Momentum and Impulse Lecture Review

DEFINITION OF MOMENTUM

Momentum is defined as continuously moving and hard to stop.
It describes the quantity of motion an object possesses.
The formula to calculate momentum is given by: p = m \times v where,
- p = momentum
- m = mass of the object
- v = velocity of the object

Question: Why does a moving train take longer to stop than a moving bicycle?
- Answer: It has more momentum.
- Reason: A train has a significantly larger mass than a bicycle; thus, even if both are moving at the same speed, the momentum of the train is greater because momentum is directly proportional to mass.

Impulse relates to the concept that more momentum makes an object harder to stop, necessitating an impulse to slow it down.
Definition: It refers to the force applied over time to change an object's momentum.
Formula: I = F \times t where,
- I = impulse
- F = force applied
- t = time duration the force is applied

A ball of 2 kg is moving at a speed of 3 m/s. A force of 10 N is applied to the ball for 4 seconds in the opposite direction of its motion.
To calculate the impulse applied to the ball, we follow these steps:
- Given:
- Force: 10 N
- Time: 4 s
- Using the formula, we substitute the values:
  I = F \times t
  I = 10 \times 4
- Therefore, the impulse:
  I = 40 N.s
- Full Answer:
- The impulse applied to the ball is 40 N⋅s.

A 1500 kg car travels at an initial speed of 25 m/s. After applying the brakes, the car slows down to a final speed of 5 m/s in 4 seconds. The braking force applied by the brakes is 3000 N.
Given:
- Mass: 1500 kg
- Final Velocity: 5 m/s
- Initial Velocity: 25 m/s
- Time: 4 seconds
Solution leads to the force acting on the object being -7500 N.

This law asserts that in an isolated system, the momentum remains constant if no external forces act on the system.

A collision occurs when two objects come into contact and exchange forces, disrupting their motions.
Example Situation:
- A baseball player follows through with their swing to hit the ball farther.
- Reasoning: The conservation of momentum is at play. By following through with the swing, the player applies force over a longer period, increasing the impulse. This results in a greater change in momentum, allowing the ball to travel farther.
Formula:
m1 v{1i} + m2 v{2i} = m1 v{1f} + m2 v{2f}

In an elastic collision, both momentum and kinetic energy are conserved.
Objects involved bounce off each other without permanent deformation (e.g., billiard balls colliding).

Two billiard balls of equal mass (0.5 kg each) collide head-on. The first ball moves at 4 m/s toward the second ball, which is stationary. After the collision, the first ball stops, and the second ball moves forward. What is the velocity of the second ball after the collision?
Solution of this problem will follow the formulas provided.

In an inelastic collision, momentum is conserved, while kinetic energy is not; instead, it is transformed into sound and heat (e.g., clay sticking together after impact).
Given:
- Mass of first ball: m_1 = 0.5 kg
- Mass of second ball: m_2 = 0.5 kg
- Initial velocity of the first ball: v_{1i} = 4 m/s
- Initial velocity of the second ball: v_{2i} = 0 m/s
- Final velocity of the first ball: v_{f1} = 0 m/s
- Final velocity of the second ball: v_{f2}
Formula:
m1 v{1i} + m2 v{2i} = m1 v{f1} + m2 v{f2}
Substitute the known values:
(0.5)(4) + (0.5)(0) = (0.5)(0) + (0.5)v_{f2}
Processing: (2) + (0) = (0) + 0.5 v_{f2}
- Thus, 2 = 0.5 v_{f2}
- Dividing both sides by 0.5 gives:
  v_{f2} = 4 m/s
Full Answer: The second ball moves at 4 m/s after the collision.

A 1000 kg car moving at 20 m/s collides with a 500 kg stationary car. If the two cars stick together post-collision, what will be their final velocity?
Solution of the Problem:
- Given:
- Mass of the first car: m_1 = 1000 kg
- Mass of the second car: m_2 = 500 kg
- Initial velocity of the first car: v_{1i} = 20 m/s
- Initial velocity of the second car: v_{2i} = 0 m/s
- Final velocity of the first car: v_{f1}
- Final velocity of the second car: v_{f2}
Formula for total momentum before and after collision:
m1 v{1i} + m2 v{2i} = (m1 + m2)v_{f}
Substitute the known values:
(1000)(20) + (500)(0) = (1000 + 500)v_{f}
Processing gives: (20000) + (0) = 1500 v_{f}
- Therefore, 20000 = 1500 v_f
- Dividing both sides by 1500 yields:
  v_f = \frac{20000}{1500} \approx 13.33 m/s
Full Answer: After the collision, both cars move together at approximately 13.33 m/s.

Reminder: Do not rely solely on this guide; refer to any additional material provided by your instructors.
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