This worksheet provides a comprehensive overview of the principle of conservation of momentum, including practice problems with detailed solutions. Understanding conservation of momentum is crucial in physics, providing a fundamental framework for analyzing collisions and interactions between objects.
What is Conservation of Momentum?
The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In simpler terms, the total momentum before an event (like a collision) equals the total momentum after the event. This principle is incredibly useful for predicting the outcome of interactions between objects.
Momentum (p) is defined as the product of an object's mass (m) and its velocity (v):
p = mv
Where:
- p is momentum (measured in kg·m/s)
- m is mass (measured in kg)
- v is velocity (measured in m/s) Remember that velocity is a vector quantity, meaning it has both magnitude and direction.
For a system with multiple objects, the total momentum is the vector sum of the individual momenta.
Practice Problems
Here are some practice problems to help solidify your understanding of conservation of momentum. Remember to consider both magnitude and direction of velocity.
Problem 1:
A 2 kg cart traveling at 5 m/s collides with a stationary 1 kg cart. After the collision, the 2 kg cart moves at 2 m/s in the same direction. What is the velocity of the 1 kg cart after the collision?
Solution:
-
Before collision:
- Momentum of 2 kg cart: p₁ = (2 kg)(5 m/s) = 10 kg·m/s
- Momentum of 1 kg cart: p₂ = (1 kg)(0 m/s) = 0 kg·m/s
- Total momentum before collision: Pᵢ = p₁ + p₂ = 10 kg·m/s
-
After collision:
- Momentum of 2 kg cart: p₃ = (2 kg)(2 m/s) = 4 kg·m/s
- Let v be the velocity of the 1 kg cart after the collision.
- Momentum of 1 kg cart: p₄ = (1 kg)(v) = v kg·m/s
- Total momentum after collision: Pƒ = p₃ + p₄ = 4 kg·m/s + v kg·m/s
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Conservation of momentum: Pᵢ = Pƒ => 10 kg·m/s = 4 kg·m/s + v kg·m/s
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Solving for v: v = 10 kg·m/s - 4 kg·m/s = 6 kg·m/s. Therefore, the velocity of the 1 kg cart after the collision is 6 m/s in the same direction as the initial velocity of the 2 kg cart.
Problem 2:
Two billiard balls, each with a mass of 0.17 kg, collide head-on. Ball 1 is initially moving at 2 m/s to the right, and Ball 2 is initially moving at 1 m/s to the left. After the collision, Ball 1 moves at 0.5 m/s to the left. What is the final velocity of Ball 2?
Solution: (Remember to assign positive values to velocities to the right and negative values to velocities to the left)
-
Before collision:
- Momentum of Ball 1: p₁ = (0.17 kg)(2 m/s) = 0.34 kg·m/s
- Momentum of Ball 2: p₂ = (0.17 kg)(-1 m/s) = -0.17 kg·m/s
- Total momentum before collision: Pᵢ = p₁ + p₂ = 0.17 kg·m/s
-
After collision:
- Momentum of Ball 1: p₃ = (0.17 kg)(-0.5 m/s) = -0.085 kg·m/s
- Let v be the velocity of Ball 2 after the collision.
- Momentum of Ball 2: p₄ = (0.17 kg)(v) = 0.17v kg·m/s
- Total momentum after collision: Pƒ = p₃ + p₄ = -0.085 kg·m/s + 0.17v kg·m/s
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Conservation of momentum: Pᵢ = Pƒ => 0.17 kg·m/s = -0.085 kg·m/s + 0.17v kg·m/s
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Solving for v: 0.17v = 0.255 kg·m/s => v = 1.5 m/s. Therefore, the final velocity of Ball 2 is 1.5 m/s to the right.
Problem 3: (More challenging – requires considering momentum as a vector)
(This problem requires a diagram for proper visualization. Draw two objects colliding at an angle. Assign masses, initial velocities and angles to illustrate the problem. Then solve for the final velocity of one of the objects given the final velocity of the other.)
These examples demonstrate the application of the conservation of momentum principle. Remember to always account for the vector nature of momentum. Further practice problems can be found in your physics textbook or online resources. Understanding this fundamental concept will significantly aid your understanding of more complex physics problems.
