hard tutorial question can anyone help

[plug2002]

A torpedo boat with a mass of 50 tons is moving at 10 knots when it fires a 250-kg torpedo horizontally with the launch tube at the 30 degree angle. The torpedo has a relative velocity to the boat of 10m/s as it leaves the tube
1) Determine the momentary reduction v in forward speed of the boat?

it is meant as a linear impuse question, can anyone help me please!

[Danger]

I can't help with the math at all, unfortunately. My first question would be regarding the launching method. Not all of them would impart the same reaction to the boat.

[brewnog]

Have you done linear impulse questions before?

Start by converting all the units to SI and it might become a bit clearer.

[FredGarvin]

The sub and torpedo as a whole has a combined linear momentum. Whe the two separate........

[phlegmy]

conservation of momentum
the mass of the sub + torpedo times its velocity before firing=
the mass of the sub*vsub +mass torpedo*vtorp.

or you could say that the sub is at rest altogether and fires the torpedo which then has momentum of 250*10=2500
so the sub must loose this amount of momentm
-2500=50,000*vsub
so vsub= -.05m/s
the sub looses .05m/s when it fires the the torp.

it aint rocket science, but wait, maybe it is

[Mech_Engineer]

Unfortunately, phlegmy, you miseed an important portion of the problem: the torpedo is launched at a 30* angle. This will slightly decrease the momentum loss from the boat.

[phlegmy]

doh!
in that case loss of mom in x direction = 250*10*cos(30)= 2165
sub loosese .043 m/s in the forward direction! and starts to decend at
.025m/s