Conservation of Momentum Questions and Answers | PDF - Free Printable

The image contains a worksheet on the Conservation of Momentum. Below, I will solve each problem step by step and explain the solution.

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Problem 1:

Two heavy cars of different masses collide and stick together. What is their velocity after the collision?

#### Given:
- Mass of Car 1 (\( m_1 \)): 30 tons = \( 30 \times 10^3 \) kg
- Velocity of Car 1 (\( v_1 \)): 20 km/h = \( 20 \times \frac{1000}{3600} \) m/s = \( \frac{50}{9} \) m/s
- Mass of Car 2 (\( m_2 \)): 40 tons = \( 40 \times 10^3 \) kg
- Velocity of Car 2 (\( v_2 \)): 0 km/h = 0 m/s

#### To Find:
Final velocity (\( v_f \)) after the collision.

#### Solution:
The principle of conservation of momentum states:
\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
\]

1. Substitute the given values:
\[
(30 \times 10^3) \left( \frac{50}{9} \right) + (40 \times 10^3)(0) = (30 \times 10^3 + 40 \times 10^3) v_f
\]

2. Simplify the equation:
\[
(30 \times 10^3) \left( \frac{50}{9} \right) = (70 \times 10^3) v_f
\]

3. Calculate the left-hand side:
\[
30 \times 10^3 \times \frac{50}{9} = \frac{1500 \times 10^3}{9} = \frac{1500000}{9} \approx 166666.67 \text{ kg·m/s}
\]

4. Solve for \( v_f \):
\[
\frac{1500000}{9} = (70 \times 10^3) v_f
\]
\[
v_f = \frac{\frac{1500000}{9}}{70 \times 10^3} = \frac{1500000}{9 \times 70 \times 10^3} = \frac{1500000}{630000} = \frac{1500}{630} \approx 2.38 \text{ m/s}
\]

#### Final Answer:
\[
\boxed{2.38 \text{ m/s}}
\]

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Problem 2:

A 2.0 kg putty ball moving at 4 m/s collides with a 6 kg block of putty at rest. What is the speed of the two blocks together after the collision?

#### Given:
- Mass of Putty Ball (\( m_1 \)): 2.0 kg
- Velocity of Putty Ball (\( v_1 \)): 4 m/s
- Mass of Block (\( m_2 \)): 6.0 kg
- Velocity of Block (\( v_2 \)): 0 m/s

#### To Find:
Final velocity (\( v_f \)) after the collision.

#### Solution:
Using the conservation of momentum:
\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
\]

1. Substitute the given values:
\[
(2.0)(4) + (6.0)(0) = (2.0 + 6.0) v_f
\]

2. Simplify the equation:
\[
8.0 = 8.0 v_f
\]

3. Solve for \( v_f \):
\[
v_f = \frac{8.0}{8.0} = 1.0 \text{ m/s}
\]

#### Final Answer:
\[
\boxed{1.0 \text{ m/s}}
\]

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Problem 3:

A football player runs full tilt and plows into a 60 kg referee standing on the field causing the referee to be moved backward. If the mass of the football player is 100 kg and his speed was 5 m/s before impact, what is the speed of the referee immediately after impact?

#### Given:
- Mass of Football Player (\( m_1 \)): 100 kg
- Velocity of Football Player (\( v_1 \)): 5 m/s
- Mass of Referee (\( m_2 \)): 60 kg
- Velocity of Referee (\( v_2 \)): 0 m/s

#### To Find:
Final velocity (\( v_f \)) of the system (player + referee) after the collision.

#### Solution:
Using the conservation of momentum:
\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
\]

1. Substitute the given values:
\[
(100)(5) + (60)(0) = (100 + 60) v_f
\]

2. Simplify the equation:
\[
500 = 160 v_f
\]

3. Solve for \( v_f \):
\[
v_f = \frac{500}{160} = 3.125 \text{ m/s}
\]

#### Final Answer:
\[
\boxed{3.125 \text{ m/s}}
\]

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Summary of Answers:

1. \( \boxed{2.38 \text{ m/s}} \)
2. \( \boxed{1.0 \text{ m/s}} \)
3. \( \boxed{3.125 \text{ m/s}} \)

Each problem uses the principle of conservation of momentum, where the total momentum before the collision equals the total momentum after the collision.