Momentum worksheet|KS4 Physics|Teachit - Free Printable

Let's solve each question step by step, using the principle of conservation of momentum. This principle states that in a closed system with no external forces (like frictionless ice), the total momentum before a collision equals the total momentum after the collision.

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Question 1: Two boxes colliding and sticking together

#### Given:
- Mass of first box, $ m_1 = 1\ \text{kg} $
- Initial velocity of first box, $ v_1 = 5\ \text{m/s} $
- Mass of second box, $ m_2 = 1\ \text{kg} $
- Initial velocity of second box, $ v_2 = 0\ \text{m/s} $
- After collision, they stick together → perfectly inelastic collision

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#### a. Calculate the total momentum before the collision

Momentum $ p = mv $

- Momentum of first box:
$ p_1 = m_1 \cdot v_1 = 1\ \text{kg} \times 5\ \text{m/s} = 5\ \text{kg·m/s} $

- Momentum of second box:
$ p_2 = m_2 \cdot v_2 = 1\ \text{kg} \times 0\ \text{m/s} = 0\ \text{kg·m/s} $

- Total momentum before collision:
$ p_{\text{total before}} = p_1 + p_2 = 5 + 0 = \boxed{5\ \text{kg·m/s}} $

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#### b. State the total momentum after the collision

Since there are no external forces (frictionless ice), momentum is conserved.

So, total momentum after collision = total momentum before collision
$ \Rightarrow \boxed{5\ \text{kg·m/s}} $

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#### c. Calculate the total mass of the “double box” after the collision

After sticking together, total mass:
$ m_{\text{total}} = 1\ \text{kg} + 1\ \text{kg} = \boxed{2\ \text{kg}} $

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#### d. Work out the velocity of the “double box” after the collision

Use the formula:
$ p = mv $ → rearrange to $ v = \frac{p}{m} $

We know:
- $ p = 5\ \text{kg·m/s} $
- $ m = 2\ \text{kg} $

So,
$$
v = \frac{5}{2} = \boxed{2.5\ \text{m/s}}
$$

✔ So, the "double box" moves at 2.5 m/s in the same direction as the original moving box.

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✔ Summary for Question 1:

- a) Total momentum before: 5 kg·m/s
- b) Total momentum after: 5 kg·m/s
- c) Total mass after: 2 kg
- d) Final velocity: 2.5 m/s

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Question 2: Speedboat hits fishing boat

#### Given:
- Speedboat + passengers: mass $ m_1 = 1500\ \text{kg} $, velocity $ v_1 = 4\ \text{m/s} $ north
- Fishing boat: mass $ m_2 = 2000\ \text{kg} $, velocity $ v_2 = 0\ \text{m/s} $
- They stick together → inelastic collision

Let’s assume north is positive, so:
- $ v_1 = +4\ \text{m/s} $
- $ v_2 = 0 $

#### Step 1: Total momentum before collision
$$
p_{\text{before}} = m_1 v_1 + m_2 v_2 = (1500)(4) + (2000)(0) = 6000\ \text{kg·m/s}
$$

#### Step 2: Total momentum after collision
Same due to conservation: $ p_{\text{after}} = 6000\ \text{kg·m/s} $

#### Step 3: Total mass after collision
$$
m_{\text{total}} = 1500 + 2000 = 3500\ \text{kg}
$$

#### Step 4: Final velocity
$$
v = \frac{p}{m} = \frac{6000}{3500} = \frac{60}{35} = \frac{12}{7} \approx \boxed{1.71\ \text{m/s}}
$$

Direction: North (since momentum was positive)

✔ Answer: They move together at 1.71 m/s north.

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Question 3: Minibus vs Challenger 2 tank

#### Given:
- Minibus: mass $ m_1 = 3000\ \text{kg} $, velocity $ v_1 = -10\ \text{m/s} $ (west → negative)
- Tank: mass $ m_2 = 62000\ \text{kg} $, velocity $ v_2 = +2\ \text{m/s} $ (east → positive)

Assume east is positive, west is negative.

#### Step 1: Total momentum before collision
$$
p_{\text{before}} = m_1 v_1 + m_2 v_2
= (3000)(-10) + (62000)(2)
= -30000 + 124000 = \boxed{94000\ \text{kg·m/s}}
$$

Positive → momentum is eastward

#### Step 2: Total mass after collision
$$
m_{\text{total}} = 3000 + 62000 = 65000\ \text{kg}
$$

#### Step 3: Final velocity
$$
v = \frac{p}{m} = \frac{94000}{65000} = \frac{94}{65} \approx \boxed{1.446\ \text{m/s}}
$$

Positive → eastward

✔ Answer: They move together at 1.45 m/s east (approximately).

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✔ Final Answers:

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Question 1:

a) Total momentum before: 5 kg·m/s
b) Total momentum after: 5 kg·m/s
c) Total mass: 2 kg
d) Velocity: 2.5 m/s

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Question 2:

They move together at 1.71 m/s north

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Question 3:

They move together at 1.45 m/s east

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🔍 Key Concept:

All these problems use conservation of momentum:
$$
\sum p_{\text{before}} = \sum p_{\text{after}}
$$
For perfectly inelastic collisions (objects stick), use:
$$
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
$$

Always assign directions (positive/negative) consistently!

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