Linear Momentum
Unit 4 of AP Physics 1, worth 10–15% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Momentum, impulse, change in momentum, conservation of momentum, elastic and inelastic collisions.
How this unit is tested
Start by treating momentum as a vector, not a number: $p = mv$ always carries a direction, and in one-dimensional problems you must pick a positive direction and stick with signs throughout. Most problems in this unit reduce to one core move — write down total momentum before the interaction and set it equal to total momentum after, because momentum is conserved for any isolated system (no net external force) whether the collision is elastic or not.
Separate two families of questions. The first is impulse questions, which connect force and time to a change in momentum through $J = F\Delta t = \Delta p$; these often give you a force-vs-time graph and ask for impulse as the area underneath it, or ask you to compare forces for the same impulse delivered over different times (this is the physics behind airbags and padded surfaces). The second is collision questions, which give you masses and velocities before and after (or before only, for explosions) and ask you to apply conservation of momentum, then classify the collision by checking whether kinetic energy was also conserved.
Before writing any equation, sketch a before/after diagram with velocity arrows and labeled directions. Decide your positive direction once, convert every velocity to a signed number, then write the conservation equation $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$. Only after solving should you check energy: if kinetic energy before equals kinetic energy after, the collision is elastic; if some is lost, it's inelastic; if the objects end up moving together, it's perfectly inelastic and you use a single final velocity for the combined mass.
What you have to know
Momentum
$p = mv$. Momentum is a vector with the same direction as velocity, measured in kg·m/s.
Impulse-Momentum Theorem
$J = F_{net}\Delta t = \Delta p = m\Delta v$. The impulse delivered by the net force on an object equals its change in momentum. Graphically, impulse equals the area under a force-vs-time graph.
Conservation of Momentum
If the net external force on a system is zero (an isolated system), the total momentum of the system is constant: $\sum p_{i} = \sum p_{f}$. Internal forces between objects in the system (e.g., during a collision) cannot change the total momentum, only redistribute it.
Elastic Collision
A collision in which both total momentum and total kinetic energy are conserved. Objects separate after colliding.
Inelastic Collision
A collision in which total momentum is conserved but total kinetic energy is not; some KE converts to heat, sound, or deformation. A perfectly inelastic collision is the extreme case where the objects stick together and move with one common final velocity, losing the maximum possible kinetic energy consistent with momentum conservation.
14 practice questions
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A 3 kg object moves at 4 m/s. What is its momentum?
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Answer. 12 kg·m/s in the direction of motion
Momentum is $p=mv=(3\text{ kg})(4\text{ m/s})=12\text{ kg·m/s}$, directed the same way the object is moving. -
A net force of 15 N acts on a 5 kg object at rest for 2 seconds. What is the object's final speed?
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Answer. 6 m/s
Impulse equals $J=F\Delta t=(15)(2)=30\text{ N·s}$, which equals $\Delta p = m\Delta v$. Since the object starts at rest, $30=5v_f$, so $v_f=6\text{ m/s}$. -
Which quantity is conserved in an isolated system for both elastic and inelastic collisions?
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Answer. Total momentum
Momentum conservation follows directly from Newton's third law acting between the colliding objects; it holds regardless of whether kinetic energy is conserved. Kinetic energy is only conserved in elastic collisions. -
A force-vs-time graph shows a triangular pulse rising from 0 N to a peak of 20 N at t = 0.02 s, then back to 0 N at t = 0.04 s. What impulse does this deliver?
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Answer. 0.4 N·s
Impulse is the area under the F-t graph. For a triangle, area = (1/2)(base)(height) = (1/2)(0.04 s)(20 N) = 0.4 N·s. -
Two carts on a frictionless track: a 2 kg cart moving right at 3 m/s collides and sticks to a 1 kg cart at rest. Find the final velocity of the combined carts.
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Answer. 2 m/s in the original direction of motion
Momentum before: $p_i=(2)(3)+(1)(0)=6\text{ kg·m/s}$. Since they stick together, use combined mass: $6=(2+1)v_f$, so $v_f=2\text{ m/s}$. -
In an elastic head-on collision between two objects of equal mass, one moving at 4 m/s and the other initially at rest, what are their velocities after the collision?
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Answer. The moving object stops (0 m/s) and the object initially at rest moves off at 4 m/s
For an elastic collision between equal masses, the objects exchange velocities entirely; this is a standard result derivable from simultaneously conserving momentum and kinetic energy for equal m1 = m2. -
A 4 kg object at rest explodes into two pieces. A 1 kg piece flies off to the right at 6 m/s. What is the velocity of the 3 kg piece?
