# Simple Momentum Question

1. Jun 16, 2009

### ppl k1ll3r

1. The problem statement, all variables and given/known data
A 60 tonne submarine, intially travelling forward at 1.5 m/s, fires a 500 kg torpedo striaght ahead with a velocity of 21 m/s in relation to the submarine. What is the velocity of the submarine after it fires the torpedo?

2. Relevant equations
1 ton = 1000 kg
p = p'

3. The attempt at a solution

p = p'
(60000)(1.5) = (500)(22.5) - (59500)(Vs')
Vs' = -1.32 m/s

I don't understand why I am getting that answer, as the answer is actually forward. Because doesn't the torpedo going forwards cause the submarine to move backwards.

Other than that, can you guys tell me whether I set it up right? Like the speed of the torpedo should be 22.5 m/s right, since the submarine is already travelling at 1.5 so it gets added? Also, the final mass of submarine shud be 59 500 kg, since the torpedo is no longer attached to it?

Thanks.

2. Jun 16, 2009

### ppl k1ll3r

Anyone!?

3. Jun 16, 2009

### Bhajun

Why do you have a minus sign?

$p = mv$
$p_{sub_i} = p_{sub_f} + p_{torpedo}$
$60000 \times 1.5 = (59500 \times v_{sub_f}) + (500 \times 21)$
$v_{sub_f} = (60000 \times 1.5) - (500 \times 21) = 1.336$

Forgive the lack of units.

4. Jun 16, 2009

### ppl k1ll3r

Minus because the momentum of the torpedo being shot forward causes the sub to move back. Also, shouldn't the torpedo have a speed of 22.5 m/s, since its 21 m/s relative to the sub, so it has a speed of 22.5? Like say if you're running at 2m/s and you throw a tennis ball at 5 m/s, the speed of the tennis ball would be 5 relative to you, but 7 m/s in total.

5. Jun 16, 2009

### ppl k1ll3r

?...

6. Jun 16, 2009

### glueball8

momentum of before = momentum of after

(60000)(1.5)=(59500)v+(500*22.5)

Yes, its 22.5. But there's no negative sign since their both doing in forward (Same direction.).

7. Jun 16, 2009

### ideasrule

To avoid confusion with signs in the future, define a direction to be positive, then always use positive signs in the momentum equations (ex. m1v1=m2v2 + m3v3). Get the signs of the velocities correct and the sign of the final answer will be correct.

8. Jun 16, 2009

### flatmaster

You are confusing change in momentum with final momentum. Yes indeed the CHANGE in the sub's momentum is in the negative dirrection. However, the final momentum for the ship is still in the positive direction.