Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha
1 answer:
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Answer:
a)
b)
Explanation:
Given:
mass of the lighter block,
velocity of the lighter block,
mass of the heavier block,
velocity of the heavier block,
a)
Using conservation of linear momentum:
where:
final velocity of the lighter block
final velocity of the heavier block
........................(1)
Since kinetic energy is conserved in elastic collision:
divide the above equation by eq. (1)
.............................(2)
now we substitute the value of v from eq. (2) in eq. (1)
(negative sign denotes that the direction is towards left)
b)
now we substitute the value of v' from eq. (2) in eq. (1)
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v = v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or, = (
)
=(
)
or, y = 3.56 cm.