A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th
1 answer:
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Explanation:
When Michelson-Morley apparatus is turned through then position of two mirrors will be changed. The resultant path difference will be as follows.
Formula for change in fringe shift is as follows.
n =
v =
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
c = m/s
Hence, putting the given values into the above formula as follows.
v =
=
=
Thus, we can conclude that velocity deduced is .
Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s