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Grade 12 physics problem!

by musicfan31
Tags: help asap, help asap please!!!
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musicfan31
#1
Nov1-09, 02:30 PM
P: 6
1. The problem statement, all variables and given/known data

2 men with mass of 100kg each, are on a stationary cart with mass of 300 kg.
The cart has no friction withthe tracks. The cart can only go north and south.
1st man starts runnin at speed of 5 m/s(reletive to cart) towards north and jumps off the cart. Otherman runs with same speed (reletive to cart) towards the south after the 1st man what is the cart final velocity??


2. Relevant equations


use the equation of

Conservation of energy
Coservation of momentum



3. The attempt at a solution


North is positive

M1 = 100
M2 = 100
M3 = 300

MAN 1
m1v1 +m2v2 + m3v3 = m1v1' +m2v2' + m3v3'
100(5)+ 0+0 = V'(100+300)
500/400 = v'
10.25 m/s =v'
MAN 2
m2v2 + m3v3 = m2v2' + m3v3'
100(-5) + 300(1.25) = 300v'
-500+375 = 300v'
-0.41666 m/s = v'

NOW I need someone to proof read this and tell me wht to do next please!!

Teh final asnwer is 0.25 m/s North!!!
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Redbelly98
#2
Nov1-09, 08:13 PM
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Redbelly98's Avatar
P: 12,069
Quote Quote by musicfan31 View Post
m1v1 +m2v2 + m3v3 = m1v1' +m2v2' + m3v3'
100(5)+ 0+0 = V'(100+300)
Sorry, this isn't right.

Things start out all at rest, with the two men both standing on the cart. So the initial momentum is zero.

Then the 1st man jumps off the cart, and is going 5 m/s relative to the cart. You need to write an expression for the total momentum an instant after the man leaves the cart, and set that equal to the initial momentum of zero.

500/400 = v'
10.25 m/s =v'
MAN 2
m2v2 + m3v3 = m2v2' + m3v3'
100(-5) + 300(1.25) = 300v'
-500+375 = 300v'
-0.41666 m/s = v'

NOW I need someone to proof read this and tell me wht to do next please!!

Teh final asnwer is 0.25 m/s North!!!


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