Impulse and Force: Seat Belt vs Dashboard Impact

Physics Motion 11th-12th Grade
PROBLEM
Identical twins, each with a mass of 51 kg, are riding in a car traveling at 20 m/s. One of the girls is wearing a seat belt and the other is not. The car is in an accident where the girl wearing the seat belt is brought to a rest in 0.120 seconds and the dashboard brings the other girl to a stop in only 0.0080 seconds. a. What is the impulse each girl experiences? b. What is the force exerted on each girl?

Visualizing the Problem

Identical twins, each with a mass of 51 kg, are riding in a car traveling at 20 m/s. One of the girls is wearing a...

By the End of This Page

  • Master the impulse-momentum theorem and its relationship to force
  • Understand why stopping time dramatically affects collision forces
  • Calculate impulse from momentum changes in real-world scenarios
  • Connect physics equations to life-saving safety engineering
  • Analyze multi-part problems involving both impulse and force calculations

Solution: Method 1 — The Impulse-Momentum Approach

Step 1 — Apply the impulse-momentum theorem

Impulse equals change in momentum: J = Δp = m(vf - vi). Since both girls have identical mass and undergo the same velocity change, they experience identical impulse. Force is then found using F = J/Δt.

J = m(vf - vi)
F = J/Δt

Step 2 — Calculate the impulse for both girls

Both girls start at vi = 20 m/s and end at vf = 0 m/s, with mass m = 51 kg.

J = m(vf - vi)
J = 51 kg × (0 - 20) m/s
J = 51 kg × (-20 m/s)
J = -1020 kg⋅m/s

The negative sign indicates the impulse opposes the initial motion direction. The magnitude is 1020 kg⋅m/s for both girls.

Step 3 — Find force on the seat-belted girl

The seat belt extends the stopping time to Δt = 0.120 s.

F = J/Δt
F = (-1020 kg⋅m/s)/(0.120 s)
F = -8500 N

The magnitude is 8500 N.

Step 4 — Find force on the unbelted girl

The dashboard impact has a much shorter stopping time of Δt = 0.0080 s.

F = J/Δt
F = (-1020 kg⋅m/s)/(0.0080 s)
F = -127,500 N

The magnitude is 127,500 N — fifteen times larger than the seat belt force.

Solution: Method 2 — Kinematics with Newton's Second Law

Step 1 — Find the deceleration for each girl

Using kinematics: vf = vi + aΔt, so a = (vf - vi)/Δt.

For the seat-belted girl:

a₁ = (0 - 20 m/s)/(0.120 s) = -166.7 m/s²

For the unbelted girl:

a₂ = (0 - 20 m/s)/(0.0080 s) = -2500 m/s²

Step 2 — Apply Newton's second law to find forces

Using F = ma:

Seat-belted girl:

F₁ = ma₁ = 51 kg × (-166.7 m/s²) = -8500 N

Unbelted girl:

F₂ = ma₂ = 51 kg × (-2500 m/s²) = -127,500 N

Step 3 — Calculate impulse from force

Using J = F × Δt:

J₁ = (-8500 N) × (0.120 s) = -1020 kg⋅m/s
J₂ = (-127,500 N) × (0.0080 s) = -1020 kg⋅m/s

Both methods confirm identical impulse values, as expected from physics principles.

Answer:
a. Impulse: 1020 kg⋅m/s for both girls (magnitude)
b. Forces: Seat-belted girl: 8,500 N; Unbelted girl: 127,500 N

Verification

Let's verify our impulse calculation using the definition J = F⋅Δt:

Seat-belted girl: J = 8500 N × 0.120 s = 1020 kg⋅m/s ✓

Unbelted girl: J = 127,500 N × 0.0080 s = 1020 kg⋅m/s ✓

We can also verify using momentum change directly:

Initial momentum: p₁ = mv₁ = 51 kg × 20 m/s = 1020 kg⋅m/s
Final momentum: pf = mv_f = 51 kg × 0 m/s = 0 kg⋅m/s
Change: Δp = 0 - 1020 = -1020 kg⋅m/s ✓

The magnitude matches our calculated impulse, confirming our solution.

The Life-Saving Physics

This problem reveals the fundamental physics behind automotive safety design. Both passengers experience identical impulse because they undergo the same momentum change. However, the force—which determines injury severity—depends entirely on the time over which this change occurs.

The seat belt increases stopping time by a factor of 15 (from 0.008 s to 0.120 s), which decreases the force by the same factor. This transforms a potentially fatal impact force of 127,500 N into a survivable 8,500 N.

Engineering Insight: Modern cars extend this principle further with crumple zones, airbags, and seat belt pretensioners—all designed to maximize stopping time and minimize peak forces on passengers.

Watch Out for These Mistakes

✗ Thinking impulse depends on stopping time
Wrong: "The seat-belted girl has larger impulse because she stops more slowly."
Right: Impulse depends only on momentum change. Both girls have identical impulse because they have the same mass and velocity change, regardless of stopping time.
✗ Confusing force and impulse units
Wrong: Reporting force as "8500 kg⋅m/s" or impulse as "1020 N"
Right: Force has units of Newtons (N), while impulse has units of kg⋅m/s. These are equivalent to N⋅s, but the distinction matters for understanding.
✗ Using positive and negative signs inconsistently
Wrong: Mixing up the direction conventions halfway through the problem
Right: Choose a positive direction at the start (e.g., forward motion = positive) and stick with it throughout. The negative impulse and force indicate they oppose the initial motion.

