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Chapter 7: Problem 20

A submarine of mass \(2.5 \times 10^{6} \mathrm{kg}\) and initially at rest fires a torpedo of mass \(250 \mathrm{kg} .\) The torpedo has an initial speed of \(100.0 \mathrm{m} / \mathrm{s} .\) What is the initial recoil speed of the submarine? Neglect the drag force of the water.

Short Answer

Expert verified
Answer: The initial recoil speed of the submarine is \(0.01\;\mathrm{m/s}\) in the opposite direction of the fired torpedo.

Step by step solution

01

Define variables and given values

Let's define the given variables: Mass of submarine (M1) = \(2.5\times10^6 \;\mathrm{kg}\) Mass of torpedo (M2) = \(250\;\mathrm{kg}\) Initial speed of submarine (V1i) = \(0 \;\mathrm{m/s}\) Initial speed of torpedo (V2i) = \(0 \;\mathrm{m/s}\) Final speed of torpedo (V2f) = \(100\;\mathrm{m/s}\) We need to find the final speed of submarine (V1f).
02

Apply the conservation of momentum principle

According to the conservation of momentum principle, the total momentum before the event equals the total momentum after the event: M1*V1i + M2*V2i = M1*V1f + M2*V2f The initial speeds of both the submarine and the torpedo are 0, so the equation simplifies to: M1*V1f + M2*V2f = 0
03

Solve for the final speed of the submarine

Now, let's solve for the final speed of the submarine (V1f): V1f = -(M2*V2f) / M1 V1f = -(\(250\;\mathrm{kg}\times 100\;\mathrm{m/s}\)) / \((2.5\times10^6 \;\mathrm{kg})\) V1f = \(-25000\;\mathrm{kg\cdot m/s}\) / \((2.5\times 10^6\;\mathrm{kg})\) V1f = \(-0.01\;\mathrm{m/s}\) The negative sign indicates that the direction of the submarine's initial recoil speed is opposite to that of the torpedo. So the initial recoil speed of the submarine is \(0.01\;\mathrm{m/s}\) in the opposite direction of the fired torpedo.

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