SME
Dynamics · Problem type
Direct impact and the coefficient of restitution
Use this for “A 2 kg cart at 6 m/s runs into a 3 kg cart moving the same way at 1 m/s, with e = 0.5. How fast is each cart moving after the impact, and how much kinetic energy is lost?”
Diagram
Before: m₁v₁ + m₂v₂ = 2 × 6 + 3 × 1 = 15 kg·m/s
After: m₁v₁′ + m₂v₂′ = 2 × 1.5 + 3 × 4 = 15 kg·m/s
Kinetic energy: 37.5 J before, 26.3 J after: 11.3 J lost 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)
×
v₁′ (m/s)
+
m₂ (kg)
×
v₂′ (m/s)
v₂′ (m/s)
−
v₁′ (m/s)
=
e
(
v₁ (m/s)
−
v₂ (m/s)
)
ΔKE
Kinetic energy lost
worked out (not typed)
Formulas
v₁
′
=
((
m₁
−
e
×
m₂
)
×
v₁
+
(1
+
e
)
×
m₂
×
v₂
)
÷
(
m₁
+
m₂
)
+1.5 = ((2 − 0.5 × 3) × 6 + (1 + 0.5) × 3 × 1) ÷ (2 + 3)
v₂
′
=
((
m₂
−
e
×
m₁
)
×
v₂
+
(1
+
e
)
×
m₁
×
v₁
)
÷
(
m₁
+
m₂
)
+4 = ((3 − 0.5 × 2) × 1 + (1 + 0.5) × 2 × 6) ÷ (2 + 3)
e
=
v₂
′ −
v₁
′
v₁
−
v₂
0.5
=
4 − 1.5
6 − 1
ΔKE
=
1
2
×
m₁
×
m₂
m₁
+
m₂
×
(1
−
e
2
)
×
(
v₁
−
v₂
)
2
11.25
=
1
2
×
2
×
3
2 + 3
×
(1
−
0.5
2
)
×
(6 − 1)
2
Assumptions
•
+ is the way cart 1 moves; a velocity the other way is negative.
•
No outside force acts along the line during the impact, so momentum is kept: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′.
•
Restitution e = (v₂′ − v₁′) ÷ (v₁ − v₂): 1 is perfectly elastic, and 0 means the carts stick together.
Step-by-step
WE KNOW
m₁ = 2 kg, v₁ = +6 m/s, m₂ = 3 kg, v₂ = +1 m/s, e = 0.5
FIND
velocity 1 after (v₁′), velocity 2 after (v₂′), kinetic energy lost (ΔKE)
Step 1 · Find velocity 1 after (v₁′)
Use v₁′ = ((m₁ − e × m₂) × v₁ + (1 + e) × m₂ × v₂) ÷ (m₁ + m₂)
Solve
momentum
and
the
restitution
rule
together
for
v₁
′:
substitute
v₂
′
=
v₁
′
+
e
(
v₁
−
v₂
)
into
momentum.
v₁
′
=
((2
−
0.5
×
3)
×
6
+
(1
+
0.5)
×
3
×
1)
÷
(2
+
3)
v₁
′
=
(2 − 1.5) × 6 + 1.5 × 3
2 + 3
v₁
′
=
0.5 × 6 + 1.5 × 3
2 + 3
v₁
′
=
3 + 4.5
5
v₁
′
=
7.5
5
v₁
′
=
+1.5
m/s
Step 2 · Find velocity 2 after (v₂′)
Use v₂′ = ((m₂ − e × m₁) × v₂ + (1 + e) × m₁ × v₁) ÷ (m₁ + m₂)
The
same
for
cart
2:
momentum
with
v₁
′
=
v₂
′
−
e
(
v₁
−
v₂
)
put
in.
v₂
′
=
((3
−
0.5
×
2)
×
1
+
(1
+
0.5)
×
2
×
6)
÷
(2
+
3)
v₂
′
=
3 − 0.5 × 2 + 1.5 × 2 × 6
2 + 3
v₂
′
=
3 − 1 + 3 × 6
5
v₂
′
=
3 − 1 + 18
5
v₂
′
=
20
5
v₂
′
=
+4
m/s
Step 3 · Find kinetic energy lost (ΔKE)
Use ΔKE = ½ × m₁ × m₂ ÷ (m₁ + m₂) × (1 − e²) × (v₁ − v₂)²
The
kinetic
energy
before
less
after,
simplified:
the
closing
speed
squared,
the
reduced
mass
m₁
m₂
m₁
+
m₂
and
1
−
e
2
.
It
is
0
when
e
=
1.
ΔKE
=
1
2
×
2
×
3
2 + 3
×
(1
−
0.5
2
)
×
(6 − 1)
2
ΔKE
=
1
2
×
2
×
3
5
×
(1
−
0.25)
×
5
2
ΔKE
=
1
2
×
2
×
3
5
×
0.75
×
5
2
ΔKE
=
1
2
×
2
×
3
5
×
0.75
×
25
ΔKE
=
3
5
×
0.75
×
25
ΔKE = 0.6 × 0.75 × 25
ΔKE = 0.45 × 25
ΔKE = 11.25 J
Check
Put the numbers back in. Both sides should match.
+1.5 = ((2 − 0.5 × 3) × 6 + (1 + 0.5) × 3 × 1) ÷ (2 + 3) ✓
+4 = ((3 − 0.5 × 2) × 1 + (1 + 0.5) × 2 × 6) ÷ (2 + 3) ✓
0.5
=
4 − 1.5
6 − 1
✓
11.25
=
1
2
×
2
×
3
2 + 3
×
(1
−
0.5
2
)
×
(6 − 1)
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Work–energy and impulse–momentum
Main lesson
Power and efficiency of a hoist
Use this for “A hoist raises a 500 kg crate 12 m at a steady 0.8 m/s, and its motor is 85% efficient (η = 0.85). What output power lifts the crate, and what input power does the motor draw?”
Average force from impulse and momentum
Use this for “A 0.145 kg ball reaches the bat at 40 m/s and leaves at 30 m/s the other way after 2 ms of contact. What impulse does the bat give, and what average force?”