Impulse and Momentum Problems- Worked Examples

Impulse and Momentum Problems: What You Actually Need to Know

Most students stumble on impulse and momentum because they memorize formulas without understanding the connection between them. This guide fixes that. You'll work through real problems, see exactly how the equations connect, and learn where people consistently go wrong.

No motivational filler. Just physics.

The Core Definitions First

Before touching any problem, these two concepts must be solid.

What Momentum Actually Is

Momentum (p) is mass times velocity. That's it.

p = mv

It's a vector quantity, which means direction matters. A car moving east at 20 m/s has different momentum than the same car moving west at 20 m/s, even though the speed is identical.

Units: kg·m/s

What Impulse Actually Is

Impulse (J) is force multiplied by the time interval it acts. It also equals the change in momentum.

J = FΔt = Δp

This is the Impulse-Momentum Theorem, and it's the key to solving most collision and impact problems.

Units: N·s or kg·m/s (same thing, different label)

The Impulse-Momentum Theorem Explained

When a force acts on an object over time, it changes the object's momentum. The change equals the impulse.

J = FΔt = p_final - p_initial = mv_f - mv_i

That's the entire relationship. Force times time equals the change in momentum.

Why does this matter? Because in many problems you won't have all the information directly. You might know the masses and velocities before and after a collision, but not the force or time. Or vice versa.

Conservation of Momentum: When It Applies

Here's where students get sloppy. Momentum is conserved only in closed systems with no external forces.

What counts as a closed system?

• Two billiard balls colliding on a frictionless table ✓
• A rocket expelling fuel in empty space ✓
• A car crashing into a wall (external force from wall) ✗
• Two hockey players colliding on ice (friction matters, but often ignored for simplicity) ✓

In a closed system: p_initial = p_final

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

(Primes indicate final values after interaction)

Worked Example 1: Basic Impulse Calculation

Problem: A 0.5 kg baseball approaches a bat at 30 m/s. After being hit, it leaves at 40 m/s in the opposite direction. The contact time is 0.02 seconds. Find the average force exerted by the bat.

Solution:

First, establish direction. Call the initial direction positive.

Initial momentum: p_i = (0.5)(30) = 15 kg·m/s

Final momentum: p_f = (0.5)(-40) = -20 kg·m/s

Change in momentum: Δp = p_f - p_i = -20 - 15 = -35 kg·m/s

Impulse equals change in momentum: J = -35 N·s

Force: J = FΔt → F = J/Δt = -35/0.02 = -1750 N

The negative sign indicates the force direction is opposite to the initial motion. Magnitude is 1750 N.

Worked Example 2: Conservation of Momentum in a Collision

Problem: A 2 kg cart moving at 3 m/s collides with a stationary 4 kg cart. They stick together after impact. What is their combined velocity?

Solution:

This is a perfectly inelastic collision—objects stick together.

Apply conservation of momentum:

m₁v₁ + m₂v₂ = (m₁ + m₂)v_f

(2)(3) + (4)(0) = (2 + 4)v_f

6 = 6v_f

v_f = 1 m/s

Notice kinetic energy is not conserved here. Some is lost to heat, sound, deformation. Momentum is still conserved.

Worked Example 3: Elastic vs Inelastic Collision

Problem: A 3 kg ball moving at 4 m/s hits a stationary 2 kg ball. After an elastic collision, the 3 kg ball slows to 2 m/s. Find the velocity of the 2 kg ball.

Solution:

In an elastic collision, both momentum and kinetic energy are conserved.

Step 1: Conservation of momentum

m₁v₁ = m₁v₁' + m₂v₂'

(3)(4) = (3)(2) + (2)v₂'

12 = 6 + 2v₂'

2v₂' = 6

v₂' = 3 m/s

Step 2: Verify with kinetic energy (good practice)

Initial KE: ½(3)(16) = 24 J

Final KE: ½(3)(4) + ½(2)(9) = 6 + 9 = 15 J

Wait—that's not matching. Let me recalculate.

Final KE: ½(3)(4) = 6 J for first ball

½(2)(9) = 9 J for second ball

Total = 15 J

That's not equal to 24 J. Something's wrong with the problem setup.

Actually, this is a common trap. If the problem states it's elastic, the given final velocity must satisfy both conservation laws. Let me recalculate assuming I made an arithmetic error.

