# Third Law Problem

1. Oct 9, 2013

### oMovements

The problem statement, all variables and given/known data
A diver with a weight of 833 N dives from a boat with a mass of 375kg. If he leaves the boat with a velocity of 2.25m/s [W] after accelerating for 0.50s, what will be the final velocity of the boat?

The attempt at a solution
=2.25-0/0.5

Fg=mdg
833=m(9.8)
md=85kg

833+Fd=-450.5

Fb+Fd=0
Fb= -Fd
Fb= 450.5N

Fb=mbab
450.5=(375)(ab)
ab=1.26m/s^2

ab=Vfb-Vib/t
1.26=Vf-0/0.50
Vf= 0.63m/s

Is this correct?

2. Oct 9, 2013

### D H

Staff Emeritus
No, that's not correct. You should ignore gravitational force here. Assume the diver leaves the boat with a horizontal velocity of 2.25 m/s. The only thing for which you need gravitational acceleration is the relationship between weight and mass.

3. Oct 9, 2013

### oMovements

=(85)(4.5)
Fd=382.5 N

Fb= -382.5 N

Fb=mb(-ab)
-382.5=(375)(-ab)
(ab)= 1.02 m/s^2

ab= Vf-Vi/t
1.02=Vf-0/0.50
Vfb= 0.51m/s

So I would do this then?

Also, if the direction of the velocity was not given to be [W] then would I use the gravitational force?

Last edited: Oct 9, 2013
4. Oct 9, 2013

### D H

Staff Emeritus
You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.

5. Oct 9, 2013

### oMovements

Thanks, fixed my solution. But if the diver was not travelling horizontally and the direction west was not stated. Then would gravitational force apply?

6. Oct 9, 2013

### D H

Staff Emeritus
You have to know the direction. Velocity, acceleration, and force are vectors. Moreover, this is a 3rd law problem. The diver's interaction with the boat and with the Earth are two different interactions. You shouldn't mix them up.