this. F(T) = 0, but the object has been receiving impulse in the +x direction throughout, so at t = T it is moving at 2F₀T/(3m).
Working J = ∫₀ᵀ F₀(1 − t²/T²) dt = F₀(T − T/3) = 2F₀T/3. J = Δp = mv − 0, so v = 2F₀T/(3m). (Mean of the end values: (F₀ + 0)/2 × T/m = F₀T/(2m); force held at its initial value: F₀T/m; F(T) = 0.) The check reproduces the leading coefficient 2 of the key as displayed.
A constant net force acts for a time interval Δt on a cart that starts at rest. In a second trial, the cart again starts at rest, and a constant net force half as large acts on it for 3Δt. The impulse delivered in the second trial is how many times the impulse delivered in the first trial?
Answer and reasoning
A0.50 A student who takes the impulse to be set by the size of the force alone picks this. The impulse is force times time, so the threefold longer time must be included.
B0.17 A student who divides the force by the time picks this: (1/2)/3 ≈ 0.17. Impulse is force multiplied by time, J = Fnet Δt; force divided by time has units N/s.
C1.50Correct For a constant net force the impulse is J = Fnet Δt. Halving the force halves J and tripling the time triples it, so J₂/J₁ = (1/2)(3) = 1.50.
D2.25 A student who equates impulse with the kinetic energy given to the cart picks this. From rest, the kinetic energy gained is (Fnet Δt)²/(2m), which grows by (1.5)² = 2.25; the impulse, Fnet Δt, grows by 1.5.
Working For a constant net force J = Fnet Δt. J₂/J₁ = (F/2)(3Δt)/(FΔt) = 1.50.
The graph shows the net force F exerted in the +x direction on an object of mass m as a function of time t, where F₁ and t₁ are positive constants. What is the change in the object's velocity from t = 0 to t = 2t₁?
Answer and reasoning
A2F₁t₁/m A student who treats the force as constant at its peak value, F₁, for the whole 2t₁ picks this. During the first t₁ the force is less than F₁, so the area is less than the rectangle F₁ × 2t₁.
B3F₁t₁/(2m)Correct The impulse is the area under the force–time graph from 0 to 2t₁: a triangle of area (1/2)F₁t₁ plus a rectangle of area F₁t₁, a total of 3F₁t₁/2. Since J = mΔv, Δv = 3F₁t₁/(2m).
CF₁t₁/m A student who averages the force's end values, (0 + F₁)/2, over the whole 2t₁ picks this. The force is not linear over that interval: it rises for t₁ and then stays at F₁, so its time average is 3F₁/4.
DF₁/m A student who thinks the force fixes the change in velocity directly, as F/m, picks this. F₁/m is the acceleration while the force is F₁; the change in velocity depends on how long the force acts, through the area under the graph.
Working Impulse = area under the graph from 0 to 2t₁ = triangle (1/2)F₁t₁ + rectangle F₁t₁ = 3F₁t₁/2. J = mΔv, so Δv = 3F₁t₁/(2m). (Peak force for the whole time: 2F₁t₁/m; mean of end values (0 + F₁)/2 for 2t₁: F₁t₁/m; F₁/m is an acceleration.) The check reproduces the leading coefficients 3 and 2 as displayed.
The graph shows the momentum p of a cart as a function of time t as the cart slows down on a level track. A student claims that the magnitude of the net force on the cart decreases from t = 0 to t = 6 s. Which evidence and reasoning support the student's claim?
Answer and reasoning
AThe graph gets lower, and a momentum graph's height gives the force. A student who reads the force from the height of a momentum–time graph picks this. A momentum that falls steadily, along a straight line, has a constant net force even though it gets lower; only the slope shows the force.
BThe area under the graph grows more slowly; that area is the impulse. A student who applies 'area gives impulse' to the wrong graph picks this. Impulse is the area under a net force–time graph; the area under a momentum–time graph is m times the displacement, and says nothing directly about the force.
CThe cart keeps moving forward, so the force keeping it moving is being used up. A student who thinks a moving object carries a forward force picks this. The cart is slowing, so the net force on it points backward, opposite to its motion; the claim about the force's magnitude rests on the decreasing slope.
