Question

A 1450 kg torpedo strikes a 670 kg target that is initially at rest. If the combined torpedo and target move forward with a speed of 10.6 m/s, what is the initial velocity of the torpedo? Assume that no resistance is provided by the water.

Answer 1

By conservation of momentum, we will see that the initial velocity of the torpedo is 15.4 m/s

How to apply conservation of momentum?

The conservation of momentum says that the total momentum before the collision must be the same as after the collision.

Before the collision we have a momentum:

P₀ = torpedo momentum + target momentum.

Where the momentum is given by:

Momentum = Mass×Velocity.

Then the initial momentum is:

P₀ = (1450kg)*V + (670 kg)*0m/s

Where V is the initial velocity of the torpedo.

For the final momentum, the torpedo and the target travel together with a velocity of 10.6 m/s, so we can see that the final momentum is:

P₁ = (1450kg + 670kg)*10.6m/s

Because of the conservation, we must have:

P₀ = P₁

(1450kg)*V = (1450kg + 670kg)*10.6m/s

(1459kg)*V = 22,472 kg*m/s

V = (22,472 kg*m/s)/(1459kg) = 15.4 m/s

So the initial velocity of the torpedo was 15.4 m/s

If you want to learn more about momentum, you can read:

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