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

AI Thread Summary
A child with a mass of 2 kg jumps onto a cart moving at 2 m/s, causing the cart's velocity to decrease to 1 m/s. Using conservation of momentum, the mass of the cart is calculated to be 20 kg. The discussion highlights confusion regarding the thrust force's direction and the need to resolve the child's velocity into x and y components. It is noted that the thrust force in the y direction is zero since the cart does not move vertically. The calculations for both the mass of the cart and the thrust force require careful consideration of the inelastic collision principles.
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Homework Statement



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://www.slide.com/s/JuaPVKdK7j_16GybLEom3mA1r3ATs8QQ?referrer=hlnk

Homework Equations



p= mv
F= ma

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?
 
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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.
 
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?
 
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?
 
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