Worksheet Conservation Of Momentum - Free Printable

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

The formula for momentum is:
$$ p = mv $$

For two objects, the conservation equation is:
$$
(m_1v_1 + m_2v_2)_{\text{before}} = (m_1v_1 + m_2v_2)_{\text{after}}
$$

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Problem 1: Freight Cars Collide and Couple

Given:
- Car 1: mass = 10 tons, velocity = 4 m/s
- Car 2: mass = 30 tons, velocity = 0 m/s
- After collision, they couple (stick together), so same final velocity $ v_f $

We need to find the resultant velocity after collision.

#### Step 1: Use Conservation of Momentum

$$
m_1v_1 + m_2v_2 = (m_1 + m_2)v_f
$$

Plug in values:

$$
(10 \text{ tons})(4 \text{ m/s}) + (30 \text{ tons})(0 \text{ m/s}) = (10 + 30)v_f
$$

$$
40 + 0 = 40v_f
$$

$$
v_f = \frac{40}{40} = 1 \text{ m/s}
$$

✔ Answer: The resultant velocity is 1 m/s.

> Note: The diagram shows "7 m/s" after — that appears to be incorrect based on the given data. The correct answer is 1 m/s, not 7 m/s. Possibly a typo in the image.

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Problem 2: Putty Blobs Stick Together

Given:
- Blob 1: mass = 2 kg, velocity = 4 m/s
- Blob 2: mass = 6 kg, velocity = 0 m/s (at rest)
- They stick together → inelastic collision

Find: Final speed $ v_f $ of combined blob

Use conservation of momentum:

$$
m_1v_1 + m_2v_2 = (m_1 + m_2)v_f
$$

$$
(2)(4) + (6)(0) = (2 + 6)v_f
$$

$$
8 + 0 = 8v_f
$$

$$
v_f = \frac{8}{8} = 1 \text{ m/s}
$$

✔ Answer: The speed of the stuck-together blobs is 1 m/s.

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Problem 3: Football Player vs Referee – Elastic Collision

Given:
- Football player: mass = ? (let’s call it $ m $), initial velocity = 8 m/s
- Referee: mass = 80 kg, initially at rest → $ v_{r,i} = 0 $
- After collision, referee moves at 5.0 m/s forward
- Assume perfectly elastic collision → both momentum and kinetic energy are conserved

We are to find the mass of the football player.

Let:
- $ m $ = mass of football player (unknown)
- $ v_{p,i} = 8 $ m/s (initial velocity)
- $ v_{r,i} = 0 $
- $ v_{r,f} = 5.0 $ m/s
- $ v_{p,f} = ? $ → we’ll find this using elastic collision formulas

In an elastic collision, we can use the following formula for final velocity of object 1 (player):

$$
v_{p,f} = \frac{(m - M)v_{p,i} + 2M v_{r,i}}{m + M}
$$

But since $ v_{r,i} = 0 $, this simplifies to:

$$
v_{p,f} = \frac{(m - 80)(8)}{m + 80}
$$

Also, from conservation of momentum:

$$
m \cdot 8 + 80 \cdot 0 = m \cdot v_{p,f} + 80 \cdot 5
$$

So:

$$
8m = m v_{p,f} + 400
$$

Now substitute $ v_{p,f} $ from above:

$$
8m = m \left( \frac{8(m - 80)}{m + 80} \right) + 400
$$

Multiply both sides by $ m + 80 $ to eliminate denominator:

$$
8m(m + 80) = m \cdot 8(m - 80) + 400(m + 80)
$$

Expand both sides:

Left:
$$
8m^2 + 640m
$$

Right:
$$
8m(m - 80) + 400m + 32000 = 8m^2 - 640m + 400m + 32000 = 8m^2 - 240m + 32000
$$

Now set equal:

$$
8m^2 + 640m = 8m^2 - 240m + 32000
$$

Subtract $ 8m^2 $ from both sides:

$$
640m = -240m + 32000
$$

Add $ 240m $:

$$
880m = 32000
$$

Solve:

$$
m = \frac{32000}{880} = \frac{3200}{88} = \frac{400}{11} \approx 36.36 \text{ kg}
$$

Wait! That seems too light for a football player.

Let’s double-check our assumption.

But here's a better approach: In elastic collisions, we can use the fact that relative velocity reverses direction.

For elastic collisions:
$$
v_{p,i} - v_{r,i} = -(v_{p,f} - v_{r,f})
$$

Given:
- $ v_{p,i} = 8 $
- $ v_{r,i} = 0 $
- $ v_{r,f} = 5 $
- So:
$$
8 - 0 = -(v_{p,f} - 5)
\Rightarrow 8 = -v_{p,f} + 5
\Rightarrow v_{p,f} = 5 - 8 = -3 \text{ m/s}
$$

So the football player rebounds at 3 m/s backward.

Now apply conservation of momentum:

$$
m \cdot 8 + 80 \cdot 0 = m \cdot (-3) + 80 \cdot 5
$$

$$
8m = -3m + 400
$$

$$
8m + 3m = 400
\Rightarrow 11m = 400
\Rightarrow m = \frac{400}{11} \approx 36.36 \text{ kg}
$$

Still the same result.

But 36.36 kg is about 80 pounds, which is too light for a football player.

This suggests something might be wrong with the problem setup or assumptions.

Wait — let’s check if the referee flies forward at 5 m/s — that’s very fast. A football player running at 8 m/s colliding with a stationary 80 kg person causing them to go 5 m/s?

Let’s verify whether such a mass makes sense.

Try plugging back in:

Player mass $ m = 400/11 \approx 36.36 $ kg
Initial momentum: $ 36.36 \times 8 = 290.88 $ kg·m/s
Final momentum:
- Player: $ 36.36 \times (-3) = -109.08 $
- Referee: $ 80 \times 5 = 400 $
Total final momentum: $ -109.08 + 400 = 290.92 $ → matches!

So mathematically correct.

But physically unrealistic? Maybe the scenario is hypothetical.

Alternatively, perhaps the question meant inelastic collision? But it says perfectly elastic.

Wait — could the referee’s final velocity be 5 m/s only if the player has more mass?

But according to physics, if the player is lighter than the referee, he can’t make the referee go faster than himself unless there’s a large mass difference.

Let’s suppose instead the football player has mass $ m $, and after elastic collision, the referee goes at 5 m/s.

We already used the relative velocity method correctly.

So the only way for the referee to gain 5 m/s from a player going 8 m/s is if the player is lighter — which is what we found.

So even though it's unrealistic, the physics is consistent.

✔ Answer: The mass of the football player would be approximately $ \boxed{36.4} $ kg.

But wait — maybe we misread the problem.

Let me re-read:

> "...causing the referee to fly forward at 5.0 m/s. If this were a perfectly elastic collision, what would the mass of football player be?"

So it's hypothetical: if it were elastic, what would the mass be?

But in reality, it was inelastic, but now we're asked: what mass would the player have to be if the collision were perfectly elastic and the referee ended up at 5 m/s?

Yes — so the answer is still $ \boxed{\frac{400}{11} \approx 36.4} $ kg.

Even though it's unrealistic, it's the correct solution under the given assumptions.

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✔ Final Answers:

1. Resultant velocity of freight cars: $ \boxed{1 \text{ m/s}} $
2. Speed of stuck putty blobs: $ \boxed{1 \text{ m/s}} $
3. Mass of football player (if elastic): $ \boxed{36.4 \text{ kg}} $

> Note: Problem 3 gives a physically unlikely result, but mathematically it's correct under the stated conditions.