Solving 6kg Cat on Wagon Mass Problem

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The discussion revolves around a physics problem involving a 6 kg cat walking on a stationary cart, with the cart moving backward at 3 m/s after the cat walks at 1 m/s. Participants agree that momentum conservation applies, leading to the equation m_cat * v_cat + m_cart * v_cart = 0. It's emphasized that directionality of velocities must be considered, as they can be positive or negative based on movement. The focus is on correctly applying momentum principles to find the cart's mass. Understanding the relationship between the cat's and cart's velocities is crucial for solving the problem accurately.
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Homework Statement



If a 6 kilogram cat walks on a stationary cart 1 m/s straight and then the wagon moves back at a speed of 3 m/s what's the mass of the cart?

Homework Equations


The Attempt at a Solution



I was thinking it would be a momentum problem, so mv of cat = mv of cart

I'm not sure I got this right...
 
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I Like Pi said:
I was thinking it would be a momentum problem, so mv of cat = mv of cart
Assuming that both the cat and the cart are at rest when the cat is not walking, yes, I agree with you. :approve:
 
Just for clarity though, Be careful about direction. What I should clarified in my last post is

mcatvcat + mcartvcart = 0,

where the velocities can be positive or negative, depending on the direction.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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