Calculating Momentum of Firing Marksman with X and Y Components

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To calculate the momentum of the marksman firing at an angle, it's essential to recognize that momentum is conserved in both the horizontal and vertical directions. Initially, the total momentum is zero, and after firing, the horizontal momentum must remain zero due to the absence of external forces. The bullet's horizontal momentum can be calculated using its speed and the angle of projection, while the marksman and car's horizontal movement must counterbalance this momentum. The vertical momentum of the bullet does not affect the horizontal calculations since the car cannot move vertically. Understanding these principles allows for solving the problem effectively.
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These problems with x and y components always leave me confused...

A marksman standing on a motionless railroad car fires a gun into the air at an angle of 30° from the horizontal (the figure below ). The bullet has a speed of 173 m/s (relative to the ground) and a mass of 0.02 kg. The man and car move to the left at a speed of 1.2 " × 10-3 m/s after he shoots. What is the mass of the man and car? [Hint A component of a system's momentum along an axis is conserved if the net external force acting on the system has no component along that axis.]

http://img248.imageshack.us/img248/1495/fig026vo8.gif

I know that momentum is conserved in the form of Pi = Pf. What I don't know is how to calculate the momentums when there are angles involved... Can someone help me get started on this problem please? Thanks
 
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Before the bullet is fired, the total momentum is zero. After the bullet is fired, the horizontal momentum is zero. The bullet also has vertical momentum, but the car cannot move downward to cancel the vertical bullet momentum, so the total vertical momentum after firing is not zero. All you need to find the solution is to work out the horizontal problem.
 
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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