Unit 4 of AP Physics 1: Linear Momentum and Collisions.
This unit covers impulse, conservation of momentum and elastic and inelastic collisions — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers impulse, conservation of momentum and elastic and inelastic collisions — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Impulse
Impulse is the product of the net force and the time interval over which it acts, equal to the change in momentum of an object (J = FΔt = Δp). Students must be able to interpret force-time graphs, where the area under the curve equals impulse. The impulse-momentum theorem connects force, time, and velocity change and is central to free-response analysis.
Key Points
- J = FΔt = Δp = mΔv; units are N·s or kg·m/s (equivalent)
- Area under a Force vs. Time graph = impulse delivered
- A larger contact time means a smaller average force for the same impulse (e.g., airbags, padding)
- Impulse is a vector — direction matches the direction of the net force
A 0.5 kg ball moving at 6 m/s to the right hits a wall and bounces back at 4 m/s. The collision lasts 0.02 s. Find the average force exerted by the wall on the ball.
First, define rightward as positive. The initial momentum is +3 kg·m/s and the final momentum is −2 kg·m/s, so Δp = −5 kg·m/s. Using J = FΔt, the average force is F = Δp/Δt = −5/0.02 = −250 N. The negative sign indicates the force is directed to the left, away from the wall.
2 Conservation of Momentum
In a closed, isolated system (no net external force), total momentum is conserved: Σp_initial = Σp_final. Students must identify whether a system is isolated before applying this principle, and recognize that internal forces (e.g., between two colliding carts) do not change total system momentum. This law applies to all collision and explosion scenarios on the AP exam.
Key Points
- p_total is conserved only when net external force = 0 (isolated system)
- m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f is the standard form for two-object systems
- Applies to explosions (e.g., recoil) as well as collisions — total initial p is often zero
- Friction from an external surface IS an external force and violates isolation; flag this on FRQs
A 3 kg cart moving right at 4 m/s collides with a stationary 1 kg cart on a frictionless track. After the collision, the 3 kg cart moves right at 1 m/s. Find the final velocity of the 1 kg cart.
Apply conservation of momentum: (3)(4) + (1)(0) = (3)(1) + (1)v₂f. This gives 12 = 3 + v₂f, so v₂f = 9 m/s to the right. The track is frictionless, confirming an isolated system where the principle is valid.
3 Elastic Collisions
An elastic collision conserves both total momentum and total kinetic energy. On the AP exam, students must be able to verify whether a collision is elastic by checking if KE_total is the same before and after, not just assume it. Perfectly elastic collisions are an idealization; real collisions are never perfectly elastic but some problems treat them as such.
Key Points
- Both momentum AND kinetic energy are conserved: ΣKE_i = ΣKE_f
- To verify elasticity, compute ½mv² for each object before and after and compare totals
- In a 1D elastic collision between equal masses where one is at rest, the moving object stops and the stationary one moves at the original speed (classic result)
- Do NOT assume a collision is elastic unless the problem states it or you verify it numerically
A 2 kg ball moving at 5 m/s strikes a stationary 2 kg ball on a frictionless surface. After the collision, the first ball is at rest and the second moves at 5 m/s. Is this collision elastic?
Check kinetic energy: KE_initial = ½(2)(5²) = 25 J; KE_final = ½(2)(0²) + ½(2)(5²) = 0 + 25 = 25 J. Since kinetic energy is conserved, the collision is elastic. This also matches the equal-mass elastic collision rule where the first object transfers all its kinetic energy to the second.
4 Inelastic Collisions
In an inelastic collision, momentum is conserved but kinetic energy is not — some KE is converted to heat, sound, or deformation. A perfectly inelastic collision is the special case where the two objects stick together and move as one, which is the most common collision type tested on the AP exam. Students must calculate the loss in kinetic energy and explain where it goes.
Key Points
- Momentum is still conserved; use m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f for perfectly inelastic
- Kinetic energy is NOT conserved — ΔKE = KE_f − KE_i is negative (energy is lost to internal energy)
- The amount of KE lost is a common FRQ calculation: compute KE_i and KE_f separately, then subtract
- Perfectly inelastic collisions produce the maximum possible loss of KE for a given momentum constraint
A 4 kg block moving at 6 m/s to the right collides and sticks to a stationary 2 kg block. Find their combined velocity and the kinetic energy lost.
Using conservation of momentum: (4)(6) + (2)(0) = (4 + 2)v_f, so v_f = 24/6 = 4 m/s to the right. Initial KE = ½(4)(6²) = 72 J; final KE = ½(6)(4²) = 48 J. The kinetic energy lost is 72 − 48 = 24 J, which was converted to internal energy (heat and deformation) during the collision.
Questions, answered.
What is Linear Momentum and Collisions?
Linear Momentum and Collisions is Unit 4 of AP Physics 1, covering impulse, conservation of momentum and elastic and inelastic collisions.
How to study for AP Physics 1 Unit 4?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.