SME
Dynamics · Problem type
Average force from impulse and momentum
Use this for “A 0.145 kg ball reaches the bat at 40 m/s and leaves at 30 m/s the other way after 2 ms of contact. What impulse does the bat give, and what average force?”
Diagram
Change in momentum: J = m(v − v₀) = 0.145 × (−30 − 40) = −10.1 kg·m/s
Average force: Favg = J/Δt = −10.1/0.002 = −5,070 N
The area under the force–time graph, Favg × Δt, is the impulse: the change in momentum.
This is an example. Type your own values to start a new problem.
J (N·s)
=
m (kg)
(
v (m/s)
−
v₀ (m/s)
)
Favg (N)
=
J (N·s)
÷
Δt (ms)
Formulas
J
=
m
×
(
v
−
v₀
)
−10.15 = 0.145 × (−30 − 40)
Favg
=
J
Δt
−5,075
=
−
10.15
0.002
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
+ is the way the ball first moves: a velocity back the other way, or a force against it, is negative.
•
Impulse–momentum: the impulse J, the area under the force–time curve, equals the change in momentum m(v − v₀).
•
Favg is the steady force giving the same impulse over the contact time Δt; the peak force is higher.
Step-by-step
WE KNOW
m = 0.145 kg, v₀ = +40 m/s, v = −30 m/s, Δt = 2 ms
FIND
impulse (J), average force (Favg)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s).
Δt = 2 ms = 0.002 s (1 s = 1000 ms)
Step 2 · Find impulse (J)
Use J = m × (v − v₀)
The impulse is the change in momentum: the momentum after less the momentum before.
J
=
0.145
×
(−30
−
40)
J
=
0.145
×
(−70)
J
=
−10.15
N·s
Step 3 · Find average force (Favg)
Use Favg = J ÷ Δt
Spread the impulse evenly over the contact time; the sign says which way the force acts.
Favg
=
−
10.15
0.002
Favg = −5,075 N
Check
Put the numbers back in. Both sides should match.
−10.15 = 0.145 × (−30 − 40) ✓
−5,075
=
−
10.15
0.002
✓
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