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Zanzabum
2 years ago
7

Compare the catching of two different water balloons.

Physics
1 answer:
Stels [109]2 years ago
4 0

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

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