Calculating Momentum of Firing Marksman with X and Y Components

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The discussion focuses on calculating the momentum of a marksman firing a bullet at a 30° angle with a speed of 173 m/s. The bullet has a mass of 0.02 kg, and the marksman and the railroad car move left at 1.2 × 10-3 m/s after firing. The conservation of momentum principle is applied, emphasizing that horizontal momentum is conserved while vertical momentum is not affected by the car's movement. The key takeaway is that the initial total momentum is zero, and the horizontal momentum must be calculated to find the mass of the man and car.

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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.
 

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