A boy that jumps off a skate, v of the skate=?

[frank1]

A boy of 30kg is riding a skateboard of 4.9kg and both are going on with a velocity of 2.2m/s. In a moment the boy jumps forward, pushing the skate to the other direction. Immediatelly after the jump, the horizontal velocity of the boy in relation to the skate is 3m/s. Whats the velocity of the skate in this moment?

What I've tried to do:

Momentum before=Momentum after
(mass of the system skate+boy)(initial velocity)=(mass of boy)(velocity of boy)+(mass of skate)(velocity of skate)

I think what I am probably missthinking is the velocity of the boy, because in the question, he says "velocity of the boy in relation to skate", so i suppose, it would be: (mass of the boy)(3 + velocity of the skate)?

It sounds wrong, i don't know why XD

Can somebody give me a light here?

 

[Infinitum]

Hi frank! Welcome to PF!

The question has given you the velocity of the boy relative to the skate. That means,

$$V_B - V_S = V_{BS} = 3$$

So yes, you got it right!

 

[frank1]

So can you check if I am doing it correctly?

look:

(34.9)x(2.2)=30(3+Vs)+4.9Vs
76.78=90+30Vs+4.9Vs
-13.22=34.9Vs
Vs=-0.3787m/s

EDIT: thanks for helping so quickly :)

 

[Infinitum]

So can you check if I am doing it correctly?

look:

(34.9)x(2.2)=30(3+Vs)+4.9Vs
76.78=90+30Vs+4.9Vs
-13.22=34.9Vs
Vs=-0.3787m/s

Yep, your method looks correct!