Identical 50 μC charges are fixed on an x axis at x = ±3.0 m. A particle of charge q = -15 μC is then released from rest at a po
1 answer:
Answer:
14 J
Explanation:
Let the angle of charge be θ
Then the distance from the x-axis = 3 m
The angle is given as
Similarly, the charge is on the left side, so the distance will be - 3 m
The angle therefore will be = 45°
So we need to find the net force given by the law:
The distance between the points will be
= 3
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Answer:
The speed decreases 75%.
Explanation:
- Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
- For the first collission, only mass 1 is moving before it, so we can write the following equation:
- Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
- From (1) and (2) we get:
- v₁ = v₀/2 (3)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
- Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
- From (4) and (5) we get:
- v₂ = v₀/3 (6)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
- Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
- From (7) and (8) we get:
- v₃ = v₀/4
- This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.