Source: University of Washington General Physics notes
Tags: momentum, acceleration, impulse, Newton's second law, constant acceleration, variable acceleration, integration, kg m/s, p = mv, p = mat
Difficulty: Intermediate | Prerequisites: basic kinematics (velocity, acceleration, Newton's second law), algebra. Familiarity with integration is helpful for the variable-acceleration case but not essential for the core material.
Momentum and acceleration are two of the most fundamental quantities in classical mechanics, but they measure different things. Momentum (p = mv) captures how much "oomph" a moving object carries; acceleration (a) tells you how quickly its velocity is changing. The link between them runs through velocity: if you know an object's acceleration and the time it acts, you can find the velocity, and from there the momentum. This sits at the heart of dynamics and feeds directly into topics like impulse, collisions, and Newton's second law in its momentum form (F = dp/dt). You will need a solid grasp of basic kinematics before this material clicks.
Momentum equals mass times velocity (p = mv). If you only have acceleration, multiply mass by the product of acceleration and time (for constant acceleration) or by the integral of acceleration over time (for variable acceleration). The mass of the object always stays in the equation.
Momentum (p)
The product of an object's mass and its velocity: p = mv. It is a vector quantity, measured in kg·m/s. Think of it as a measure of how hard it would be to stop something that is moving.
Acceleration (a)
The rate of change of velocity with respect to time: a = dv/dt, measured in m/s². In simple terms, this tells you how quickly something is speeding up, slowing down, or changing direction.
Velocity (v)
The rate of change of position with respect to time, measured in m/s. It is a vector: speed with a direction attached.
Impulse
The product of a force and the time interval over which it acts, equal to the change in momentum (J = Δp = FΔt). Think of it as the "push over time" that changes an object's momentum.
Integration (∫a dt)
A calculus operation that sums up a continuously changing quantity over an interval. In this context, integrating acceleration over time gives you velocity. In simple terms, it is the mathematical way of adding up lots of tiny slices of acceleration to find the total change in velocity.
When acceleration is constant, velocity increases linearly with time: v = a × t (assuming the object starts from rest, i.e. v₀ = 0).
Momentum then follows directly: p = m × v = m × a × t.
If the object has an initial velocity v₀, the full expression is: p = m × (v₀ + a × t).
This is the simpler and more common exam scenario.
When acceleration changes over time, you cannot simply multiply a single value of a by t.
Instead, velocity is found by integrating acceleration with respect to time: v = ∫a(t) dt.
Momentum becomes: p = m × ∫a(t) dt.
Graphically, the integral of a(t) over a time interval equals the area under the acceleration-time curve for that interval.
This approach requires calculus and appears more often in advanced or calculus-based physics courses.
Given: mass m = 2 kg, constant acceleration a = 5 m/s², duration t = 3 s, starting from rest.
Step 1: Find velocity. v = a × t = 5 × 3 = 15 m/s.
Step 2: Find momentum. p = m × v = 2 × 15 = 30 kg·m/s.
Equivalently: p = m × a × t = 2 × 5 × 3 = 30 kg·m/s.
Momentum is measured in kilogram-metres per second (kg·m/s), sometimes written as N·s (newton-seconds), since 1 N = 1 kg·m/s².
Always check dimensional consistency: mass (kg) × acceleration (m/s²) × time (s) = kg·m/s. The seconds cancel correctly.
p = m \times vv = v_0 + a \times t \quad \text{(constant acceleration)}p = m(v_0 + a \times t) \quad \text{(constant acceleration, from rest: } v_0 = 0\text{)}p = m \int a(t)\, dt \quad \text{(variable acceleration)}F = \frac{dp}{dt} \quad \text{(Newton's second law in momentum form)}Car safety engineers use the relationship between acceleration, time, and momentum to design crumple zones and airbags. By extending the time over which a passenger decelerates during a crash, these systems reduce the peak force while producing the same change in momentum.
Rocket scientists calculate the momentum of a spacecraft by integrating its (variable) acceleration over the burn time of the engine, accounting for the fact that the rocket's mass decreases as fuel is consumed.
Students often think momentum and acceleration are the same thing. They are not. Acceleration is about how velocity changes; momentum is about how much motion an object carries at a given instant.
Students sometimes write p = m × a and stop there. That expression has units of newtons (kg·m/s²), which is a force, not a momentum. You need the time factor: p = m × a × t.
Forgetting initial velocity is a frequent mistake. If the object was already moving before the acceleration began, you must include v₀ in the calculation: p = m(v₀ + at).
Students occasionally try to use the constant-acceleration formula when acceleration is changing. If a(t) varies, you must integrate; plugging in a single value of a will give the wrong answer.
⚠️ Expect problems that give you mass, acceleration, and time, and ask for momentum. The trap is writing p = ma instead of p = mat.
⚠️ Unit analysis questions are common. You may be asked to show that kg × m/s² × s simplifies to kg·m/s.
⚠️ Problems may give a non-zero initial velocity. Read the question carefully for phrases like "initially travelling at" or "already moving at."
⚠️ In calculus-based courses, be ready to integrate a given a(t) function over a time interval and then multiply by mass.
True or False: Momentum and force have the same SI units. (Answer: False. Momentum is kg·m/s; force is kg·m/s², i.e. newtons.)
Fill in the blank: For an object starting from rest under constant acceleration, p = m × ___ × ___. (Answer: a × t)
True or False: If acceleration varies with time, you can still use p = mat. (Answer: False. You must integrate a(t) over time.)
A 5 kg box accelerates from rest at 2 m/s² for 4 s. Its momentum is ___ kg·m/s. (Answer: 40)
True or False: Doubling the mass while keeping acceleration and time the same will double the momentum. (Answer: True.)
Q: A 3 kg object accelerates uniformly from rest at 4 m/s² for 6 seconds. What is its momentum at the end of the 6 seconds?
A: v = at = 4 × 6 = 24 m/s. p = mv = 3 × 24 = 72 kg·m/s.
Q: An object of mass 1.5 kg is already moving at 10 m/s when a constant acceleration of 3 m/s² is applied for 5 seconds. What is the final momentum?
A: v = v₀ + at = 10 + (3 × 5) = 25 m/s. p = mv = 1.5 × 25 = 37.5 kg·m/s.
Q: Why can you not simply write p = ma to find momentum?
A: The expression ma has units of kg·m/s² (newtons), which is force, not momentum. You need to multiply by time as well: p = m × a × t, which gives units of kg·m/s.
Q: A particle's acceleration is given by a(t) = 6t m/s². If the particle has a mass of 2 kg and starts from rest, what is its momentum at t = 3 s?
A: v = ∫₀³ 6t dt = 3t² evaluated from 0 to 3 = 27 m/s. p = mv = 2 × 27 = 54 kg·m/s.
Q: Explain why kg·m/s and N·s are equivalent units for momentum.
A: 1 N = 1 kg·m/s². Multiplying by seconds: 1 N·s = 1 kg·m/s² × s = 1 kg·m/s. Both express momentum.
This material connects directly to impulse and the impulse-momentum theorem (J = FΔt = Δp), which is the bridge between force, time, and changes in momentum.
It also underpins collision analysis (elastic and inelastic), where conservation of momentum is the central principle. If you can calculate momentum from acceleration and time, you can set up the "before" side of any collision problem.
Newton's second law in its original form (F = dp/dt) is the deeper version of F = ma and shows up in variable-mass problems such as rockets and conveyor belts.
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