SME
Grade 11 Science · Problem type
Impulse: force × time = change in momentum
Use this for “A 0.2 kg ball at 25 m/s bounces straight back at 15 m/s after touching a wall for 0.05 s. What average force acts on it? Why would a softer wall push less hard?”
Diagram
Change in momentum: Δp = m(v − v₀) = 0.2 × (−15 − 25) = −8 kg·m/s
Average force: F = Δp/Δt = −8/0.05 = −160 N
Spread over 0.2 s, the same Δp needs only −40 N: a slower stop, a smaller force.
The area under the force–time graph, FΔt, is the impulse: the change in momentum.
This is an example. Type your own values to start a new problem.
F (N)
×
Δt (s)
=
m (kg)
×
(
v (m/s)
−
v₀ (m/s)
)
Δp
Change in momentum
calculated · kg·m/s
Formulas
Δp
=
m
×
(
v
−
v₀
)
−8 = 0.2 × (−15 − 25)
F
=
Δp/Δt
−160 = −8/0.05
Assumptions
•
+ is the way the object first moves; a force against it is negative.
•
F is the average force over the time Δt.
•
The dashed rectangle is the same change in momentum spread over 0.2 s: a longer push is a smaller force.
Step-by-step
WE KNOW
m = 0.2 kg, v₀ = 25 m/s, v = −15 m/s, Δt = 0.05 s
FIND
change in momentum (Δp), average force (F)
Step 1 · Find change in momentum (Δp)
Use Δp = m × (v − v₀)
The momentum after less the momentum before.
Δp = 0.2 × (−15 − 25)
Δp = 0.2 × (−40)
Δp
=
−8
kg·
m
/s
Step 2 · Find average force (F)
Use F = Δp/Δt
The change in momentum per second.
F
=
−8/0.05
F
=
−160
N
Check
Put the numbers back in. Both sides should match.
−8 = 0.2 × (−15 − 25) ✓
−160 = −8/0.05 ✓
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