Sign convention problem in momentum calulations

AI Thread Summary
The discussion revolves around a momentum calculation problem involving two colliding particles, A and B, with differing masses and velocities. The initial calculations yield a velocity of 1.6x10^8 m/s for particle A after the collision, raising questions about the sign convention used for direction. Participants note that the problem may contain an error, as both particles appear to move to the right post-collision, contradicting the expected leftward movement of particle A. Clarifications about directional notation highlight that both "left at 7 m/s" and "-7 m/s" can represent the same velocity but may cause confusion. Ultimately, the issue underscores the importance of clear communication in physics problems.
Fariaz Haque
Messages
5
Reaction score
0

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

 
Physics news on Phys.org
Fariaz Haque said:
i get a value of 1.6x10^8 for v ...

This looks like a relativistic collision because the projectile particle travels at (2/3)c. Are you studying relativity?
 
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...
 
Fariaz Haque said:
i get a value of 1.6x10^8 for v
Quite so. Both particles will move to the right. There must be an error in the question.
 
  • Like
Likes Fariaz Haque
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.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top