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Answer. 2 m/s to the left
Total momentum before the explosion is zero, so it must remain zero after: $0=(1)(6)+(3)v_2$, giving $v_2=-2\text{ m/s}$, meaning 2 m/s in the direction opposite the 1 kg piece. -
Why do padded dashboards and airbags reduce injury in a car crash, in terms of momentum concepts?
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Answer. They increase the time over which the change in momentum occurs, which reduces the average force on the occupant for the same impulse
Since $J=F\Delta t=\Delta p$ is fixed by the collision (the person's momentum change is the same either way), stretching out $\Delta t$ reduces the required average force $F$. -
When a perfectly inelastic collision occurs, where does the 'lost' kinetic energy go?
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Answer. It converts into other forms of energy such as heat, sound, and permanent deformation of the objects
Momentum is always conserved in a collision, but kinetic energy is only conserved if the collision is elastic; in a perfectly inelastic collision, the maximum possible amount of kinetic energy is transformed into other energy forms. -
Two ice skaters, initially at rest, push off from each other. Skater A (50 kg) moves away at 2.8 m/s. Skater B has mass 70 kg. What is skater B's speed?
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Answer. 2 m/s (in the opposite direction from skater A)
Total momentum starts at zero, so it must remain zero: $0=(50)(2.8)+(70)v_B$, giving $v_B=-2\text{ m/s}$, meaning 2 m/s opposite to skater A's direction. -
A 0.5 kg ball moving at 6 m/s toward a wall bounces straight back off the wall at 4 m/s. Taking the initial direction toward the wall as positive, what impulse does the wall deliver to the ball?
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Answer. -5 N·s (5 N·s directed away from the wall)
$\Delta p = m(v_f-v_i) = 0.5\text{ kg}(-4-6)\text{ m/s} = -5\text{ kg·m/s}$. By the impulse-momentum theorem, the impulse from the wall equals this change in momentum, directed opposite to the ball's original motion. -
A 2 kg ball moving east at 3 m/s collides and sticks with a 2 kg ball moving north at 3 m/s. What is the magnitude of their combined velocity right after the collision?
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Answer. About 2.1 m/s
Conserve momentum along each axis separately: $p_x=(2)(3)=6$, $p_y=(2)(3)=6$, total mass 4 kg, so $v_x=v_y=1.5\text{ m/s}$. Magnitude is $\sqrt{1.5^2+1.5^2}\approx2.1\text{ m/s}$, directed 45° between east and north. -
A small ball moving at speed v collides elastically head-on with a much more massive object initially at rest. What happens to the small ball immediately after the collision?
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Answer. It rebounds with a speed close to its original speed v, but in the opposite direction, while the massive object barely moves
In the limit where the target mass is much greater than the projectile mass, momentum and energy conservation together predict the light object essentially reverses its velocity while the heavy object's velocity change approaches zero, similar to a ball bouncing off a wall. -
For a fixed change in momentum delivered to an object, if the time of interaction is doubled, what happens to the average force required?
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Answer. The average force is cut in half
From $F\Delta t=\Delta p$, if $\Delta p$ is constant and $\Delta t$ doubles, then $F$ must be halved to keep the product the same.
What people get wrong
- Treating momentum as a scalar and dropping signs. Fix: pick one positive direction before writing any equation, and give every velocity (including 'at rest' as zero) a signed value consistent with that choice.
- Assuming kinetic energy is conserved in every collision. Fix: only assume KE conservation if the problem explicitly states 'elastic' or you have verified it by calculation; otherwise treat KE as unknown and solve using momentum alone.
- Applying conservation of momentum to a system that isn't isolated, such as ignoring friction, an incline, or an external push during the interaction. Fix: check that the net external force is zero (or that the interaction time is short enough that external impulses are negligible) before conserving momentum.
- Confusing impulse with force. Fix: remember impulse is $F\Delta t$, a quantity that depends on both the force and how long it acts; a small force over a long time can deliver the same impulse as a large force over a short time.
- In perfectly inelastic collisions, forgetting to combine the masses into one term on the final-momentum side. Fix: write $m_1v_{1i}+m_2v_{2i} = (m_1+m_2)v_f$ explicitly rather than reusing two separate final velocities.
- Misreading a bounce-back problem by using the same sign for incoming and outgoing velocity. Fix: if an object reverses direction, its final velocity must have the opposite sign from its initial velocity.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.