Two electrons, each with mass mmm and charge qqq, are released from positions very far from each other. With respect to a certai
n reference frame, electron A has initial nonzero speed vvv toward electron B in the positive x direction, and electron B has initial speed 3v3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
What is the minimum separation r_min that the electrons reach?
What is the minimum separation r_min that the electrons reach?
1 answer:
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Answer:
θ = 62.72°
Explanation:
The projectile describes a parabolic path:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = x₀+ vx*t Formula (1)
vx = v₀x
Where:
x: horizontal position in meters (m)
x₀: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
v₀x: Initial speed in x in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)
vfy= v₀y -gt Formula (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 120 m/s , at an angle θ above the horizontal
v₀x= 55 m/s
x-y components of the initial velocity ( v₀)
v₀x = v₀*cosθ Equation (1)
v₀y = v₀*sinθ Equation (2)
Calculating of the angle θ
We replace data in the Equation (1)
55 = 120*cosθ
cosθ = 55 / 120
θ = 62.72°
The x -component of the object's acceleration is 2 m/s².
<h3>What's the resultant force along x- direction?</h3>- Forces along x axis direction are as follows
- 4N along +x axis, so it's taken as +4 N
- 2N along -x axis , so it's taken as -2N.
- Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.
- As per Newton's second law, Force = mass × acceleration of the object
- Force along x axis= mass × acceleration along x axis= 2N
- Acceleration = 2/ mass = 2/1 = 2 m/s²
Thus, we can conclude that the acceleration along x axis is 2 m/s².
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?
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