SME
University Physics I: Mechanics · Problem type
Impulse from a triangular force pulse
Use this for “A kick pushes a 0.43 kg ball at rest with a force that rises evenly to 2,400 N and falls back to 0 over 8 ms. What is the impulse, and how fast does the ball leave?”
Diagram
J
=
∫F
dt
=
1
2
×
Fₘₐₓ
×
Δt
J
=
1
2
×
1,200
×
0.01
=
6
N·s
Average
force
=
J
Δt
=
6
0.01
=
600
N:
the
dashed
rectangle
has
the
same
area.
Δv
=
J
m
=
6
0.15
=
40
m/s
This is an example. Type your own values to start a new problem.
J (N·s)
=
½
×
Fₘₐₓ (N)
×
Δt (s)
Favg
Average force
worked out (not typed)
m
Mass
example
Δv
Change in speed
worked out (not typed)
Formulas
J
=
1
2
×
Fₘₐₓ
×
Δt
6
=
1
2
×
1,200
×
0.01
Favg
=
J
Δt
600
=
6
0.01
Δv
=
J
m
40
=
6
0.15
Assumptions
•
The impulse is the area under the force–time graph, J = ∫F dt. Here the force rises in a straight line to its peak and falls back to 0, so the area is a triangle.
•
The average force is the steady force that gives the same impulse over the same contact time.
•
Only this force acts along the motion during the contact, so the impulse is the change in momentum, J = mΔv.
Step-by-step
WE KNOW
Fₘₐₓ = 1,200 N, Δt = 0.01 s, m = 0.15 kg
FIND
impulse (J), average force (Favg), change in speed (Δv)
Step 1 · Find impulse (J)
Use J = ½ × Fₘₐₓ × Δt
J
=
∫F
dt
is
the
area
under
the
pulse:
a
triangle,
half
its
base
Δt
times
its
height
Fₘₐₓ.
J
=
1
2
×
1,200
×
0.01
J
=
600
×
0.01
J
=
6
N·s
Step 2 · Find average force (Favg)
Use Favg = J ÷ Δt
Spread the impulse evenly over the contact time. For a triangle this is half the peak.
Favg
=
6
0.01
Favg = 600 N
Step 3 · Find change in speed (Δv)
Use Δv = J ÷ m
The
impulse
is
the
change
in
momentum,
J
=
mΔv:
divide
it
by
the
mass.
Δv
=
6
0.15
Δv = 40 m/s
Check
Put the numbers back in. Both sides should match.
6
=
1
2
×
1,200
×
0.01
✓
600
=
6
0.01
✓
40
=
6
0.15
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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