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The electric field is zero at x = -16.45cm
Data;
- q1 = 3.4 μC
- q2 = -2.0 μC
- distance = 5cm
As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.
Let the point be at x.
For an electric field to be equal to zero;
Let's solve for x using mathematical methods.
Solving the above quadratic equation;
The electric field is zero at x = -16.45cm
Learn more on electric field at a point here;
Answer:
1) elastic shock, the velocity of the center of mass does not change
2) inelastic shock, he velocity of the mass center change
Explanation:
The position of the center of mass of your system is defined by
=
in this case we have two bodies
x_{cm} = (x₁m₁ + x₂ m₂)
the velocity of the center of mass is
x_{cm} = dx_{cm} / dt =
x_{cm} =
where M is the total mass of the system.
Therefore to answer this question we have to find the velocity of the body after the collision.
Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.
Let's solve each case separately.
2) inelastic shock
initial instant. Before the crash
p₀ = m₁ v₀ + 0
final instant. After the collision with the cars together
p_f = (m₁ + m₂) v
p₀ = p_f
m₁ v₀ = (m₁ + m₂) v
v = v₀
let's find the velocity of the center of mass
M = m₁ + m₂
initial.
=
(m₁ vo)
final
=
( v) = v
v_{cm f} =
Let's find the ratio of the velocities of the center of mass
vcmf / vcmo =
therefore the velocity of the mass center change
1) elastic shock
initial instant.
p₀ = m₁ v₀
final moment
p_f = m₁ v_{1f} + m₂ v_{2f}
p₀ = p_f
m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}
m₁ (v₀ - v_{2f}) = m₂ v_{2f}
in this case the kinetic energy is conserved
K₀ = K_f
½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²
m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}
we write our system of equations
m₁ (v₀ - v_{1f}) = m₂ v_{2f} (1)
m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²
we solve the system
v₀ + v_{1f} = v_{2f}
we substitute and look for the final speeds
v_{1f} =
v_{2f} =
now let's find the velocity of the center of mass
initial
=
m₁ v₀
final
=
(m₁ v_{1f} + m₂ v_{2f} )
v_{cm f} = [
+
] v₀
v_{cm f} = ( m₁² - m₁m₂ +2 m₁m₂) v₂
v_{cm f} = (m₁² + m₁ m₂) v₀
let's look for the relationship
v_{cm f} / v_{cm o} = M
v_{cm f} / v_{cm o} = 1
therefore the velocity of the center of mass does not change
we see in either case the velocity of the center of mass does not change.