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A 1450 kg torpedo strikes a 780 kg target that is initially at rest. If the combined torpedo and target move forward with a speed of 8.65 m/s, what is the initial velocity of the torpedo? Assume that no resistance is provided by the water

User Pusoy
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1 Answer

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If the torpedo's initial speed is v, then by conservation of momentum,

(1450 kg) v + 0 = (1450 kg + 780 kg) (8.65 m/s)

Solve for v :

(1450 kg) v = 19,116.5 kg•m/s

v ≈ 13.2 m/s

User Calvillo
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