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**Momentum question--just checking my method**

Two men, of mass 100 kg each, stand on a cart of mass 300 kg. The cart can roll with negligible friction along a north-south track, and everything is initially at rest. One man runs towards the north and jumps off the cart at a speed of 5.0 m/s, relative to the cart. After he has jumped, the second man runs towards the south and jumps off the cart, again with a speed of 5.0 m/s relative to the cart. Calculate the speed and direction of the cart after both men have jumped off.

Since there is no net force acting, the momentum is conseved

p of system(s)= Ps

ps will be 0

hence Ps is 0.

So now the final momentum Ps is equal to P of man 1+P of man 2+ P of the cart

Hence Ps=500(N)-500(S)+P3

0=500(N)-500(S)+P3

500(S)=500(N)+P3

Now by phythagorous theorem,

P3= root of [(500^2)+(500^2)]

P3= 702.106kgm/s

And we know that m=300Kg

so we can find v

Does that seem right?