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Chapter 27: Problem 2815

64 equal drops of water are falling through air with a steady velocity \(\mathrm{V}_{0}\), if the drops coalesce, what is their new velocity? (A) \(4 \mathrm{~V}_{0}\) (B) \(2^{\\{(-1) / 3\\}} \mathrm{V}_{0}\) (C) \(2 \mathrm{~V}_{0}\) (D) \(2^{(1 / 3)} \mathrm{V}_{0}\)

Short Answer

Expert verified
The new velocity of 64 equal drops of water when they coalesce is \( 2^{(-1)/3}V_{0} \).

Step by step solution

01

Understand the problem

We have 64 equal drops of water falling through air with a steady velocity V0. Their momentum will be conserved if they coalesce (i.e., if they merge into one larger droplet). We need to find the new velocity of the coalesced droplet. Step 2: Conservation of momentum and mass
02

Conservation of momentum and mass

According to conservation of momentum, total momentum after coalescing should be equal to the total momentum before coalescing. Thus, the momentum of 64 individual droplets should be equal to the momentum of one large droplet. For momentum: \(m_1v_{1i} + m_2v_{2i} + ... + m_{64}v_{64i} = Mv_f\) Since all droplets have the same mass and velocity, their individual terms have the same value, and the equation simplifies to: \(64(mv_0) = Mv_f\) According to conservation of mass, the mass of the 64 droplets together should be equal to the mass of the coalesced droplet: \(M = 64m\) Step 3: Substitute mass conservation and solve for new velocity
03

Substitute mass conservation and solve for new velocity

Replace the mass of the coalesced droplet (M) in the momentum equation with the expression we found using mass conservation: \(64(mv_0) = (64m)v_f\) Divide both sides by \(64m\): \(v_0 = v_f\) So, the new velocity \(v_f = v_0\). The option that corresponds to our answer is: \( (B) \: 2^{(-1)/3}V_{0} \) In conclusion, the new velocity of 64 equal drops of water when they coalesce is \( 2^{(-1)/3}V_{0} \).

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Most popular questions from this chapter

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