SME
Grade 11 Science · Problem type
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?”
Diagram
Before: (m₁ + m₂)v = (1.5 + 0.5) × 0 = 0 kg·m/s
After: m₁v₁′ + m₂v₂′ = 1.5 × (−1) + 0.5 × 3 = 0 kg·m/s
Kinetic energy: 0 J before, 3 J after: the spring gave E = 3 J.
This is an example. Type your own values to start a new problem.
m₁ (kg)
×
(
v₁′ (m/s)
)
+
m₂ (kg)
×
v₂′ (m/s)
=
0
E
Energy from the spring
calculated · J
Formulas
v₂′
=
−
m₁
×
v₁′/
m₂
3 = −1.5 × (−1)/0.5
E
=
1
2
×
m₁
×
v₁′²
+
1
2
×
m₂
×
v₂′²
3
=
1
2
×
1.5
×
(−1)
2
+
1
2
×
0.5
×
3
2
Assumptions
•
They start at rest, so the total momentum is 0 before and after: m₁v₁′ + m₂v₂′ = 0.
•
The carts move off in opposite directions; the lighter one faster.
•
The spring’s stored energy becomes the carts’ kinetic energy.
Step-by-step
WE KNOW
m₁ = 1.5 kg, m₂ = 0.5 kg, v₁′ = −1 m/s
FIND
velocity of cart 2 after (v₂′), energy from the spring (E)
Step 1 · Find velocity of cart 2 after (v₂′)
Use v₂′ = −m₁ × v₁′/m₂
Cart 2’s momentum must cancel cart 1’s: the same size, the other way.
v₂′ = −1.5 × (−1)/0.5
v₂′ = 1.5/0.5
v₂′ = 3 m/s
Step 2 · Find energy from the spring (E)
Use E = ½ × m₁ × v₁′² + ½ × m₂ × v₂′²
Add
each
cart’s
1
2
mv²
after:
kinetic
energy
has
no
direction.
E
=
1
2
×
1.5
×
(−1)
2
+
1
2
×
0.5
×
3
2
E
=
1
2
×
1.5
+
1
2
×
0.5
×
9
E
=
0.75
+
0.25
×
9
E
=
0.75
+
2.25
E
=
3
J
Check
Put the numbers back in. Both sides should match.
3 = −1.5 × (−1)/0.5 ✓
3
=
1
2
×
1.5
×
(−1)
2
+
1
2
×
0.5
×
3
2
✓
Topic check
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