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

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Homework Help Overview

The problem involves a cart moving on tracks and a child jumping onto it, affecting the cart's velocity. The subject area includes concepts of momentum conservation and inelastic collisions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and the implications of an inelastic collision. There are attempts to calculate the mass of the cart based on initial and final velocities, with some questioning the correctness of the calculations. Additionally, there is confusion regarding the direction of the thrust force and how to approach its calculation.

Discussion Status

Some participants have provided guidance on the conservation of momentum, suggesting that the original poster's calculations may need adjustments. There is ongoing exploration of the thrust force, with differing opinions on how to resolve the velocity components and the implications for force calculation.

Contextual Notes

Participants note the need to resolve velocities into x and y components and question the assumptions regarding the direction of forces involved in the problem.

mmoadi
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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?
 
Last edited by a moderator:
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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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