SME
University Physics I: Mechanics
Momentum
Use this for “A 1200 kg car going east at 25 m/s hits a 1800 kg van going north at 10 m/s, and they lock together. How fast and in what direction do they slide?”
Diagram
p₁: 30,000 kg·m/s at 0°, so p₁ = ⟨30,000, 0⟩.
p₂: 37,500 kg·m/s at 90°, so p₂ = ⟨0, 37,500⟩.
Tip to tail: p = ⟨30,000, 37,500⟩
|p|
=
7500
√
41
≈
48,020
kg·m/s
at
51.34°.
This is an example. Type your own values to start a new problem.
m₁ (kg)
×
v₁ (m/s)
=
(
m₁
+
m₂
)
×
v (m/s)
cos(
θ
°)
m₂ (kg)
×
v₂ (m/s)
=
(
m₁
+
m₂
)
×
v
sin(
θ
°)
The arrows are the momenta: p₁ = m₁v₁ east and p₂ = m₂v₂ north. Their sum p carries the wreck, so its speed is v = p ÷ (m₁ + m₂).
pₓ
Momentum east
worked out (not typed) · kg·m/s
pᵧ
Momentum north
worked out (not typed) · kg·m/s
p
Total momentum
worked out (not typed) · kg·m/s
Kₗₒₛₜ
Kinetic energy lost
worked out (not typed) · J
Formulas
pₓ
=
m₁
×
v₁
30,000 = 1,500 × 20
pᵧ
=
m₂
×
v₂
37,500 = 2,500 × 15
p
=
√
pₓ² +
pᵧ
2
48,023.4318
=
√
30,
000
2
+ 37,
500
2
θ
=
tan⁻¹(
pᵧ
pₓ
)
51.3402
=
tan⁻¹(
37,500
30,000
)
v
=
p
m₁
+
m₂
12.0059
=
48,023.4318
1,500 + 2,500
Kₗₒₛₜ
=
m₁
×
m₂
×
v₁
2
+
v₂
2
2 × (
m₁
+
m₂
)
292,968.75
=
1,500
×
2,500
×
20
2
+
15
2
2 × (1,500 + 2,500)
Assumptions
•
No outside force acts during the crash (road friction is small over that short time), so momentum is kept east and north separately.
•
Object 1 moves east and object 2 moves north before the crash; afterward they stick and move as one mass m₁ + m₂.
•
Kinetic energy is not kept: the energy lost goes into bending, heat and sound.
•
The direction θ is measured from east toward north.
Step-by-step
WE KNOW
m₁ = 1,500 kg, v₁ = 20 m/s, m₂ = 2,500 kg, v₂ = 15 m/s
FIND
momentum east (pₓ), momentum north (pᵧ), total momentum (p), direction north of east (θ), speed after the crash (v), kinetic energy lost (Kₗₒₛₜ)
Step 1 · Find momentum east (pₓ)
Use pₓ = m₁ × v₁
Only object 1 moves east, so all the momentum east is its own, mass times speed.
pₓ = 1,500 × 20
pₓ = 30,000 kg·m/s
Step 2 · Find momentum north (pᵧ)
Use pᵧ = m₂ × v₂
Only object 2 moves north, so all the momentum north is its own, mass times speed.
pᵧ = 2,500 × 15
pᵧ = 37,500 kg·m/s
Step 3 · Find total momentum (p)
Use p = √(pₓ² + pᵧ²)
The east and north momenta are at right angles, so the total is the hypotenuse of their right triangle.
p
=
√
30,
000
2
+ 37,
500
2
p
=
√
900,000,000 + 1,406,250,000
p
=
√
2
,306,250,000
p
=
48,023.43
kg·m/s
Step 4 · Find direction north of east (θ)
Use θ = tan⁻¹(pᵧ ÷ pₓ)
The wreck moves along the total momentum; its angle from east has tangent north over east.
θ
=
tan⁻¹(
37,500
30,000
)
θ
=
51.3402°
Step 5 · Find speed after the crash (v)
Use v = p ÷ (m₁ + m₂)
After
the
crash
the
two
move
as
one
mass
m₁
+
m₂
carrying
the
same
total
momentum.
v
=
48,023.43
1,500 + 2,500
v
=
48,023.43
4,000
v
=
12.0059
m/s
Step 6 · Find kinetic energy lost (Kₗₒₛₜ)
Use Kₗₒₛₜ = m₁ × m₂ × (v₁² + v₂²) ÷ (2 × (m₁ + m₂))
Kinetic
energy
before,
1
2
m₁
v₁
2
+
1
2
m₂
v₂
2
,
less
after,
p
2
2
(
m₁
+
m₂
),
simplifies
to
this:
the
energy
of
their
speed
toward
each
other.
Kₗₒₛₜ
=
1,500
×
2,500
×
20
2
+
15
2
2 × (1,500 + 2,500)
Kₗₒₛₜ
=
1,500
×
2,500
×
20
2
+
15
2
2 × 4,000
Kₗₒₛₜ
=
1,500
×
2,500
×
400 + 225
8,000
Kₗₒₛₜ
=
1,500
×
2,500
×
625
8,000
Kₗₒₛₜ
=
3,750,000
×
625
8,000
Kₗₒₛₜ
=
2,343,750,000
8,000
Kₗₒₛₜ = 292,968.8 J
Check
Put the numbers back in. Both sides should match.
30,000 = 1,500 × 20 ✓
37,500 = 2,500 × 15 ✓
48,023.43
=
√
30,
000
2
+ 37,
500
2
✓
51.3402
=
tan⁻¹(
37,500
30,000
)
✓
12.0059
=
48,023.43
1,500 + 2,500
✓
292,968.8
=
1,500
×
2,500
×
20
2
+
15
2
2 × (1,500 + 2,500)
✓
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An elastic collision in 1D, both moving
Topic check
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