SME
Grade 11 Science · Problem type
An elastic collision
Use this for “A 1 kg cart at 4 m/s meets a 3 kg cart coming the other way at 2 m/s. They bounce apart elastically. What are their velocities after?”
Diagram
Before: m₁v₁ + m₂v₂ = 1 × 4 + 3 × (−2) = −2 kg·m/s
After: m₁v₁′ + m₂v₂′ = 1 × (−5) + 3 × 1 = −2 kg·m/s
Kinetic energy: 14 J before, 14 J after: kept, the collision is elastic.
This is an example. Type your own values to start a new problem.
m₁
Mass of cart 1
example
m₂
Mass of cart 2
example
v₁
Velocity of cart 1 before
example
v₂
Velocity of cart 2 before
example
v₁′
Velocity of cart 1 after
calculated
v₂′
Velocity of cart 2 after
calculated
p
Total momentum
worked out (not typed) · kg·m/s
KE
Kinetic energy before
calculated · J
KE′
Kinetic energy after
calculated · J
Formulas
p
=
m₁
×
v₁
+
m₂
×
v₂
−2 = 1 × 4 + 3 × (−2)
v₁
′
=
((
m₁
−
m₂
)
×
v₁
+
2
×
m₂
×
v₂
)/(
m₁
+
m₂
)
−5 = ((1 − 3) × 4 + 2 × 3 × (−2))/(1 + 3)
v₂
′
=
((
m₂
−
m₁
)
×
v₂
+
2
×
m₁
×
v₁
)/(
m₁
+
m₂
)
1 = ((3 − 1) × (−2) + 2 × 1 × 4)/(1 + 3)
KE
=
1
2
×
m₁
×
v₁
2
+
1
2
×
m₂
×
v₂
2
14
=
1
2
×
1
×
4
2
+
1
2
×
3
×
(−2)
2
KE′
=
1
2
×
m₁
×
v₁
′²
+
1
2
×
m₂
×
v₂
′²
14
=
1
2
×
1
×
(−5)
2
+
1
2
×
3
×
1
2
Assumptions
•
Elastic: the carts bounce apart keeping both the total momentum p and the total kinetic energy.
•
+ is to the right; a cart moving left has a negative velocity.
•
Magnets or springy bumpers make a collision close to elastic.
Step-by-step
WE KNOW
m₁ = 1 kg, m₂ = 3 kg, v₁ = 4 m/s, v₂ = −2 m/s
FIND
total momentum (p), velocity of cart 1 after (v₁′), velocity of cart 2 after (v₂′), kinetic energy before (KE), kinetic energy after (KE′)
Step 1 · Find total momentum (p)
Use p = m₁ × v₁ + m₂ × v₂
Add the carts’ momenta, signs and all.
p
=
1
×
4
+
3
×
(−2)
p
=
4
−
6
p
=
−2
kg·m/s
Step 2 · Find velocity of cart 1 after (v₁′)
Use v₁′ = ((m₁ − m₂) × v₁ + 2 × m₂ × v₂)/(m₁ + m₂)
Momentum and kinetic energy both kept: this is cart 1’s velocity after.
v₁
′
=
((1
−
3)
×
4
+
2
×
3
×
(−2))/(1
+
3)
v₁
′
=
(−2
×
4
+
2
×
3
×
(−2))/(1
+
3)
v₁
′
=
(−8
+
6
×
(−2))/4
v₁
′
=
(−8
−
12)/4
v₁
′
=
−
20
4
v₁
′
=
−5
m/s
Step 3 · Find velocity of cart 2 after (v₂′)
Use v₂′ = ((m₂ − m₁) × v₂ + 2 × m₁ × v₁)/(m₁ + m₂)
The same rule with the carts swapped.
v₂
′
=
((3
−
1)
×
(−2)
+
2
×
1
×
4)/(1
+
3)
v₂
′
=
(2
×
(−2)
+
2
×
4)/(1
+
3)
v₂
′
=
(−4
+
8)/4
v₂
′
=
4
4
v₂
′
=
1
m/s
Step 4 · Find kinetic energy before (KE)
Use KE = ½ × m₁ × v₁² + ½ × m₂ × v₂²
Add
each
cart’s
1
2
mv²
before:
kinetic
energy
has
no
direction.
KE
=
1
2
×
1
×
4
2
+
1
2
×
3
×
(−2)
2
KE
=
1
2
×
16
+
1
2
×
3
×
4
KE = 8 + 1.5 × 4
KE = 8 + 6
KE = 14 J
Step 5 · Find kinetic energy after (KE′)
Use KE′ = ½ × m₁ × v₁′² + ½ × m₂ × v₂′²
Add
each
cart’s
1
2
mv²
after:
kinetic
energy
has
no
direction.
KE′
=
1
2
×
1
×
(−5)
2
+
1
2
×
3
×
1
2
KE′
=
1
2
×
25
+
1
2
×
3
KE′ = 12.5 + 1.5
KE′ = 14 J
Check
Put the numbers back in. Both sides should match.
−2 = 1 × 4 + 3 × (−2) ✓
−5 = ((1 − 3) × 4 + 2 × 3 × (−2))/(1 + 3) ✓
1 = ((3 − 1) × (−2) + 2 × 1 × 4)/(1 + 3) ✓
14
=
1
2
×
1
×
4
2
+
1
2
×
3
×
(−2)
2
✓
14
=
1
2
×
1
×
(−5)
2
+
1
2
×
3
×
1
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Momentum, impulse and collisions
Main lesson
Pushed apart from rest
Use this for “A spring pushes a 1.5 kg cart and a 0.5 kg cart apart from rest. The big cart moves left at 1 m/s. How fast does the small one go, and how much energy did the spring give?”
One velocity after, the other from momentum
Use this for “A 0.6 kg cart at 0.5 m/s meets a 0.4 kg cart rolling the other way at 0.25 m/s. The first cart slows to 0.1 m/s. How fast does the other cart move, and how much kinetic energy is lost?”
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?”