DThe graph gets less steep, so the momentum changes more and more slowly.Correct The net force is the slope of the momentum–time graph, Fnet = dp/dt. The curve is steep at t = 0 and flattens as time goes on, so the momentum changes more and more slowly and the magnitude of the net force decreases.
A 1.2 kg puck slides east at 3.0 m/s. After being struck by a hockey stick, it slides north at 4.0 m/s. What is the magnitude of the puck's change in momentum?
Answer and reasoning
A6.0 kg·m/sCorrect Δp⃗ = p⃗ − p⃗₀. The initial momentum is 3.6 kg·m/s east and the final momentum 4.8 kg·m/s north, so Δp⃗ has components 3.6 kg·m/s west and 4.8 kg·m/s north, of magnitude √(3.6² + 4.8²) = 6.0 kg·m/s.
B1.2 kg·m/s A student who subtracts the sizes of the momenta picks this: 4.8 − 3.6 = 1.2 kg·m/s. Momentum is a vector, and its direction has changed by 90°, so the vectors must be subtracted.
C8.4 kg·m/s A student who adds the sizes of the two momenta picks this: 3.6 + 4.8 = 8.4 kg·m/s. That would be right only if the puck reversed direction; perpendicular vectors combine by the Pythagorean theorem.
D4.8 kg·m/s A student who takes the change in momentum to be the final momentum picks this. The puck was already moving east, so its initial momentum must be subtracted.
Working p⃗₀ = (1.2 kg)(3.0 m/s) = 3.6 kg·m/s east; p⃗ = (1.2 kg)(4.0 m/s) = 4.8 kg·m/s north. Δp⃗ = p⃗ − p⃗₀ has components 3.6 kg·m/s west and 4.8 kg·m/s north, so |Δp⃗| = √(3.6² + 4.8²) = 6.0 kg·m/s.
A ball of mass m is dropped from rest at a height h above a floor and rebounds to a maximum height h/9. Air resistance is negligible, and the contact with the floor is so brief that the impulse of the gravitational force during the contact is negligible. If g is the acceleration due to gravity, what is the magnitude of the impulse exerted by the floor on the ball?
Answer and reasoning
Am√(2gh) A student who thinks the floor only stops the ball picks this: it is the ball's momentum on arrival. The floor also sends the ball back up at √(2gh)/3, and that rebound adds to the change in momentum.
Bm√(2gh)/3 A student who takes the change in momentum to be the final momentum picks this. The ball arrived with momentum m√(2gh) downward, which must be subtracted, as a vector, from the final momentum.
C(4/3)m√(2gh)Correct The ball reaches the floor at √(2gh) downward and leaves at √(2g·h/9) = √(2gh)/3 upward. With up positive, the impulse equals the change in momentum: m√(2gh)/3 − (−m√(2gh)) = (4/3)m√(2gh).
D(8/9)mgh A student who uses energy in place of momentum picks this: it is the mechanical energy lost in the bounce, mgh − mgh/9, in joules. Impulse equals the change in momentum, which depends on the velocities, not on their squares.
Working Speed just before contact: v₁ = √(2gh), downward. Speed just after: v₂ = √(2g(h/9)) = √(2gh)/3, upward. Taking up as positive: J = Δp = m(√(2gh)/3) − m(−√(2gh)) = (4/3)m√(2gh). (Initial momentum only: m√(2gh); final only: m√(2gh)/3; mechanical energy lost: mgh − mgh/9 = (8/9)mgh.)
An object of mass m moves in the +x direction with speed v₀ at t = 0. From t = 0, the net force on it is F(t) = −bt, where b is a positive constant. Which expression gives the object's velocity at time t, before it comes to rest?
Answer and reasoning
Av₀−bt²/(2m)Correct By the impulse–momentum theorem, m(v − v₀) = ∫₀ᵗ(−bt′) dt′ = −bt²/2. So v = v₀ − bt²/(2m), valid until the object stops.
