Recoil Speed?

Question

A submarine of mass 2.2X10^6 kg and initially at rest fires a torpedo of mass 260 kg. The torpedo has an initial speed of 100.6 m/s. What is the initial recoil speed of the submarine? Neglect the drag force of the water.

Answer 1

(260kg x 100.6m/s) / 2.2x10^6kg = 0.01889m/s

or

Just less than 2cm/s.

Answer 2

This is a simple conservation of momentum problem. The initial momentum of the system is zero (since the submarine is at rest).

The launched torpedo has an initial momentum of 260 * 100.6.

Set this equal to 2.2E6 * x, where x is the recoil speed, and solve.

Answer 3

apply the Law of Momentum conservation to the submarine-torpedo system
(M + m)v(initial) = Mv' + mv"
v(initial) = 0, hence left hand side reduces to 0
so Mv' = -mv"
2.2x10^6kg*v' = -260kgx100.6m/s
v' = -260x100.6/2.2x10^6kg
v' = 0.01m/s or 1cm/s.
This is the speed of recoil

Answer 4

quark_sa has the conservation of momentum formula right. I calculate 1.18 cm/s