- #1
SpeeDFX
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A ball of mass 0.200 kg has a velocity of 1.50i m/s; a ball of mass 0.300 kg has a velocity of -0.400i m/s (where "i" is supposed to be that unit vector along x-axis). They meet in a head-on elastic collision. (a) Find their velocities after the collision.
I know I can use the conservation of both momentum and kinetic energy to solve this problem. I've done that. I looked in the solution manual and the person who wrote the solution used a different method I don't comprehend. First, they state the conservation of linear momentum, then they related the final velocities and way I don't understand.
The following 2 equations are used to solve the problem in the solution manual:
conservation of momentum for the two-ball system gives us:
0.200 kg (1.50 m/s) + 0.300 kg(-0.400 m/s) = 0.200 kg V1f + 0.300 kg V2f
Relative Velocity Equation:
V2f - V1f = 1.90 m/s
The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it? ( don't know, that's why I'm here, asking ) So, how do they come to that conclusion?
I know I can use the conservation of both momentum and kinetic energy to solve this problem. I've done that. I looked in the solution manual and the person who wrote the solution used a different method I don't comprehend. First, they state the conservation of linear momentum, then they related the final velocities and way I don't understand.
The following 2 equations are used to solve the problem in the solution manual:
conservation of momentum for the two-ball system gives us:
0.200 kg (1.50 m/s) + 0.300 kg(-0.400 m/s) = 0.200 kg V1f + 0.300 kg V2f
Relative Velocity Equation:
V2f - V1f = 1.90 m/s
The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it? ( don't know, that's why I'm here, asking ) So, how do they come to that conclusion?