The Impulse-Momentum Connection

This problem demonstrates one of physics' most important relationships. The impulse-momentum theorem states that:

J = Δp = F⋅Δt

This seemingly simple equation connects three fundamental quantities:

  • Impulse (J): The "kick" delivered to change motion
  • Momentum change (Δp): How much the motion actually changes
  • Force and time (F⋅Δt): How that change is delivered

In collisions, the momentum change is fixed by the initial and final velocities. The key insight is that this fixed impulse can be delivered as a large force over short time, or a smaller force over longer time. Safety engineering always chooses the latter.

Real-World Applications

Automotive Safety: This exact calculation is used to design seat belts, airbags, and crumple zones. Engineers minimize peak forces while managing the space constraints of a vehicle interior.

Sports Equipment: Boxing gloves, football helmets, and gymnastics mats all work by increasing impact time to reduce forces on athletes.

Packaging Design: Bubble wrap, foam padding, and air cushions protect fragile items by extending the time over which shipping impacts occur.

What If?

1
Higher Speed Collision
The same twins are in a highway collision traveling at 35 m/s instead of 20 m/s. The seat belt still stops one girl in 0.120 s, and the dashboard stops the other in 0.0080 s. Find the impulse and forces for each girl.
Step 1 — Calculate the impulse

Both girls: J = m(vf - vi) = 51 kg × (0 - 35) = -1785 kg⋅m/s

Step 2 — Find force on seat-belted girl

F₁ = J/Δt = -1785/(0.120) = -14,875 N

Step 3 — Find force on unbelted girl

F₂ = J/Δt = -1785/(0.0080) = -223,125 N

Step 4 — Verify and compare

Answer: Impulse: 1785 kg⋅m/s each. Forces: 14,875 N (seat belt), 223,125 N (dashboard). The higher speed increases all values by the factor 35/20 = 1.75.

2
Survival Threshold
Medical research suggests humans can survive forces up to 50,000 N in such collisions. For the unbelted 51 kg girl traveling at 20 m/s, what minimum stopping time would be needed for survival?
Step 1 — Calculate the impulse

J = m(vf - vi) = 51 kg × (0 - 20) = -1020 kg⋅m/s

Step 2 — Use force limit to find time

F = J/Δt, so Δt = J/F = 1020/50,000 = 0.0204 s

Step 3 — Compare to original scenario

The dashboard impact time was 0.0080 s, which is less than the 0.0204 s survival threshold.

Step 4 — Conclusion

Answer: Minimum stopping time: 0.0204 s. The original dashboard impact (0.0080 s) would likely be fatal, while the seat belt time (0.120 s) provides a large safety margin.

3
Airbag Addition
The unbelted girl hits an airbag that increases her stopping time to 0.040 s (still much less than the seat belt). Calculate her force and compare it to both the original dashboard force and the seat belt force.
Step 1 — Use same impulse

J = -1020 kg⋅m/s (unchanged from original problem)

Step 2 — Calculate airbag force

F = J/Δt = -1020/(0.040) = -25,500 N

Step 3 — Compare all three forces

Dashboard: 127,500 N
Airbag: 25,500 N
Seat belt: 8,500 N

Step 4 — Analyze the improvement

Answer: Airbag force: 25,500 N. This is 5× smaller than the dashboard but 3× larger than the seat belt. The airbag saves lives but the seat belt remains superior.

4
Different Masses
Now the twins aren't identical. The seat-belted girl has mass 60 kg, while the unbelted girl has mass 45 kg. Both travel at 20 m/s initially. Find the impulse and force for each, using the original stopping times (0.120 s and 0.0080 s).
Step 1 — Calculate impulse for 60 kg girl

J₁ = m₁(vf - vi) = 60 kg × (0 - 20) = -1200 kg⋅m/s

Step 2 — Calculate impulse for 45 kg girl

J₂ = m₂(vf - vi) = 45 kg × (0 - 20) = -900 kg⋅m/s

Step 3 — Find forces

Seat belt (60 kg): F₁ = -1200/0.120 = -10,000 N
Dashboard (45 kg): F₂ = -900/0.0080 = -112,500 N

Step 4 — Compare results

Answer: Impulses: 1200 kg⋅m/s (60 kg girl), 900 kg⋅m/s (45 kg girl). Forces: 10,000 N (seat belt), 112,500 N (dashboard). The lighter girl experiences less impulse but still faces a potentially fatal dashboard force.

Frequently Asked Questions

What is impulse and how do you calculate it in a collision?+
Impulse is the change in momentum of an object and equals the force applied multiplied by the time interval. Use the impulse-momentum theorem: J = Δp = m(vf - vi). In this collision problem, both passengers have the same impulse of 1020 kg⋅m/s because they experience the same change in momentum from 20 m/s to 0 m/s.
Why does a longer stopping time reduce the force in a collision?+
Force equals impulse divided by time (F = J/Δt). Since impulse is fixed by the momentum change, increasing the stopping time decreases the average force. Here, the seat belt extends stopping time from 0.008 s to 0.120 s, reducing force from 127,500 N to 8,500 N - a 15-fold decrease that can save a life.
How do you solve two-part physics problems involving impulse and force?+
First calculate impulse using the momentum change: J = m(vf - vi). Then find force by dividing impulse by the time interval: F = J/Δt. Always work in consistent units and remember that impulse is the same for objects with identical mass and velocity changes, but forces differ based on stopping times.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-08-27