SME
University Physics I: Mechanics · Problem type
An elastic collision in 1D, both moving
Use this for “A 0.5 kg puck sliding east at 3 m/s catches a 1.5 kg puck sliding east at 1 m/s, and they bounce apart elastically. What are their velocities after?”
Diagram
Before: m₁v₁ + m₂v₂ = 2 × 6 + 4 × (−1.5) = 6 kg·m/s
After: m₁v₁′ + m₂v₂′ = 2 × (−4) + 4 × 3.5 = 6 kg·m/s
Kinetic energy: 40.5 J before, 40.5 J after: kept, the collision is elastic.
This is an example. Type your own values to start a new problem.
v₁′ (m/s)
=
−
+
×
v₁
+
2
×
+
×
(
v₂
)
v₂′ (m/s)
=
2
×
+
×
v₁
+
−
+
×
(
v₂
)
p
Total momentum
worked out (not typed) · kg·m/s
vcm
Velocity of the center of mass
worked out (not typed)
K
Kinetic energy before
worked out (not typed) · J
K′
Kinetic energy after
worked out (not typed) · J
Formulas
p
=
m₁
×
v₁
+
m₂
×
v₂
6 = 2 × 6 + 4 × (−1.5)
vcm
=
p
m₁
+
m₂
1
=
6
2 + 4
v₁
′
=
2
×
vcm
−
v₁
−4 = 2 × 1 − 6
v₂
′
=
2
×
vcm
−
v₂
3.5 = 2 × 1 − (−1.5)
K
=
1
2
×
m₁
×
v₁
2
+
1
2
×
m₂
×
v₂
2
40.5
=
1
2
×
2
×
6
2
+
1
2
×
4
×
(−1.5)
2
K
′
=
1
2
×
m₁
×
v₁
′²
+
1
2
×
m₂
×
v₂
′²
40.5
=
1
2
×
2
×
(−4)
2
+
1
2
×
4
×
3.5
2
Assumptions
•
The collision is head-on along one line and elastic: total momentum and total kinetic energy are both kept.
•
+ points from object 1 toward object 2; a velocity the other way is negative. No outside force acts during the short contact.
•
The center of mass moves at vcm = p ÷ (m₁ + m₂) throughout. Seen riding along with it, each object’s velocity just reverses, so v′ = 2vcm − v.
•
Written out, that is v₁′ = ((m₁ − m₂)v₁ + 2m₂v₂) ÷ (m₁ + m₂), and the same with 1 and 2 swapped for v₂′.
Step-by-step
WE KNOW
m₁ = 2 kg, v₁ = 6 m/s, m₂ = 4 kg, v₂ = −1.5 m/s
FIND
total momentum (p), velocity of the center of mass (vcm), velocity of object 1 after (v₁′), velocity of object 2 after (v₂′), kinetic energy before (K), kinetic energy after (K′)
Step 1 · Find total momentum (p)
Use p = m₁ × v₁ + m₂ × v₂
Each object’s momentum is its mass times its velocity, sign included; the total is their sum.
p
=
2
×
6
+
4
×
(−1.5)
p
=
12
+
(−6)
p
=
6
kg·m/s
Step 2 · Find velocity of the center of mass (vcm)
Use vcm = p ÷ (m₁ + m₂)
The
center
of
mass
carries
the
total
momentum
as
if
all
the
mass
were
there,
so
divide
p
by
m₁
+
m₂
.
vcm
=
6
2 + 4
vcm
=
6
6
vcm = 1 m/s
Step 3 · Find velocity of object 1 after (v₁′)
Use v₁′ = 2 × vcm − v₁
Relative
to
the
center
of
mass,
object
1
moves
at
v₁
−
vcm
before
and
the
reverse
after.
Add
vcm
back:
v₁
′
=
2vcm
−
v₁
.
v₁
′
=
2
×
1
−
6
v₁
′
=
−4
m/s
Step 4 · Find velocity of object 2 after (v₂′)
Use v₂′ = 2 × vcm − v₂
Object
2’s
velocity
reverses
about
vcm
the
same
way.
Written
out,
this
is
(
m₂
−
m₁
)
v₂
+ 2
m₁v₁
m₁
+
m₂
.
v₂
′
=
2
×
1
−
(−1.5)
v₂
′
=
3.5
m/s
Step 5 · Find kinetic energy before (K)
Use K = ½ × m₁ × v₁² + ½ × m₂ × v₂²
Add the two objects’ kinetic energies before the collision.
K
=
1
2
×
2
×
6
2
+
1
2
×
4
×
(−1.5)
2
K
=
1
2
×
2
×
36
+
1
2
×
4
×
2.25
K
=
36
+
2
×
2.25
K
=
36
+
4.5
K
=
40.5
J
Step 6 · Find kinetic energy after (K′)
Use K′ = ½ × m₁ × v₁′² + ½ × m₂ × v₂′²
Add
the
kinetic
energies
after.
In
an
elastic
collision
K
′
equals
K
:
the
check
that
the
velocities
are
right.
K
′
=
1
2
×
2
×
(−4)
2
+
1
2
×
4
×
3.5
2
K
′
=
1
2
×
2
×
16
+
1
2
×
4
×
12.25
K
′
=
16
+
2
×
12.25
K
′
=
16
+
24.5
K
′
=
40.5
J
Check
Put the numbers back in. Both sides should match.
6 = 2 × 6 + 4 × (−1.5) ✓
1
=
6
2 + 4
✓
−4 = 2 × 1 − 6 ✓
3.5 = 2 × 1 − (−1.5) ✓
40.5
=
1
2
×
2
×
6
2
+
1
2
×
4
×
(−1.5)
2
✓
40.5
=
1
2
×
2
×
(−4)
2
+
1
2
×
4
×
3.5
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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