Sign convention problem in momentum calulations

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Fariaz Haque
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Homework Statement


Particle A has a mass of 1kg and velocity 2x10^8m/s to the right and collides with a stationary particle B that has a mass of 4kg. after the collision, particle A moves to the left with a velocity(v) and particle B moves to the right with a velocity of 1x10^7 m/s. calcuate the value of 'v'.
i get a value of 1.6x10^8 for v but shouldn't it be -1.6x10^8 as it is moving to the left?

The thing is, in other collisions where two objects moving towards each other collide and then move apart after the collision, the answer justifies the direction just fine. for example...

An example: Particle C has a mass of 1kg and velocity 2x10^8 m/s to the right and collides with particle D that has a mass of 4kg and is moving to the left at 1.5x10^8 m/s. after the collision, particle C moves to the left with a velocity(v) and particle D moves to the right with a velocity of 2x10^7 m/s. calcuate the value of 'v'.
solution: (2x10^8 x 1) + [4 x (-1.5x10^8) ] = (1x v) + (4x 2x10^7)
v= -0.48x10^9(here the answer justifies the direction)

Homework Equations

:[/B]
momentum before collision= momentum after collision

The Attempt at a Solution


momentum before collision= momentum after collision
(2x10^8 x 1) + (4x0) = (1xv) + (1x10^7 x 4)
v= 1.6x10^8

 
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kuruman said:
This looks like a relativistic collision because the projectile particle travels at (2/3)c. Are you studying relativity?

nope... just a simple momentum question that was given at school...
 
haruspex said:
Quite so. Both particles will move to the right. There must be an error in the question.

Indeed. Asked my teacher today and verified the problem! :)
 
Context.

Left at 7m/s
Left, at -7m/s (note comma)

can mean the same thing. The first one is 7m/s leftwards, the second is -7m/s along an implied left-to-right horizontal axis (x) - the "left" is semi-redundant.

On the other hand proofreaders don't get paid much, so best not to make them work too hard if it can be avoided.
 
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