Bv₀−bt²/m A student who treats the force as constant at its value at time t, −bt, for the whole interval picks this: the impulse would be −bt × t. The force grows from zero, so the impulse is the area of a triangle, −bt²/2.
Cv₀−bt/m A student who thinks the force at an instant fixes the change in velocity, as F/m, picks this. −bt/m is the acceleration at time t; the change in velocity is the integral of the acceleration, −bt²/(2m).
D−bt²/(2m) A student who takes the impulse to give the final momentum picks this: it is the change in velocity. The object was already moving at v₀, so v = v₀ + Δv.
Working J(t) = ∫₀ᵗ(−bt′) dt′ = −bt²/2 = m(v − v₀), so v = v₀ − bt²/(2m). (Force held at its value at time t: v₀ − bt²/m; Δv = F/m: v₀ − bt/m; initial momentum left out: −bt²/(2m).)
In a crash test, a car hits a rigid barrier and comes to rest. In a second test, the car's initial speed is doubled, and a redesigned front end makes the collision last 1.5 times as long. The magnitude of the average net force on the car in the second test is how many times that in the first test?
Answer and reasoning
A2.7 A student who takes the change in momentum to grow as the square of the speed picks this: 4/1.5 ≈ 2.67. That is how kinetic energy grows; momentum is proportional to speed, so Δp only doubles.
B1.3Correct The average net force is Δp/Δt. The car's change in momentum, from mv₀ to zero, doubles when v₀ doubles, and the time is 1.5 times as long, so the force changes by 2/1.5 ≈ 1.3.
C3.0 A student who thinks a longer collision means a larger force picks this: 2 × 1.5 = 3. For a given change in momentum, a longer collision means a smaller average force, so the time divides.
D2.0 A student who thinks the force is fixed by the change in momentum alone picks this. The average net force is the change in momentum divided by the time, and the longer collision time reduces it.
Working Average net force = Δp/Δt. Δp = 0 − mv₀ doubles when v₀ doubles; Δt is 1.5 times as long. Ratio = 2/1.5 ≈ 1.3.
A 0.060 kg tennis ball moving at 20 m/s is struck by a racket and leaves in the opposite direction at 30 m/s. The ball is in contact with the racket for 0.0050 s. What is the magnitude of the average net force on the ball during the contact?
Answer and reasoning
A6.0 × 10² NCorrect Taking the final direction as positive, Δp = (0.060 kg)(30 m/s) − (0.060 kg)(−20 m/s) = 3.0 kg·m/s. The average net force is Δp/Δt = 3.0 kg·m/s ÷ 0.0050 s = 6.0 × 10² N.
B1.2 × 10² N A student who subtracts the speeds picks this: (0.060)(30 − 20)/0.0050 = 120 N. The ball reverses direction, so with signs the velocity changes by 50 m/s, not 10 m/s.
C3.6 × 10² N A student who uses the final momentum as the change in momentum picks this: (0.060)(30)/0.0050 = 360 N. The ball's initial momentum, in the opposite direction, must be subtracted, which adds to the change.
D3.0 × 10³ N A student who uses the change in kinetic energy in place of the change in momentum picks this: (1/2)(0.060)(30² − 20²)/0.0050 = 3000. That is a change in energy, in joules, divided by a time; the force is the change in momentum divided by the time.
Working Taking the final direction as positive: Δp = (0.060)(30) − (0.060)(−20) = 3.0 kg·m/s. Average net force = Δp/Δt = 3.0/0.0050 = 6.0 × 10² N.
Which statement correctly describes how Newton's second law in the form Fnet = ma is related to the relationship Fnet = dp/dt?
Answer and reasoning
AFnet = ma and Fnet = dp/dt are equivalent for every system, since p = mv. A student who thinks Fnet = ma holds for every system picks this. If the mass changes, dp/dt is not m dv/dt: a conveyor belt moving at constant velocity while sand lands on it needs a net force although a = 0.
BFnet = dp/dt applies during collisions, and Fnet = ma applies at other times. A student who treats momentum relationships as rules only for collisions picks this. Fnet = dp/dt applies to any system at any time; Fnet = ma is its form for constant mass.
