# What is the speed of 1.4-kg cart right after the separation?

## Homework Statement

A system consists of a 5.0-kg cart and a 1.4-kg cart attached to each other by a compressed spring. Initially, the system is at rest on a low-friction track. When the spring is released, an explosive separation occurs at the expense of the internal energy of the compressed spring. The change in the spring's internal energy during the separation is 1.0 kJ.
What is the speed of 1.4-kg cart right after the separation?

## Homework Equations

k=1/2mav2+ 1/2 mbv2

## The Attempt at a Solution

-5v1=1.4v2
v1=-1.4v2/5

1000J=(mav12)/2 +(mbv22)/2
2000J=5(-1.4v2/5)+1.4v22
v2=√2000/(24/5)
v2=20.4 m/s

## Answers and Replies

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Doc Al
Mentor
1000J=(mav12)/2 +(mbv22)/2
2000J=5(-1.4v2/5)+1.4v22
Redo that second equation, making sure to square things properly.

2000J=5(-1.4v22/5)+1.4v22

2000J=5(-1.4v22/5)+1.4v22
i caught my error the answer is 33.4 m/s

Doc Al
Mentor
2000J=5(-1.4v22/5)+1.4v22
I would write it as: 2000J=5(1.4/5)2v22 + 1.4v22