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Grade 11 Science · Physics

Momentum, impulse and collisions

Use this for “A 2 kg cart at 3 m/s hits a 1 kg cart coming at 3 m/s, and they stick. How fast do they move, and how much kinetic energy is lost?”

Diagram

Beforetotal 3 kg·m/s2 kgp1 6v1 3 m/s1 kgp2 −3v2 −3 m/sAftertotal 3 kg·m/s2 kgv′ 1 m/s1 kg
Before: m₁v₁ + m₂v₂ = 2 × 3 + 1 × (−3) = 3 kg·m/s
After: (m₁ + m₂)v′ = 3, so v′ = 3/3 = 1 m/s
Kinetic energy: 13.5 J before, 1.5 J after: 12 J turned to heat and sound.
This is an example. Type your own values to start a new problem.
m₁ (kg)
×
v₁ (m/s)
+
m₂ (kg)
×
(
v₂ (m/s)
)
=
(
m₁ (kg)
+
m₂ (kg)
)
×
v′ (m/s)
p
Total momentum
calculated · kg·m/s
KE
Kinetic energy before
calculated · J
KE′
Kinetic energy after
calculated · J

Formulas

p
=
m₁
×
v₁
+
m₂
×
v₂
3 = 2 × 3 + 1 × (−3)
v′
=
p
/(
m₁
+
m₂
)
1 = 3/(2 + 1)
KE
=
1
2
×
m₁
×
v₁
2
+
1
2
×
m₂
×
v₂
2
13.5
=
1
2
×
2
×
3
2
+
1
2
×
1
×
(−3)
2
KE′
=
1
2
×
(
m₁
+
m₂
)
×
v′²
1.5
=
1
2
×
(2
+
1)
×
1
2

Assumptions

•
No outside push along the track, so the total momentum is the same before and after.
•
+ is to the right: a cart moving left has a negative velocity and momentum.
•
The carts stick together: momentum is kept, but some kinetic energy turns to heat and sound.

Step-by-step

WE KNOW
m₁ = 2 kg, v₁ = 3 m/s, m₂ = 1 kg, v₂ = −3 m/s
FIND
total momentum (p), velocity together after (v′), 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
=
2
×
3
+
1
×
(−3)
p
=
6
−
3
p
=
3
kg·m/s
Step 2 · Find velocity together after (v′)
Use v′ = p/(m₁ + m₂)
The same momentum, now carried by both masses together.
v′ = 3/(2 + 1)
v′
=
3
3
v′ = 1 m/s
Step 3 · 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
×
2
×
3
2
+
1
2
×
1
×
(−3)
2
KE
=
1
2
×
2
×
9
+
1
2
×
9
KE = 9 + 4.5
KE = 13.5 J
Step 4 · Find kinetic energy after (KE′)
Use KE′ = ½ × (m₁ + m₂) × v′²
After, both masses move together at v′.
KE′
=
1
2
×
(2
+
1)
×
1
2
KE′
=
1
2
×
3
×
1
2
KE′
=
1
2
×
3
KE′ = 1.5 J
Check
Put the numbers back in. Both sides should match.
3 = 2 × 3 + 1 × (−3) ✓
1 = 3/(2 + 1) ✓
13.5
=
1
2
×
2
×
3
2
+
1
2
×
1
×
(−3)
2
✓
1.5
=
1
2
×
(2
+
1)
×
1
2
✓
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