CFnet = dp/dt gives the average force over an interval; Fnet = ma, the force at an instant. A student who reads dp/dt as Δp/Δt, an average over an interval, picks this. dp/dt is the instantaneous rate of change of momentum, so Fnet = dp/dt gives the net force at each instant; for constant mass it equals ma at that same instant.
DFnet = ma is what Fnet = dp/dt becomes when the system's mass is constant.Correct With p = mv and m constant, dp/dt = m dv/dt = ma, so Fnet = dp/dt reduces to Fnet = ma. When the mass changes, dp/dt includes the change in mass and the two are not equivalent.
For which of the following systems does Fnet = ma, where m is the mass of the system and a is its acceleration, give an incorrect value for the net external force on the system?
Answer and reasoning
AA satellite moving at constant speed in a circular orbit around the Earth A student who thinks constant speed means zero acceleration picks this. The satellite's velocity changes direction, so it has a centripetal acceleration, and Fnet = ma correctly gives the gravitational force on it.
BA conveyor belt and its sand, at constant velocity while more sand lands on itCorrect The system's mass increases while its velocity stays constant, so a = 0 but dp/dt = (dm/dt)v is not zero: a net external force must act to give each new portion of sand the belt's velocity. Fnet = ma holds only for constant mass and gives zero here.
CA ball thrown straight upward, at the instant it reaches its highest point A student who thinks zero velocity means zero acceleration picks this. At the top the ball's acceleration is still g downward, so Fnet = ma correctly gives the gravitational force mg.
DA car moving at constant velocity along a level road while its engine is running A student who thinks a moving object needs a net force in its direction of motion picks this. At constant velocity the forward force on the car balances the resistive forces, so the net force is zero, as Fnet = ma gives.
Sand drops vertically onto a horizontal conveyor belt at a rate of 120 kg/s, landing with no horizontal velocity, and the belt carries it away at a constant speed of 1.5 m/s. Friction in the belt's rollers is negligible. What horizontal force must the motor exert on the belt to keep its speed constant? Use g = 10 m/s².
Answer and reasoning
A90 N A student who equates the force with the rate of kinetic energy given to the sand, divided by the speed, picks this: (1/2)(120)(1.5)²/1.5 = 90. Force is the rate of change of momentum, (dm/dt)v; energy and momentum do not interchange.
B0 N A student who applies Fnet = ma to this system picks this: the belt moves at constant velocity, so a = 0. The system's mass is increasing, so Fnet = dp/dt = (dm/dt)v is not zero.
C180 NCorrect The belt's velocity is constant, but its system gains 120 kg of sand each second, and each kilogram must be given a horizontal velocity of 1.5 m/s. The force needed is the rate of change of horizontal momentum, F = (dm/dt)v = (120 kg/s)(1.5 m/s) = 180 N.
D1200 N A student who takes the force needed to be the weight of the sand added each second picks this: (120)(10) = 1200. The weight is vertical; the horizontal force depends on the rate at which horizontal momentum is given to the sand.
Working System: the belt and the sand on it. Its velocity is constant, but each second 120 kg of sand with no horizontal velocity joins it and is brought to 1.5 m/s. F = dp/dt = (dm/dt)v = (120 kg/s)(1.5 m/s) = 180 N.
Sand falls vertically onto a horizontal conveyor belt, landing with no horizontal velocity, and a motor keeps the belt moving at a constant speed v. Starting at t = 0, the rate at which sand lands on the belt increases with time t as dm/dt = βt, where β is a positive constant. No sand leaves the belt during the time considered, and friction in the belt's rollers is negligible. What is the magnitude of the horizontal impulse the motor exerts on the belt from t = 0 to t = T? g is the acceleration due to gravity.
Answer and reasoning
A1.00βvT² A student who takes the force at the end of the interval, F(T) = βvT, as acting for the whole time T gets βvT². The force grows from zero as the delivery rate grows, so the impulse is the integral ∫₀ᵀ βvt dt, half of that.
