5.2 Representations of Changes in Momentum

5.2 Representations of Changes in Momentum

Two-Object Problems

When solving two-object problems to calculate the change in momentum, it’s crucial to individually consider the initial and final momentum of each object. Apply the principle of conservation of momentum: pinitial = pfinal. When solving for the final momentum of the system, decide whether to use the individual mass value or the combined mass of all objects involved.

If solving for the final velocity of the system, always use the total mass of the system. For objects initially at rest, their initial momentum will be zero since their initial velocity is zero (p = m × v).

Momentum change is proportional to the force and time (Δp = F × t). Reducing either the force or the time will decrease the change in momentum.

Example Problems

Example Problem #1

A tennis ball of mass 0.05 kg is hit with a racket, causing it to travel at a velocity of 40 m/s. The racket has a mass of 0.5 kg and is traveling at a velocity of 10 m/s. What is the total momentum of the ball and the racket before and after the collision?

Solution:

1. Momentum of the ball before collision: • p = m × v = 0.05 kg × 40 m/s = 2 kg⋅m/s

2. Momentum of the racket before collision: • p = m × v = 0.5 kg × 10 m/s = 5 kg⋅m/s

3. Total momentum before collision: • p = 2 kg⋅m/s + 5 kg⋅m/s = 7 kg⋅m/s

4. Since no external forces act, total momentum after collision remains the same: • p = 7 kg⋅m/s

Example Problem #2

Two cars collide head-on. Car A (mass: 2000 kg) is traveling at 60 m/s, and Car B (mass: 1000 kg) is traveling at 40 m/s. After collision, they stick together. What is the total momentum before and after the collision?

Solution:

1. Momentum of Car A before collision: • p = m × v = 2000 kg × 60 m/s = 120,000 kg⋅m/s

2. Momentum of Car B before collision: • p = m × v = 1000 kg × 40 m/s = 40,000 kg⋅m/s

3. Total momentum before collision: • p = 120,000 kg⋅m/s + 40,000 kg⋅m/s = 160,000 kg⋅m/s

4. Combined mass after collision: • m = 2000 kg + 1000 kg = 3000 kg

5. Combined velocity after collision: • v = p / m = 160,000 kg⋅m/s / 3000 kg = 53.33 m/s

6. Total momentum after collision: • p = 3000 kg × 53.33 m/s = 160,000 kg⋅m/s

Example Problem #3

A ball of mass 0.1 kg is thrown at 10 m/s and caught at 5 m/s. What is the change in momentum of the ball?

Solution:

1. Momentum before catching: • pinitial = m × v = 0.1 kg × 10 m/s = 1 kg⋅m/s

2. Momentum after catching: • pfinal = m × v = 0.1 kg × 5 m/s = 0.5 kg⋅m/s

3. Change in momentum: • Δp = pfinal – pinitial = 0.5 kg⋅m/s – 1 kg⋅m/s = -0.5 kg⋅m/s

Example Problem #4

A 100 kg rocket launches at 100 m/s carrying a 50 kg payload. What is the total momentum before and after launch?

Solution:

1. Momentum before launch: • p = m × v = (100 kg + 50 kg) × 0 m/s = 0 kg⋅m/s

2. Momentum after launch: • Rocket: p = 100 kg × 100 m/s = 10,000 kg⋅m/s • Payload: p = 50 kg × 100 m/s = 5000 kg⋅m/s • Total: p = 10,000 kg⋅m/s + 5000 kg⋅m/s = 15,000 kg⋅m/s

Example Problem #5

A baseball of 0.15 kg is hit at 50 m/s with a bat traveling at -20 m/s (opposite direction). What is the total momentum before and after collision?

Solution:

1. Momentum of baseball: • p = m × v = 0.15 kg × 50 m/s = 7.5 kg⋅m/s

2. Momentum of bat: • p = m × v = 1 kg × -20 m/s = -20 kg⋅m/s

3. Total momentum before collision: • p = 7.5 kg⋅m/s – 20 kg⋅m/s = -12.5 kg⋅m/s

4. Total momentum after collision: • p = -12.5 kg⋅m/s (conservation of momentum)