Compare the catching of two different water balloons.
Case A: A 150-ml water balloon moving at 8 m/s is caught and brought to a stop.
Case B: A 600-ml water balloon moving at 8 m/s is caught with the same
technique and brought to a stop.
The collision time is the same for each case.
Case A
Case B
Before
After
Before
After
Which variable is different for these two cases?
Maths
Which Case involves the greatest momentum change?
Which Case involves the greatest impulse?
Which Case involves the greatest force?

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 = × Δt = mΔV

∴ = m·ΔV/Δt

∴ For Case A = 149.55×8/Δt =  1196.4/Δt N

For Case B   = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,    for Case B >>    for Case B

Therefore, Case B involves the greatest force.

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explanation:

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

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explanation: