Child jumps on a cart, cart slows down, find mass of the cart and thrust force

[mmoadi]

1. The problem statement, all variables and given/known data

A cart is driving on straight tracks with a velocity of 2 m/s. In the opposite direction, with an angle of 60º according to the tracks, a child with a mass of 2 kg is running with a velocity of 2m/s, he jumps on the cart and stays there.
- Find the mass of the cart, if its velocity is reduced to 1 m/s when the child jumps onto it.
- What is the thrust force (impulse) perpendicular to the tracks?

Picture:
http://widget.slide.com/rdr/1/1/4/S/2b00000015dd8474/1/68/AGZykH-6sz-IT4kV4GyynFT8ClMUk_yn.jpg (http://www.slide.com/s/JuaPVKdK7j_16GybLEom3mA1r3ATs8QQ?referrer=hlnk)

2. Relevant equations

p= mv
F= ma

3. The attempt at a solution

First part: Find the mass of the cart, if its velocity is reduced to 1 m/s when the child jumps onto it.

Conservation of the moment:

m(2)= 20 kg, v(1-initial)= v(2-initial)= 2 m/s, v(1-final)= 1 m/s, v(2-final)= 0 m/s

m(1)v(1-initial) + m(2)v(2-initial)= m(1)v(1-final) + m(2)v(2-fianl)
m(2)v(2-initial)= m(1)v(1-final) - m(1)v(1-initial)
m(2)v(2-initial)= m(1)*(v(1-final) - v(1-initial))
m(1)= m(2)v(2-initial) / (v(1-final) - v(1-initial))
m(1)= 40 kg

Are these calculations correct?

For the second part, I'm kind of confused to find the direction of the thrust force. Can someone give me a hint?

[semc]

Yup COM is correct but its inelastic collision aka they stick together so the first part of your working need to change. Didn't really punch in the numbers but did you use velocity of kid=2m/s if not should be correct.

[mmoadi]

Inelastic collision- conservation of the moment:

m(2)= 20 kg, v(1)= v(2)= 2 m/s, v(3)= 1 m/s

m(1)v(1) + m(2)v(2)= (m(1) + m(2))v(3)
2m(1) + 40= m(1) + 20
m(1)= 20 kg


Is this now correct?
What about the second part?

[semc]

Nope, you can't just use velocity=2m/s you need to resolve into the x and y direction.
Hmm... I am not very sure about the second one but since F=dp/dt and the train doesn't move in the y direction, so force =0?