g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.
1 answer:
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1) Potential difference: 1 V
2)
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
where
q is the charge's magnitude
is the potential difference between the initial and final position
In this problem, we have:
is the magnitude of the charge
is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative: