Momentum in different directions

[diciccod]

Hi, this is my first time posting here, looks like a good community and people get a lot of help here. Well here goes, i have this physics problem on momentum that i have never seen before and i can't get past the first step, i kind of know what i have to do.

Problem 1:
A boy and a dog are standing on a 110kg raft in the middle of a lake. Just as the 55kg boy dives off the raft with a horizontal velocity of 4.0m/s due EAST, the 22kg dog leaps off the raft horizontally with a velocty of 5.0m/s due north. What is the resulting velocity of the raft?

for several other problems i have been using the equation
Mi*Vi = Mf1*Vf1 + Mf2*Vf2 I think i have to manipulate that somehow though because I am not just working on positive, negative anymore. If someone could just get me started i would greatly appreciate it.

 

[pnaj]

In the equation you are using, let term(1) be the boy, (2) be the dog and add a third term that represents the raft.

You know both mass and velocity of both boy and dog.
You know mass of raft and velocity of raft is the thing to find.

Do you know about Conservation of Momentum? What should they all add up to?

See how you get on.

P.



P.S. This assumes you know how to add vectors and your equation is a vector equation.

 

[Doc Al]

Originally posted by diciccod

for several other problems i have been using the equation
Mi*Vi = Mf1*Vf1 + Mf2*Vf2 I think i have to manipulate that somehow though because I am not just working on positive, negative anymore. If someone could just get me started i would greatly appreciate it.

Treat the East-West components (call it the x-direction) and North-South components (call it the y-direction) separately. Write down the conservation of momentum for each direction. This will give you the components of the velocity of the raft; add them up to get magnitude and direction.

 

[diciccod]

i think i got it, not sure.
heres my work

p(all)x = p(raft+boy)
P(all)y = p(raft+dog)

X-Direction
P(all)x = (110kg)(Vfx1) + (55kg*4m/s)
0 = (110)(Vfx1) + (220kg*m/s)
(-220kg*m/s)/(110kg) = Vfx1 = -2m/s

Y-Direction
P(all)y = (110kg)(Vfy1) + (22kg*5m/s)
0 = (110kg)(Vfy1) + (110kg*m/s)
(-110kg*m/s)/(110kg) = Vfy1 = -1m/s

I think that's all my teacher wants, is Velocity Final in the X and in the Y. Well someone check my logic on this one please :)

 

[Doc Al]

Originally posted by diciccod
I think that's all my teacher wants, is Velocity Final in the X and in the Y. Well someone check my logic on this one please :)

Looks good to me.