Using momentum: 12 = 6 + 2v₂' → v₂' = 3 m/s. That's correct.

For elastic collision, v₂' should equal (2m₁v₁)/(m₁+m₂) when v₂ = 0. That's (2×3×4)/(3+2) = 24/5 = 4.8 m/s.

The given numbers don't represent an elastic collision. In an elastic collision with these masses, the target ball would move at 4.8 m/s, not 3 m/s.

Lesson: Always verify your inputs. Textbook problems sometimes have errors. If energy doesn't balance in an "elastic" collision, the problem statement is wrong.

Worked Example 4: Impulse from Variable Force

Problem: A 1 kg ball hits a wall at 10 m/s and bounces back at 8 m/s. The force-time graph shows a triangular impulse with peak force 500 N and base width 0.04 s. Find the actual impulse and compare to the force-time integral.

Solution:

Momentum approach:

Initial momentum: 1 × 10 = 10 kg·m/s

Final momentum: 1 × (-8) = -8 kg·m/s

Impulse = Δp = -8 - 10 = -18 N·s

Area under force-time curve (triangle):

Area = ½ × base × height = ½ × 0.04 × 500 = 10 N·s

The numbers don't match. This means either the graph doesn't represent the actual collision, or there's additional impulse from other forces (like the ball's weight during contact, though that's usually negligible for short-duration impacts).

For short collisions, we often ignore gravity. The impulse from the wall must equal the momentum change. If the force-time integral gives 10 N·s but Δp is -18 N·s, the graph parameters need adjustment.

Key point: The impulse-momentum theorem gives you the actual impulse. The force-time graph is a model. If they disagree, the model is incomplete.

Worked Example 5: Recoil Problem

Problem: A 60 kg person stands on a frictionless 20 kg skateboard. They throw a 5 kg ball backward at 8 m/s relative to the ground. What is the person's resulting velocity?

Solution:

Initial momentum = 0 (everything at rest)

Final momentum = m_person v_p + m_ball v_b = 0

(60)v_p + (5)(8) = 0

60v_p + 40 = 0

v_p = -40/60 = -0.67 m/s

The person moves backward (negative direction) at 0.67 m/s.

Note: The ball's velocity is given relative to the ground, so no velocity addition is needed here.

Collision Types Quick Reference

Common Mistakes That Cost Points

1. Ignoring direction signs.

Momentum is a vector. If you set one direction as positive, be consistent. Many students solve problems correctly except for the sign and lose half the points.

2. Mixing up elastic and inelastic assumptions.

If a problem doesn't specify, you can still use conservation of momentum. But kinetic energy is only conserved in elastic collisions. Don't assume energy conservation unless explicitly stated.

3. Forgetting to convert units.

Mass in kg, velocity in m/s. If you're given grams or km/h, convert first. A surprising number of wrong answers come from unit errors, not concept errors.

4. Using the wrong mass in recoil problems.

When a gun fires, the bullet goes forward and the gun recoils. Use the remaining mass of the system, not the original mass before firing.

5. Applying conservation incorrectly to single objects.

Momentum conservation applies to systems. If a single object experiences an external force (like gravity during a jump), you cannot use conservation for that object alone. You must include whatever is exerting the external force.

Getting Started: How to Approach Any Problem

Step 1: Identify the system.

What objects are interacting? Is it a closed system?

Step 2: Choose your coordinate system.

Pick one direction as positive. Write it down. Stick to it.

Step 3: List what you know.

Masses, velocities before and after, time intervals, forces—whatever the problem gives you.

Step 4: Decide which equation applies.

Looking for force or time? Use impulse-momentum theorem. Looking for velocities after collision? Use conservation of momentum.

Step 5: Plug in and solve.

Keep algebraic symbols as long as possible. Substitute numbers only at the end. This reduces arithmetic errors.

Step 6: Check your work.

Does the direction make sense? Is the magnitude reasonable? If a car recoils harder than the bullet that was fired into it, something is wrong.

When to Use Which Equation

• J = FΔt = Δp → You know force and time, or you know momentum change and need force/time
• m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' → You need final velocities after collision
• KE = ½mv² → Only for elastic collisions, and only as a check unless the problem specifically asks for energy

These three equations solve 90% of standard impulse and momentum problems.