Worksheet: Conservation of Momentum - Free Printable

Absolutely! Let’s solve each problem step by step using the principle of conservation of momentum, which states:

> Total momentum before collision = Total momentum after collision

We’ll use the formula:
p = mv (momentum = mass × velocity)
and for collisions:
(m₁v₁ + m₂v₂)ₜₒₜₐₗ₋ᵦₑfₒᵣₑ = (m₁v₁ + m₂v₂)ₜₒₜₐₗ₋ₐfₜₑᵣ

---

## Problem 1: Two freight cars collide and couple

Given:

- Mass of car 1, m₁ = 10,000 kg, velocity v₁ = 4 m/s
- Mass of car 2, m₂ = 30,000 kg, velocity v₂ = 0 m/s (at rest)
- They couple together → so they move with a common final velocity, call it v_f

Apply Conservation of Momentum:

Before collision:
> p_before = m₁v₁ + m₂v₂ = (10,000)(4) + (30,000)(0) = 40,000 kg·m/s

After collision:
> p_after = (m₁ + m₂)v_f = (10,000 + 30,000)v_f = 40,000 v_f

Set them equal:
> 40,000 = 40,000 v_f
→ v_f = 1 m/s

✔ Answer: Resultant velocity = 1 m/s

---

## Problem 2: Putty blobs stick together (inelastic collision)

Given:

- Mass of blob 1, m₁ = 2 kg, velocity v₁ = 4 m/s
- Mass of blob 2, m₂ = 6 kg, velocity v₂ = 0 m/s (at rest)
- They stick together → common final velocity v_f

Apply Conservation of Momentum:

Before:
> p_before = (2)(4) + (6)(0) = 8 kg·m/s

After:
> p_after = (2 + 6)v_f = 8 v_f

Set equal:
> 8 = 8 v_f
→ v_f = 1 m/s

✔ Answer: Speed after collision = 1 m/s

---

## Problem 3: Football player collides with referee (perfectly elastic collision?) — Wait!

Actually, the problem says:
> “If this were a perfectly elastic collision...”

But then it adds in parentheses:
> “(they bounce off and the player comes to rest)”

This is contradictory. In a perfectly elastic collision, both kinetic energy and momentum are conserved — and if the player comes to rest, that implies all his momentum was transferred to the referee, which is only possible if the masses are equal (like in billiard balls). But here, the referee is much heavier (80 kg), so if the player stops, that suggests an inelastic or special case.

BUT — let’s read carefully:
> “a football player runs at 8 m/s and plows into an 80 kg referee standing on the field causing the referee to fly forward at 5.0 m/s. If this were a perfectly elastic collision, what would the mass of the football player be?”

Wait — this is impossible under perfectly elastic collision unless we ignore the "player comes to rest" note. The parenthetical note seems to be a student's handwritten guess, not part of the official problem.

Let’s solve it as written:
> Player mass = ? (call it m)
> Player initial velocity = 8 m/s
> Referee mass = 80 kg, initial velocity = 0 m/s
> After collision: Referee velocity = 5.0 m/s
> We’re told to assume perfectly elastic collision

In a perfectly elastic collision, both momentum AND kinetic energy are conserved.

So we need to set up two equations:

---

Step 1: Conservation of Momentum

> m·8 + 80·0 = m·v_p + 80·5
→ 8m = m·v_p + 400
→ Equation (1): 8m - m·v_p = 400

---

Step 2: Conservation of Kinetic Energy

> (1/2)m(8)² + (1/2)(80)(0)² = (1/2)m(v_p)² + (1/2)(80)(5)²
Multiply both sides by 2 to eliminate 1/2:

> m·64 = m·v_p² + 80·25
→ 64m = m·v_p² + 2000
→ Equation (2): 64m - m·v_p² = 2000

---

Now, from Equation (1):
8m - m·v_p = 400
→ m(8 - v_p) = 400
→ m = 400 / (8 - v_p) → [Equation A]

Plug into Equation (2):

64·[400 / (8 - v_p)] - [400 / (8 - v_p)]·v_p² = 2000

Factor out 400 / (8 - v_p):

→ [400 / (8 - v_p)] · (64 - v_p²) = 2000

Divide both sides by 400:

→ [ (64 - v_p²) / (8 - v_p) ] = 5

Note: 64 - v_p² = (8 - v_p)(8 + v_p) → difference of squares!

So:

→ [ (8 - v_p)(8 + v_p) / (8 - v_p) ] = 5
→ 8 + v_p = 5
→ v_p = -3 m/s

(Negative means the player bounces backward!)

Now plug back into Equation A:

m = 400 / (8 - (-3)) = 400 / 11 ≈ 36.36 kg

✔ Answer: Mass of football player = 400/11 kg ≈ 36.4 kg

---

## ✔ Final Answers:

1. Resultant velocity = 1 m/s
2. Speed after collision = 1 m/s
3. Mass of football player = 400/11 kg ≈ 36.4 kg

---

## 📝 Note on Problem 3:
The handwritten note “(they bounce off and the player comes to rest)” is likely incorrect or misleading. In a perfectly elastic collision with a stationary heavier object, the lighter object bounces back, not stops. Our solution reflects true physics — the player rebounds at 3 m/s backward, and we find his mass to be ~36.4 kg.

Let me know if you’d like diagrams or further explanation!