B0.50βvT²Correct Taking the belt and the sand on it as the system, the motor's force is the only horizontal external force, so it equals the rate of change of the system's momentum. The belt's momentum is constant and sand gains momentum at the rate (dm/dt)v = βvt, so F = βvt. The impulse is J = ∫₀ᵀ βvt dt = (1/2)βvT² = 0.50βvT².
C0.25βvT² A student who finds the force from the rate at which the sand gains kinetic energy, Fv = (1/2)(dm/dt)v², gets F = (1/2)βvt and an impulse of 0.25βvT². The force is set by the rate of change of momentum, (dm/dt)v. The motor's power, Fv = (dm/dt)v², is twice the rate at which the sand gains kinetic energy: the sand slips on the belt as it speeds up, and the rest of the energy is dissipated.
D0.17βgT³ A student who takes the motor's force to equal the weight of the sand on the belt, (1/2)βt²g, and integrates it over time gets βgT³/6 ≈ 0.17βgT³. The weight is vertical and is balanced by the belt's upward push; the horizontal force is (dm/dt)v, which does not involve g.
Working System: belt plus the sand on it. Its only horizontal external force is the motor's force F (roller friction negligible; the arriving sand brings no horizontal momentum). The belt's momentum is constant, and sand gains horizontal momentum at the rate (dm/dt)v, so F = dp/dt = (dm/dt)v = βvt. Impulse: J = ∫₀ᵀ βvt dt = (1/2)βvT² = 0.50βvT² (units: kg/s² · m/s · s² = N·s). Checked with sympy. Distractors: end value F(T) = βvT for the whole interval → βvT² = 1.00βvT²; force from kinetic energy, Fv = dK/dt = (1/2)(dm/dt)v² → F = (1/2)βvt → J = 0.25βvT²; force equal to the weight of the sand on the belt, m(t) = βt²/2 → ∫₀ᵀ (1/2)βgt² dt = βgT³/6 ≈ 0.17βgT³.
A puck of mass m, tied to a string, moves at constant speed v in a horizontal circle of radius R on frictionless ice. What is the magnitude of the average net force on the puck during the time it takes to travel one quarter of the way around the circle?
Answer and reasoning
A1.00mv²/R A student who averages the magnitude of the net force, which stays mv²/R throughout, picks this. The force changes direction as the puck goes around, so the force vectors at different instants partly cancel, and the average net force, |Δp⃗|/Δt, is smaller.
B0.90mv²/RCorrect The momenta at the start and end of the quarter turn both have magnitude mv and are perpendicular, so |Δp⃗| = √2mv. The quarter turn takes Δt = (1/4)(2πR/v) = πR/(2v). The average net force is |Δp⃗|/Δt = (2√2/π)mv²/R ≈ 0.90mv²/R, less than mv²/R because the force keeps changing direction.
C0.64mv²/R A student who takes the change in momentum to be the final momentum, of magnitude mv, gets mv/(πR/(2v)) = (2/π)mv²/R ≈ 0.64mv²/R. The initial momentum must be subtracted as a vector, which gives |Δp⃗| = √2mv.
D0.71mv²/R A student who averages the net force vectors at the start and end of the quarter turn, each of magnitude mv²/R and perpendicular to each other, gets mv²/(√2R) ≈ 0.71mv²/R. The force changes continuously in between, so its average is |Δp⃗|/Δt, not the mean of its first and last values.
Working At the start and end of the quarter turn the momentum has magnitude mv, and the two momenta are perpendicular. Δp⃗ = p⃗ − p⃗₀ has magnitude √((mv)² + (mv)²) = √2mv. The quarter turn takes Δt = (1/4)(2πR/v) = πR/(2v). Impulse–momentum theorem: Favg Δt = |Δp⃗|, so Favg = √2mv/(πR/(2v)) = (2√2/π)mv²/R ≈ 0.90mv²/R (units N). Checked with sympy by integrating the net force vector, of constant magnitude mv²/R, over the quarter turn. Distractors: average of the force's constant magnitude → 1.00mv²/R; Δp taken as the final momentum mv → (2/π)mv²/R ≈ 0.64mv²/R; mean of the net force vectors at the start and end (perpendicular, each mv²/R) → mv²/(√2R) ≈ 0.71mv²/R.
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
